Tüm alıştırma soruları

5556 soru

Soru 2681Soru

Chords ABAB and CDCD intersect at point EE inside a circle. If AE=6AE = 6, EB=8EB = 8, and the total length of chord CDCD is 1616, what is the length of the shorter segment of chord CDCD?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The length of the shorter segment of chord CDCD is 44.
According to the Intersecting Chords Theorem, when two chords intersect inside a circle, the product of the segments of one chord is equal to the product of the segments of the other. For chords ABAB and CDCD intersecting at point EE, this relationship is expressed as AEEB=CEEDAE \cdot EB = CE \cdot ED. Substituting the given values yields 68=CEED6 \cdot 8 = CE \cdot ED, so CEED=48CE \cdot ED = 48. Since the total length of chord CDCD is 1616, we can define CE=xCE = x and ED=16xED = 16 - x. The equation becomes x(16x)=48x(16 - x) = 48, which simplifies to the quadratic equation x216x+48=0x^2 - 16x + 48 = 0. Factoring this equation gives (x12)(x4)=0(x - 12)(x - 4) = 0, meaning the two segments of chord CDCD have lengths of 1212 and 44. The length of the shorter segment is 44.

Adım Adım Çözüm

1
State the relationship between intersecting chord segments.
AEEB=CEEDAE \cdot EB = CE \cdot ED
By the Intersecting Chords Theorem, the product of the segments of one chord equals the product of the segments of the other.
2
Substitute the known lengths and define the segments of CDCD using a variable xx.
68=x(16x)6 \cdot 8 = x(16 - x), which simplifies to 48=16xx248 = 16x - x^2.
We are given AE=6AE = 6 and EB=8EB = 8. Since the total length of chord CDCD is 1616, if one segment is xx, the remaining segment must be 16x16 - x.
3
Solve the quadratic equation for xx by factoring.
x216x+48=0    (x12)(x4)=0x^2 - 16x + 48 = 0 \implies (x - 12)(x - 4) = 0, so x=12x = 12 or x=4x = 4.
Rearranging the equation into standard quadratic form allows us to find the two possible segment lengths.
4
Identify the shorter segment length from the two solutions.
44
The two segment lengths are 1212 and 44. The problem asks for the shorter segment, which is 44.

Anahtar Kavram

Intersecting Chords Theorem
Soru 2682Soru

In triangle ABCABC, the length of side aa is 55 centimeters, the length of side bb is 88 centimeters, and the measure of angle CC is 6060^\circ. What is the length, in centimeters, of side cc?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The length of side cc is 77 centimeters.
Applying the Law of Cosines directly to the given Side-Angle-Side (SAS) triangle yields c2=52+822(5)(8)cos60=25+6440=49c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ = 25 + 64 - 40 = 49, which gives c=7c = 7.

Adım Adım Çözüm

1
Identify the given values and the appropriate formula.
We are given two sides, a=5a = 5 and b=8b = 8, and the included angle C=60C = 60^\circ. To find the opposite side cc, we use the Law of Cosines: c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C.
The Law of Cosines relates three sides of a triangle to the cosine of one of its angles, which is applicable for Side-Angle-Side (SAS) configurations.
2
Substitute the known values into the Law of Cosines equation.
c2=52+822(5)(8)cos60c^2 = 5^2 + 8^2 - 2(5)(8) \cos 60^\circ
Plugging the given values into the formula allows us to solve for the unknown side cc.
3
Evaluate the trigonometric and arithmetic terms.
Since cos60=0.5\cos 60^\circ = 0.5, we get:
c2=25+6480(0.5)c^2 = 25 + 64 - 80(0.5)
c2=8940c^2 = 89 - 40
c2=49c^2 = 49
Simplifying the expression step-by-step leads to the value of c2c^2.
4
Take the square root of both sides to find the side length.
c=49=7c = \sqrt{49} = 7
Since side lengths must be positive, the square root of 4949 gives the exact length of side cc.

Anahtar Kavram

Law of Cosines
Soru 2683Soru

A coordinate grid contains a right triangle, XYZXYZ, with the right angle located at vertex ZZ. The horizontal leg XZXZ has a length of 2424 units, and the vertical leg YZYZ has a length of 77 units. What is the value of cos(X)\cos(X)?

Cevabı ve açıklamayı göster

Cevap: 2425\frac{24}{25}

Cevap

The correct answer is the option representing 2425\frac{24}{25}.
The cosine of an angle in a right triangle is the ratio of the length of the adjacent leg to the length of the hypotenuse. The hypotenuse of the triangle is calculated as 2525 units using the Pythagorean theorem. Relative to angle XX, the adjacent leg is XZ=24XZ = 24 and the hypotenuse is XY=25XY = 25. Therefore, cos(X)=2425\cos(X) = \frac{24}{25}.

Adım Adım Çözüm

1
Use the Pythagorean theorem to calculate the hypotenuse XYXY.
XY=XZ2+YZ2=242+72=576+49=625=25XY = \sqrt{XZ^2 + YZ^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25.
To calculate any primary trigonometric ratio, we need the lengths of the relevant sides, including the hypotenuse.
2
Identify the adjacent side relative to angle XX.
The leg adjacent to angle XX is XZ=24XZ = 24.
Angle XX is formed by the hypotenuse XYXY and the adjacent leg XZXZ.
3
Apply the definition of cosine (adjacent over hypotenuse).
cos(X)=XZXY=2425\cos(X) = \frac{XZ}{XY} = \frac{24}{25}.
By SOHCAHTOA, cosine is the ratio of the adjacent side length to the hypotenuse length.

Anahtar Kavram

Using SOHCAHTOA definitions to find trigonometric ratios in a right triangle.
Tahmini Süre:1m 0s
Soru 2684Soru

For an angle θ\theta such that π2<θ<π\frac{\pi}{2} < \theta < \pi, if cosθ1sinθ=3\frac{\cos\theta}{1 - \sin\theta} = -3, what is the value of sinθtanθ\sin\theta - \tan\theta?

Cevabı ve açıklamayı göster

Cevap: 3215\frac{32}{15}

Cevap

3215\frac{32}{15}
By multiplying the numerator and denominator of the given expression by 1+sinθ1 + \sin\theta and utilizing the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1, we find that secθ+tanθ=3\sec\theta + \tan\theta = -3. Since sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1, it follows that secθtanθ=13\sec\theta - \tan\theta = -\frac{1}{3}. Solving this system of equations yields secθ=53\sec\theta = -\frac{5}{3} (which means cosθ=35\cos\theta = -\frac{3}{5}) and tanθ=43\tan\theta = -\frac{4}{3}. In Quadrant II, sine is positive, so sinθ=tanθcosθ=45\sin\theta = \tan\theta \cos\theta = \frac{4}{5}. Thus, sinθtanθ=45(43)=12+2015=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{12 + 20}{15} = \frac{32}{15}.

Adım Adım Çözüm

1
Multiply the numerator and denominator of the given expression by 1+sinθ1 + \sin\theta to simplify it.
cosθ(1+sinθ)(1sinθ)(1+sinθ)=3    cosθ(1+sinθ)1sin2θ=3\frac{\cos\theta(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} = -3 \implies \frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta} = -3
To set up the Pythagorean identity in the denominator.
2
Substitute the Pythagorean identity 1sin2θ=cos2θ1 - \sin^2\theta = \cos^2\theta into the denominator.
cosθ(1+sinθ)cos2θ=1+sinθcosθ=secθ+tanθ=3\frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta = -3
To simplify the expression into basic trigonometric functions.
3
Use the reciprocal identity relationship sec2θtan2θ=1\sec^2\theta - \tan^2\theta = 1 to find the difference of secant and tangent.
secθtanθ=1secθ+tanθ=13=13\sec\theta - \tan\theta = \frac{1}{\sec\theta + \tan\theta} = \frac{1}{-3} = -\frac{1}{3}
To create a system of linear equations for secant and tangent.
4
Solve the system of equations for secθ\sec\theta and tanθ\tan\theta.
secθ=53\sec\theta = -\frac{5}{3} and tanθ=43\tan\theta = -\frac{4}{3}
Adding the two equations yields 2secθ=103    secθ=532\sec\theta = -\frac{10}{3} \implies \sec\theta = -\frac{5}{3}, and subtracting them yields 2tanθ=83    tanθ=432\tan\theta = -\frac{8}{3} \implies \tan\theta = -\frac{4}{3}.
5
Find cosθ\cos\theta and sinθ\sin\theta, and calculate the final value of sinθtanθ\sin\theta - \tan\theta.
cosθ=35\cos\theta = -\frac{3}{5}, sinθ=45\sin\theta = \frac{4}{5}, and sinθtanθ=45(43)=3215\sin\theta - \tan\theta = \frac{4}{5} - \left(-\frac{4}{3}\right) = \frac{32}{15}
Since θ\theta lies in Quadrant II, sine is positive, which is verified by sinθ=tanθcosθ=(43)(35)=45\sin\theta = \tan\theta \cdot \cos\theta = \left(-\frac{4}{3}\right)\left(-\frac{3}{5}\right) = \frac{4}{5}.

Anahtar Kavram

Fundamental Trigonometric Identities
Tahmini Süre:2m 0s
Soru 2685Soru

### Origin of the Moon

Scientists debate how the Moon was formed. Two primary hypotheses have been proposed:

Hypothesis 1
The Moon formed from the debris of a collision between the early Earth and a Mars-sized protoplanet. The impact ejected mostly the silicate-rich outer mantles of both bodies into orbit, where they accreted to form the Moon. Because the iron cores of both bodies merged to remain with Earth, the Moon was left with a very small iron core.

Hypothesis 2
The Moon formed independently in a different region of the solar nebula and was later captured by Earth's gravitational field during a close planetary flyby. This explains why the Moon has a much lower bulk density than Earth, as it formed from materials in a region of the nebula that was naturally depleted of iron.

New Evidence
Analysis of lunar rock samples reveals that the oxygen isotope ratios (17O/16O^{17}\text{O}/^{16}\text{O} and 18O/16O^{18}\text{O}/^{16}\text{O}) of lunar rocks are identical to those of Earth's mantle. Bodies that form in different regions of the solar nebula generally possess distinct, unique oxygen isotope signatures.

Based on this information, how does the new evidence affect the two hypotheses?

Cevabı ve açıklamayı göster

Cevap: It supports Hypothesis 1 and weakens Hypothesis 2.

Cevap

The correct answer states that the new evidence supports Hypothesis 1 and weakens Hypothesis 2.
The correct answer explains that the new evidence supports the collision model (Hypothesis 1) and weakens the capture model (Hypothesis 2). According to the new evidence, bodies that form in different regions of the solar nebula possess distinct oxygen isotope signatures. Because the Moon and Earth's mantle have identical oxygen isotope ratios, they must have formed from the same reservoir of material. This supports Hypothesis 1, which claims the Moon formed from Earth's ejected mantle, and weakens Hypothesis 2, which claims the Moon formed independently in a different nebular region.

Adım Adım Çözüm

1
Identify the core premise of Hypothesis 1 and Hypothesis 2 regarding the Moon's origin of material.
Hypothesis 1 states the Moon formed from Earth's mantle debris (shared material source). Hypothesis 2 states the Moon formed in a different region of the solar nebula (distinct material source).
Understanding the expected chemical/isotopic relationship between Earth and the Moon for each hypothesis is necessary to evaluate the new evidence.
2
Analyze the new evidence concerning oxygen isotope ratios.
Lunar rocks have oxygen isotope ratios identical to Earth's mantle, and planetary bodies from different regions of the nebula have distinct signatures.
This establishes that identical isotope ratios indicate a shared origin of material, while different regions of formation lead to different ratios.
3
Determine the impact of the new evidence on both hypotheses.
Because Earth and the Moon have identical ratios, they likely share a material source, which supports Hypothesis 1 (debris collision) and weakens Hypothesis 2 (independent formation in a different region).
This directly answers how the evidence supports or weakens each hypothesis based on the logical consistency of their claims with the new data.

Anahtar Kavram

Evaluating how new experimental or observational findings support or weaken existing scientific hypotheses.
Tahmini Süre:1m 30s
Soru 2686Soru

In the standard (x,y)(x, y) coordinate plane, a line passes through the points (3,1)(3, -1) and (1,y)(-1, y). If the slope of the line is 2-2, what is the value of yy?

Cevabı ve açıklamayı göster

Cevap: 77

Cevap

The correct value of yy is 77.
To find the value of yy, apply the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the coordinates (3,1)(3, -1) and (1,y)(-1, y) alongside the slope of 2-2 yields the equation 2=y(1)13-2 = \frac{y - (-1)}{-1 - 3}. Simplifying the denominator gives 2=y+14-2 = \frac{y + 1}{-4}. Multiplying both sides of the equation by 4-4 results in 8=y+18 = y + 1. Finally, subtracting 11 from both sides gives y=7y = 7.

Adım Adım Çözüm

1
Set up the slope formula using the given points (x1,y1)=(3,1)(x_1, y_1) = (3, -1) and (x2,y2)=(1,y)(x_2, y_2) = (-1, y).
The formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}, which becomes 2=y(1)13-2 = \frac{y - (-1)}{-1 - 3}.
This establishes the algebraic relationship between the coordinates and the given slope.
2
Simplify the numerator and denominator of the fraction.
2=y+14-2 = \frac{y + 1}{-4}
Subtracting a negative number is equivalent to addition, and 13-1 - 3 simplifies to 4-4.
3
Solve for yy by isolating the variable.
Multiply both sides by 4-4 to get 8=y+18 = y + 1, then subtract 11 from both sides to get y=7y = 7.
This isolates the variable yy to find its numerical value.

Anahtar Kavram

Calculating the slope of a line using the coordinates of two points on the line.
Tahmini Süre:1m 0s
Soru 2687Soru

In the figure, point BB lies on the line segment ACAC, and segment BDBD is perpendicular to ACAC. Triangle ABDABD is a right triangle with hypotenuse AD=8AD = 8 and ADB=30\angle ADB = 30^\circ. If the length of segment BCBC is 1111, what is the length of segment CDCD?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The length of segment CDCD is 1313.
The length of segment CDCD is 1313. Since BDBD is perpendicular to segment ACAC at point BB, ABD\triangle ABD and DBC\triangle DBC are both right triangles. In the 30609030^\circ-60^\circ-90^\circ right triangle ABD\triangle ABD, the hypotenuse is AD=8AD = 8, so the longer leg opposite the 6060^\circ angle is BD=43BD = 4\sqrt{3}. In right triangle DBC\triangle DBC, using the Pythagorean Theorem: CD2=BD2+BC2=(43)2+112=48+121=169CD^2 = BD^2 + BC^2 = (4\sqrt{3})^2 + 11^2 = 48 + 121 = 169. Taking the square root gives CD=13CD = 13.

Adım Adım Çözüm

1
Find the length of the shared perpendicular segment BDBD using the properties of the special 30609030^\circ-60^\circ-90^\circ right triangle ABD\triangle ABD.
BD=43BD = 4\sqrt{3}
In a 30609030^\circ-60^\circ-90^\circ right triangle, the side opposite the 6060^\circ angle is 32\frac{\sqrt{3}}{2} times the hypotenuse.
2
Apply the Pythagorean Theorem to right triangle DBC\triangle DBC to calculate the length of hypotenuse CDCD.
CD=13CD = 13
The Pythagorean Theorem states that the square of the hypotenuse is equal to the sum of the squares of the legs (CD2=BD2+BC2CD^2 = BD^2 + BC^2).

Anahtar Kavram

Solving for unknown sides in adjacent right triangles by combining special right triangle ratios (30609030^\circ-60^\circ-90^\circ) and the Pythagorean Theorem.
Soru 2688Soru

A straight skateboard ramp has a vertical height of 99 feet and a horizontal length of 1212 feet along the ground. The vertical height and horizontal length meet at a right angle. What is the sine of the angle that the ramp makes with the ground?

Cevabı ve açıklamayı göster

Cevap: 35\frac{3}{5}

Cevap

The value of the sine of the angle is 35\frac{3}{5}.
The sine of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the hypotenuse. The opposite side is the vertical height of 99 feet. Using the Pythagorean theorem, the hypotenuse (the length of the ramp) is calculated as 92+122=15\sqrt{9^2 + 12^2} = 15 feet. Therefore, the sine of the angle is 915\frac{9}{15}, which simplifies to 35\frac{3}{5}.

Adım Adım Çözüm

1
Identify the lengths of the legs of the right triangle and calculate the length of the hypotenuse using the Pythagorean theorem.
The hypotenuse (ramp length) is 1515 feet.
The vertical height (99 feet) and horizontal length (1212 feet) form the perpendicular legs of a right triangle. The length of the ramp is the hypotenuse: 92+122=81+144=225=15\sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15 feet.
2
Identify the opposite side and hypotenuse relative to the angle the ramp makes with the ground, then apply the sine ratio definition.
The sine of the angle is 35\frac{3}{5}.
The angle the ramp makes with the ground is at the bottom vertex. The side opposite this angle is the vertical height (99 feet), and the hypotenuse is the ramp length (1515 feet). The sine ratio is sin(θ)=oppositehypotenuse=915\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{15}, which simplifies to 35\frac{3}{5}.

Anahtar Kavram

Right Triangle Trigonometry (SOHCAHTOA)
Tahmini Süre:1m 0s
Soru 2689Soru

In the early summer of 19301930, off the coast of Nonsuch Island, Bermuda, a bizarre vessel was lowered from the deck of the barge Ready into the sapphire waters of the Atlantic. This was the Bathysphere, a spherical steel deep-sea submersible designed by Otis Barton and championed by the renowned naturalist William Beebe. At a time when deep-sea exploration was limited to dredging nets that pulled up crushed, lifeless specimens, Beebe sought to observe marine organisms in their native, undisturbed habitats. The design of the Bathysphere was a marvel of utilitarian engineering. Barton realized that only a sphere could uniformly distribute the crushing hydrostatic pressure of the abyss, which increases by approximately 11 atmosphere for every 3333 feet of depth. Cast by the Watson-Stillman Hydraulic Machinery Company, the sphere measured exactly 44 feet 99 inches in diameter and possessed steel walls 1141\frac{1}{4} inches thick. Its weight, a formidable 5,4005,400 pounds, required a heavy-duty winch and a specialized steel cable capable of supporting several tons without twisting.

To allow Beebe to look out into the darkness, Barton installed 33 window ports, though only 22 were eventually fitted with windows while the third was plugged with steel. These windows were not made of standard glass, which would have shattered under the intense pressure, but of fused quartz cylinders. Each cylinder was 88 inches in diameter and 33 inches thick, ground and polished to optical perfection by the General Electric Company. To seal these windows against the steel frame, Barton utilized a mixture of red lead and putty, baking the assemblies in an oven to harden the seal. The interior of the sphere was cramped and hostile. Beebe and Barton had to crawl through a circular hatch just 1414 inches in diameter, which was then secured by a heavy steel lid fastened with 1010 large brass bolts. Inside, the researchers sat on the cold steel floor, surrounded by essential life-support equipment.

Oxygen was supplied from 22 high-pressure cylinders, each containing oxygen at 150150 atmospheres, which was released through a reducing valve at a constant rate of 22 liters per minute—sufficient for both men for up to 88 hours. To prevent the accumulation of toxic gases, 22 wire mesh baskets were suspended from the ceiling. One basket contained 33 pounds of soda lime to absorb carbon dioxide, while the other held 33 pounds of calcium chloride to remove moisture exhaled by the divers. A single, custom-designed cable connected the Bathysphere to the surface. This cable housed both a telephone line and a power wire that fed a 250250-watt spotlight. On the deck of the Ready, Gloria Hollister, a research associate and an accomplished zoologist in her own right, sat with a telephone headset, recording Beebe’s breathless, real-time descriptions as they descended.

On their historic dive of June 11, 19301930, the pair reached a depth of 800800 feet, far deeper than any human had survived in a submersible before. Over the next 44 years, they pushed the limits of the vessel, culminating in a record-breaking dive to 3,0283,028 feet on August 15, 19341934. In the pitch-black depths, illuminated only by the faint glow of the spotlight and the bioluminescence of passing creatures, Beebe witnessed a world never before seen by human eyes. He observed species that were entirely new to science, such as the *Bathysphaera intacta* (the giant dragonfish) and the *Bathysidus pentagrammus* (the five-lined constellation fish). Because these observations were fleeting and could not be verified by physical specimens, some contemporary ichthyologists reacted with intense skepticism. They argued that the optical distortions of the thick quartz windows, combined with the psychological effects of sensory deprivation and nitrogen narcosis, had caused Beebe to misidentify known species or imagine entirely new ones. Nevertheless, Beebe’s meticulous notes, preserved through his telephone conversations with Hollister, laid the groundwork for modern deep-sea biology.

Based on the passage, is the statement that the Watson-Stillman Hydraulic Machinery Company ground and polished the Bathysphere’s 33-inch-thick quartz window cylinders true or false?

Cevabı ve açıklamayı göster

Cevap: False

Cevap

The statement is false because the quartz window cylinders were ground and polished by the General Electric Company, not the Watson-Stillman Hydraulic Machinery Company.
The correct answer is False because the passage explicitly states that the quartz window cylinders were ground and polished by the General Electric Company, whereas the Watson-Stillman Hydraulic Machinery Company was responsible for casting the steel sphere itself.

Adım Adım Çözüm

1
Locate the mentions of the Watson-Stillman Hydraulic Machinery Company and the quartz window cylinders in the text.
The Watson-Stillman Hydraulic Machinery Company is mentioned in the first paragraph in connection with casting the steel sphere. The quartz window cylinders are mentioned in the second paragraph.
Finding the specific sentences that mention these components allows us to verify their exact properties and origins.
2
Verify which company was responsible for the quartz window cylinders.
The second paragraph explicitly states that the quartz cylinders were 'ground and polished to optical perfection by the General Electric Company.'
This details the actual manufacturer of the windows to compare against the claim in the statement.
3
Compare the passage's explicit statements with the prompt's statement to determine its truth value.
Since the General Electric Company polished the windows while the Watson-Stillman Hydraulic Machinery Company cast the sphere, the statement misattributes the window polishing to the sphere caster and is therefore false.
A direct comparison of the details reveals the error in the statement, confirming the correct answer is False.

Anahtar Kavram

Identifying and verifying explicitly stated details in a text by distinguishing between different entities and their associated actions.
Tahmini Süre:2m 0s
Soru 2690Soru

A point on the unit circle starts at (1,0)(1, 0) and undergoes a counterclockwise rotation of 570570^\circ about the origin. What is the radian measure of the angle in the interval [0,2π)[0, 2\pi) that corresponds to the point's final position?

Cevabı ve açıklamayı göster

Cevap: 7π6\frac{7\pi}{6}

Cevap

The correct answer is the option representing 7π6\frac{7\pi}{6} radians.
The correct answer represents 7π6\frac{7\pi}{6} radians. To find the position on the unit circle after a counterclockwise rotation of 570570^\circ, we first determine the coterminal angle within one full rotation (360360^\circ). Subtracting 360360^\circ from 570570^\circ gives 210210^\circ. We then convert this angle to radians by multiplying by π180\frac{\pi}{180^\circ}, which simplifies to 7π6\frac{7\pi}{6} radians.

Adım Adım Çözüm

1
Find a coterminal angle for 570570^\circ within the standard [0,360)[0^\circ, 360^\circ) range.
570360=210570^\circ - 360^\circ = 210^\circ
Since a full rotation is 360360^\circ, subtracting 360360^\circ yields an angle in the same position on the unit circle but within one full rotation.
2
Convert the coterminal angle from degrees to radians by multiplying by π180\frac{\pi}{180^\circ}.
210×π180=210π180=7π6210^\circ \times \frac{\pi}{180^\circ} = \frac{210\pi}{180} = \frac{7\pi}{6} radians
To convert degrees to radians, multiply by the conversion factor π180\frac{\pi}{180^\circ}.
3
Verify that the resulting angle is in the interval [0,2π)[0, 2\pi).
Since 07π6<2π0 \le \frac{7\pi}{6} < 2\pi, the angle is in the correct interval.
The question requires the final angle to be in the interval [0,2π)[0, 2\pi).

Anahtar Kavram

Finding coterminal angles and converting degree measures to radian measures on the unit circle.

Alternatif Yöntem

Convert the rotation to radians first: 570×π180=19π6570^\circ \times \frac{\pi}{180^\circ} = \frac{19\pi}{6} radians. Then, subtract 2π2\pi to find the coterminal angle: 19π62π=7π6\frac{19\pi}{6} - 2\pi = \frac{7\pi}{6} radians.
Tahmini Süre:1m 15s
Soru 2691Soru

In the standard (x,y)(x, y) coordinate plane, a parallelogram ABCDABCD has vertices A(6,3)A(-6, -3), B(1,3)B(1, -3), C(6,2)C(6, 2), and D(1,2)D(-1, 2). What is the length of the diagonal ACAC?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The correct answer is 13.
The length of the diagonal ACAC is found by calculating the distance between the vertices A(6,3)A(-6, -3) and C(6,2)C(6, 2). Using the distance formula: AC=(6(6))2+(2(3))2=122+52=144+25=169=13AC = \sqrt{(6 - (-6))^2 + (2 - (-3))^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.

Adım Adım Çözüm

1
Identify the coordinates of the endpoints of diagonal ACAC.
A(6,3)A(-6, -3) and C(6,2)C(6, 2)
The diagonal of a polygon connects non-adjacent vertices, so the diagonal ACAC connects vertex AA to vertex CC.
2
Use the distance formula to find the length of the segment ACAC.
AC=(6(6))2+(2(3))2AC = \sqrt{(6 - (-6))^2 + (2 - (-3))^2}
The distance formula calculates the straight-line distance between two points on the coordinate plane.
3
Simplify the expression to find the final integer length.
1313
Calculating the squares and summing them gives 122+52=144+25=169=13\sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.

Anahtar Kavram

Calculating the length of a geometric segment on the coordinate plane using the distance formula.
Soru 2692Soru

An angle θ\theta in standard position measures 750-750^\circ. Let ϕ\phi be the coterminal angle of θ\theta such that 0ϕ<2π0 \le \phi < 2\pi radians. If ϕ=aπb\phi = \frac{a\pi}{b}, where aa and bb are positive integers with no common factors, what is the value of a+ba + b?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

17
To find the coterminal angle ϕ\phi in the range [0,2π)[0, 2\pi) radians, we first determine the coterminal angle in degrees by adding multiples of 360360^\circ. Adding 3×360=10803 \times 360^\circ = 1080^\circ to 750-750^\circ results in 330330^\circ. We then convert 330330^\circ to radians by multiplying by π180\frac{\pi}{180^\circ}, which simplifies to 11π6\frac{11\pi}{6}. Since 1111 and 66 are positive integers with no common factors, a=11a = 11 and b=6b = 6. The sum a+ba + b is 11+6=1711 + 6 = 17.

Adım Adım Çözüm

1
Find the positive coterminal angle of 750-750^\circ within one full rotation.
330330^\circ
Adding 10801080^\circ (three full rotations of 360360^\circ) to 750-750^\circ brings the angle within the standard range of [0,360)[0^\circ, 360^\circ).
2
Convert the angle from degrees to radians.
11π6\frac{11\pi}{6} radians
Multiplying the degree measure by π180\frac{\pi}{180^\circ} and simplifying the fraction converts it to radians.
3
Identify the values of aa and bb and compute their sum.
1717
Comparing 11π6\frac{11\pi}{6} to aπb\frac{a\pi}{b} shows a=11a = 11 and b=6b = 6, which share no common factors. Their sum is 11+6=1711 + 6 = 17.

Anahtar Kavram

Finding coterminal angles and converting angle measures between degrees and radians.
Tahmini Süre:1m 30s
Soru 2693Soru

A line graphed on the standard (x,y)(x, y) coordinate plane has a slope of 0.50.5 and passes through the points (p,3)(p, 3) and (7,2p1)(7, 2p - 1). What is the value of pp?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The value of pp is 33.
Applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} to the points (p,3)(p, 3) and (7,2p1)(7, 2p - 1) with m=0.5m = 0.5 yields the linear equation 0.5=2p47p0.5 = \frac{2p - 4}{7 - p}. Cross-multiplying and solving for pp gives p=3p = 3.

Adım Adım Çözüm

1
Set up the slope formula using the given points and slope.
0.5=(2p1)37p0.5 = \frac{(2p - 1) - 3}{7 - p}
The slope of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is given by the formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Simplify the numerator of the fraction.
0.5=2p47p0.5 = \frac{2p - 4}{7 - p}
Combine the constant terms 1-1 and 3-3 in the numerator.
3
Multiply both sides by the denominator to clear the fraction.
0.5(7p)=2p40.5(7 - p) = 2p - 4
Eliminating the denominator allows solving the linear equation.
4
Distribute the 0.50.5 on the left side of the equation.
3.50.5p=2p43.5 - 0.5p = 2p - 4
Multiply each term inside the parentheses by 0.50.5.
5
Isolate the variable pp by rearranging terms.
2.5p=7.52.5p = 7.5
Add 0.5p0.5p to both sides and add 44 to both sides.
6
Divide by the coefficient of pp to solve for pp.
p=3p = 3
Dividing 7.57.5 by 2.52.5 yields the final value of the parameter.

Anahtar Kavram

Calculating a parameter using the slope formula

Alternatif Yöntem

Instead of using the decimal 0.50.5, write it as the fraction 12\frac{1}{2}. The equation becomes 12=2p47p\frac{1}{2} = \frac{2p - 4}{7 - p}. Cross-multiplying gives 1(7p)=2(2p4)7p=4p815=5pp=31(7 - p) = 2(2p - 4) \Rightarrow 7 - p = 4p - 8 \Rightarrow 15 = 5p \Rightarrow p = 3.
Tahmini Süre:1m 30s
Soru 2694Soru

Deep-sea hydrothermal vents, first discovered in 19771977, are fissures on the seafloor that release geothermally heated water. Because sunlight cannot penetrate the ocean depths where these vents exist, photosynthesis is impossible. Instead, these ecosystems rely on chemosynthesis, a process in which specialized microbes utilize chemical compounds, primarily hydrogen sulfide, to produce energy. The sudden emergence of these mineral-rich hot springs results in a highly localized deposition of copper, iron, and zinc sulfides onto the ocean floor, forming towering chimneys known as "black smokers." As the superheated water, reaching temperatures up to 400400 degrees Celsius, meets the near-freezing ocean water, the dissolved minerals precipitate out of solution immediately. This rapid precipitation not only constructs the physical structures of the chimneys but also establishes a dense chemical gradient. Crucially, the high concentration of hydrogen sulfide emitted by the vents serves as a vital nutrient source for symbiotic bacteria. These bacteria colonize the tissues of giant tube worms, which lack mouths and digestive tracts. Consequently, the tube worms are entirely dependent on these internal microbes for survival, as the bacteria convert the chemicals into organic molecules that nourish the worms.

According to the passage, the immediate precipitation of dissolved minerals on the ocean floor is caused by which of the following?

Cevabı ve açıklamayı göster

Cevap: The contact between superheated vent water and near-freezing ocean water

Cevap

The correct answer states that the contact between superheated vent water and near-freezing ocean water causes the minerals to precipitate.
The passage explicitly states that as the superheated water meets the near-freezing ocean water, the dissolved minerals precipitate out of solution immediately. This directly links the contact between these two water sources of extreme temperatures to the precipitation of the minerals.

Adım Adım Çözüm

1
Locate the mention of mineral precipitation in the passage.
The phrase "dissolved minerals precipitate out of solution immediately" is identified.
Finding the specific effect mentioned in the text allows us to look for its direct cause.
2
Identify the cause linked to this precipitation in the same sentence.
The text states that this occurs "As the superheated water... meets the near-freezing ocean water."
Analyzing the immediate context reveals the explicit cause-and-effect relationship.

Anahtar Kavram

Recognizing Explicit Cause and Effect
Soru 2695Soru

Passage

The evolution of the modern bicycle is a story of iterative mechanical improvements designed to increase speed, stability, and rider comfort. The journey began in 1817 when German baron Karl von Drais built a steerable, two-wheeled wooden vehicle. Patented in 1818 as the 'draisienne' (and often called the 'running machine' or 'dandy horse'), this device lacked pedals. Instead, riders sat astride a wooden frame and propelled themselves by pushing their feet against the ground in a walking motion. Although it demonstrated that a rider could balance on two wheels, the physical exertion required to operate it on rough carriage roads limited its widespread adoption.

For decades, the design remained a novelty until the early 1860s in Paris. Carriage makers Pierre and Ernest Michaux modified the design by attaching cranks and pedals directly to the front wheel's axle. This machine, called the 'velocipede,' allowed riders to propel themselves without touching the ground. However, because its frame was made of wrought iron and its wheels were wooden with iron bands, the vehicle transmitted every bump in the road directly to the rider. The resulting rough ride earned the velocipede the popular nickname of the 'boneshaker.'

In the 1870s, British engineers sought to make the bicycle faster. Because the pedals were still directly attached to the wheel axle, the only way to increase speed was to make the drive wheel larger. This led to the creation of the 'ordinary bicycle,' commonly known as the 'high-wheeler' or 'penny-farthing' because its wheels resembled the large penny and small farthing coins of the era. The front wheel grew to diameters of up to five feet, while the rear wheel shrank to a fraction of that size. These bicycles featured solid rubber tires and wire-spoked wheels, which slightly cushioned the ride. However, their high center of gravity made them extremely dangerous; hitting a stone could throw the rider forward over the handlebars in a crash known as a 'header.'

Recognizing the need for a safer design, English inventor John Kemp Starley introduced the Rover Safety Bicycle in 1885. The Rover featured two wheels of nearly equal size and a chain drive mechanism that connected the pedals to the rear wheel. This meant the bicycle could travel fast without requiring a dangerously large front wheel. The final major improvement came in 1888 when Scottish inventor John Boyd Dunlop patented the pneumatic (air-filled) rubber tire for bicycles. Dunlop’s invention replaced the jarring solid rubber tires, providing a smooth ride and securing the bicycle's place as a practical mode of daily transportation.

Based on the passage, in what chronological order did the following developments in bicycle design occur, from earliest to latest?

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct chronological order of the developments is: the steerable wooden vehicle without pedals (1817), followed by the velocipede with pedals attached directly to the axle (1860s), followed by the equal-wheeled safety bicycle with a rear chain drive (1885), and finally the pneumatic rubber tire (1888).
The correct chronological order of the developments is: the steerable wooden vehicle without pedals (1817), followed by the velocipede with pedals attached directly to the axle (1860s), followed by the equal-wheeled safety bicycle with a rear chain drive (1885), and finally the pneumatic rubber tire (1888).

Adım Adım Çözüm

1
Identify the earliest invention mentioned in the text.
The steerable, pedal-less wooden vehicle (draisienne) developed in 1817.
The text explicitly states this was the earliest precursor to the bicycle, built in 1817.
2
Identify the next chronologically occurring development.
The machine with pedals attached directly to the front wheel's axle (velocipede) in the early 1860s.
The passage describes this as occurring decades after the 1817 draisienne.
3
Identify the subsequent design evolution.
The safety bicycle with equal-sized wheels and a rear chain drive introduced in 1885.
The passage dates this safety-oriented design to 1885, which is after the velocipede and high-wheeler.
4
Identify the final technological addition described in the text.
The air-filled (pneumatic) rubber tire patented in 1888.
The text lists this as the final major improvement in 1888, replacing the solid rubber tires of the safety bicycle.

Anahtar Kavram

Identifying Explicit Details
Soru 2696Soru

For decades, biologists studying the high-altitude forests of the Andes observed a mysterious decline in the local population of the golden-backed tree frog. Standard hypotheses pointed to shifting temperatures or increased solar radiation at high elevations. However, a 2018 study revealed a far more direct ecological chain reaction. The construction of a new highway through the valley below had introduced a non-native predatory beetle to the region. These beetles, which thrived in the cleared roadside brush, gradually migrated up the slopes. Because the golden-backed tree frog relies on a specific species of mountain bromeliad for shelter, the frogs were highly vulnerable. The predatory beetles did not eat the frogs; instead, they fed voraciously on the bromeliad leaves, destroying the plants. Deprived of their only nesting sites and protection from predators, the frogs suffered a rapid population collapse. This simple disruption of the local flora highlights how human infrastructure projects can have devastating downstream effects on specialized montane species.

Based on the passage, the population of the golden-backed tree frog collapsed because of which direct cause?

Cevabı ve açıklamayı göster

Cevap: The beetles destroyed the mountain bromeliad plants that the frogs needed for nesting and protection.

Cevap

The beetles destroyed the mountain bromeliad plants that the frogs needed for nesting and protection.
The correct option is supported directly by the text, which explains that the predatory beetles fed on the bromeliad leaves and destroyed the plants. Because the frogs relied on these plants for nesting and protection, their destruction directly caused the frog population to collapse.

Adım Adım Çözüm

1
Locate where the passage describes the collapse of the frog population.
The passage states that the frogs 'suffered a rapid population collapse' because they were 'deprived of their only nesting sites and protection.'
Identifying the target effect described in the question.
2
Identify the cause of this deprivation as explicitly stated in the sentence immediately prior.
The text explains that the beetles 'fed voraciously on the bromeliad leaves, destroying the plants' and that the frogs 'relies on a specific species of mountain bromeliad for shelter.'
Connecting the destruction of the plants by the beetles directly to the collapse of the frog population.

Anahtar Kavram

Recognizing Explicit Cause and Effect
Soru 2697Soru

In triangle PQRPQR, the measure of angle PP is 3030^\circ, the measure of angle QQ is 4545^\circ, and the length of side QRQR is 1010 units. Which of the following expressions represents the length, in units, of side PRPR?

Cevabı ve açıklamayı göster

Cevap: 10sin(45)sin(30)\frac{10\sin(45^\circ)}{\sin(30^\circ)}

Cevap

The length of side PRPR is represented by the expression 10sin(45)sin(30)\frac{10\sin(45^\circ)}{\sin(30^\circ)}.
The correct answer is derived by setting up the Law of Sines proportion: PRsin(45)=10sin(30)\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}. Multiplying both sides by sin(45)\sin(45^\circ) yields PR=10sin(45)sin(30)PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}.

Adım Adım Çözüm

1
Identify the known values and corresponding angle-side pairs in triangle PQRPQR.
Angle P=30P = 30^\circ is opposite to side QR=10QR = 10, and angle Q=45Q = 45^\circ is opposite to side PRPR.
This allows us to set up the appropriate trigonometric relationship to solve for the unknown side.
2
Apply the Law of Sines to relate the ratios of the side lengths to the sines of their opposite angles.
PRsin(Q)=QRsin(P)\frac{PR}{\sin(Q)} = \frac{QR}{\sin(P)}, which becomes PRsin(45)=10sin(30)\frac{PR}{\sin(45^\circ)} = \frac{10}{\sin(30^\circ)}.
The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is constant for all three sides.
3
Isolate the variable representing the length of side PRPR.
PR=10sin(45)sin(30)PR = \frac{10\sin(45^\circ)}{\sin(30^\circ)}.
Multiply both sides of the equation by sin(45)\sin(45^\circ) to solve for PRPR.

Anahtar Kavram

The Law of Sines relates the side lengths of a triangle to the sines of its angles: asin(A)=bsin(B)=csin(C)\frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)}.
Tahmini Süre:1m 0s
Soru 2698Soru

For the trigonometric equations on the left, which description on the right correctly matches the graphical features of each equation?

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Öğeler

y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1
y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1
y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2
y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2

Eşleşmeler

Cevabı ve açıklamayı göster

Cevap

The correct matches are: the equation y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1 matches the description with a period of π\pi, a range of [2,4][-2, 4], and a positive yy-intercept; the equation y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1 matches the description with a period of 4π4\pi, a range of [3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units to the left; the equation y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2 matches the description with a period of π3\frac{\pi}{3} that is undefined at the yy-axis; and the equation y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2 matches the description with a period of 3π3\pi, a range of [5,1][-5, 1], and a yy-intercept of (0,2)(0, -2).
Each equation is correctly paired with the unique graphical properties determined by calculating its amplitude, period, range, phase shift, and yy-intercept. The first equation features a period of π\pi, a range of [2,4][-2, 4], and a positive yy-intercept. The second equation has a period of 4π4\pi, a range of [3,1][-3, 1], and a phase shift of π2\frac{\pi}{2} units left. The third equation has a period of π3\frac{\pi}{3} and is undefined at the yy-axis since x=0x=0 creates a vertical asymptote. The fourth equation has a period of 3π3\pi, a range of [5,1][-5, 1], and a yy-intercept of (0,2)(0, -2).

Adım Adım Çözüm

1
Determine the period, range, and y-intercept for the function y=3sin(2xπ3)+1y = -3\sin\left(2x - \frac{\pi}{3}\right) + 1.
The period is π\pi, range is [2,4][-2, 4], and yy-intercept is 1+332>01 + \frac{3\sqrt{3}}{2} > 0.
The period is 2π2=π\frac{2\pi}{2} = \pi, the range is [13,1+3]=[2,4][1-3, 1+3] = [-2, 4], and evaluating at x=0x=0 yields y=3sin(π/3)+1>0y = -3\sin(-\pi/3) + 1 > 0.
2
Determine the period, range, and phase shift for the function y=2cos(12x+π4)1y = 2\cos\left(\frac{1}{2}x + \frac{\pi}{4}\right) - 1.
The period is 4π4\pi, range is [3,1][-3, 1], and phase shift is π2\frac{\pi}{2} units left.
The period is 2π1/2=4π\frac{2\pi}{1/2} = 4\pi, the range is [12,1+2]=[3,1][-1-2, -1+2] = [-3, 1], and factoring the argument yields 12(x+π2)\frac{1}{2}(x + \frac{\pi}{2}), giving a shift of π2\frac{\pi}{2} units left.
3
Determine the period and domain restriction for the function y=tan(3xπ2)+2y = -\tan\left(3x - \frac{\pi}{2}\right) + 2.
The period is π3\frac{\pi}{3} and the function is undefined at x=0x = 0 (the yy-axis).
For tangent, the period is π3\frac{\pi}{3}. At x=0x=0, the argument is π2-\frac{\pi}{2}, where tangent is undefined, meaning the function has a vertical asymptote at the yy-axis.
4
Determine the period, range, and y-intercept for the function y=3sin(23x+π)2y = 3\sin\left(\frac{2}{3}x + \pi\right) - 2.
The period is 3π3\pi, range is [5,1][-5, 1], and yy-intercept is (0,2)(0, -2).
The period is 2π2/3=3π\frac{2\pi}{2/3} = 3\pi, the range is [23,2+3]=[5,1][-2-3, -2+3] = [-5, 1], and at x=0x=0, y=3sin(π)2=2y = 3\sin(\pi) - 2 = -2.

Anahtar Kavram

Identifying graphs of trigonometric functions from their equations by determining amplitude, period, phase shift, midline, range, and asymptotes.
Soru 2699Soru

A surveyor stands at point AA on flat ground and measures the angle of elevation to the top of a vertical tower, TT, at point CC to be 3030^\circ. Another surveyor at point BB, which is 100100 meters away from AA on the same flat ground, measures CAB=40\angle CAB = 40^\circ and CBA=65\angle CBA = 65^\circ. Which of the following expressions represents the height, in meters, of the tower?

Cevabı ve açıklamayı göster

Cevap: 100sin(65)tan(30)sin(75)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}

Cevap

The correct expression is 100sin(65)tan(30)sin(75)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
The correct expression is derived by first applying the Law of Sines to find the length of the ground segment ACAC, and then using right-triangle trigonometry to determine the height of the tower. In triangle ABCABC, the third angle ACB\angle ACB is 180(40+65)=75180^\circ - (40^\circ + 65^\circ) = 75^\circ. The Law of Sines gives ACsin(65)=100sin(75)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)}, which simplifies to AC=100sin(65)sin(75)AC = \frac{100\sin(65^\circ)}{\sin(75^\circ)}. Since the tower is vertical, triangle ACTACT is a right triangle with tan(30)=hAC\tan(30^\circ) = \frac{h}{AC}. Substituting ACAC yields h=100sin(65)tan(30)sin(75)h = \frac{100\sin(65^\circ)\tan(30^\circ)}{\sin(75^\circ)}.

Adım Adım Çözüm

1
Calculate the measure of the third angle in the ground triangle ABC\triangle ABC.
ACB=180(40+65)=75\angle ACB = 180^\circ - (40^\circ + 65^\circ) = 75^\circ
The sum of angles in any triangle must equal 180180^\circ.
2
Apply the Law of Sines to find the distance from point AA to the base of the tower at point CC (ACAC).
ACsin(65)=100sin(75)AC=100sin(65)sin(75)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)} \Rightarrow AC = \frac{100 \sin(65^\circ)}{\sin(75^\circ)}
The Law of Sines states that the ratio of a side length to the sine of its opposite angle is constant in a triangle.
3
Use the right-triangle trigonometric ratio for the vertical tower height hh from point AA.
tan(30)=hACh=ACtan(30)\tan(30^\circ) = \frac{h}{AC} \Rightarrow h = AC \tan(30^\circ)
In right triangle ACT\triangle ACT, the tangent of the angle of elevation is the ratio of the opposite side (height hh) to the adjacent side (ACAC).
4
Substitute the expression for ACAC into the equation for hh.
h=100sin(65)tan(30)sin(75)h = \frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
Replacing ACAC with its equivalent algebraic expression yields the final height in terms of the given parameters.

Anahtar Kavram

Applying the Law of Sines to find a missing side length in a non-right triangle and then using right-triangle trigonometric ratios to solve a 3D geometry problem.
Tahmini Süre:2m 0s
Soru 2700Soru

This passage is adapted from an article about the history of modern astronomy.

In the late nineteenth and early twentieth centuries, the Harvard College Observatory was a center of astronomical innovation, though much of its most tedious work was performed by a dedicated group of women. Known colloquially as the "Harvard Computers," these women were hired by the observatory director, Edward Charles Pickering, to analyze and catalog thousands of photographic plates of the night sky. Among these researchers was Henrietta Swan Leavitt, a graduate of Radcliffe College who joined the observatory in 1895. Despite facing progressive hearing loss and receiving little public recognition during her lifetime, Leavitt would make a discovery that fundamentally transformed our understanding of the scale of the universe.

Leavitt was assigned to study variable stars—stars whose brightness changes over time. In particular, she focused her attention on the Small Magellanic Cloud, a dwarf galaxy visible in the Southern Hemisphere. Using a magnifying glass to examine the glass photographic plates, Leavitt identified thousands of variable stars, including a specific class known as Cepheid variables. Cepheids are pulsating stars that brighten and dim in a highly predictable, cyclical pattern. Leavitt meticulously recorded the minimum and maximum brightness of each variable, as well as the precise length of its cycle, or period.

In 1908, Leavitt published her initial catalog of 1,777 variable stars in the Annals of the Astronomical Observatory of Harvard College. In this paper, she noted a curious trend: a handful of the brightest Cepheid variables appeared to have the longest periods. Recognizing the significance of this observation, Leavitt continued her investigations. By 1912, she had compiled data for 25 Cepheid variables in the Small Magellanic Cloud, confirming a direct mathematical relationship: the longer a star’s period of pulsation, the greater its intrinsic brightness, or luminosity.

The crucial element of Leavitt’s discovery lay in the location of the stars she observed. Because all the Cepheids in her study were located within the Small Magellanic Cloud, they were all roughly the same distance from Earth. Therefore, any difference in their apparent brightness as seen from Earth was not a function of their distance, but rather reflected a real difference in their actual light output. By establishing this "period-luminosity relation," Leavitt provided astronomers with a revolutionary tool. If the period of a distant Cepheid could be measured, its absolute luminosity could be calculated. By comparing this absolute luminosity to the star’s apparent brightness, astronomers could determine exactly how far away the star—and the galaxy hosting it—was.

Prior to Leavitt’s breakthrough, astronomers relied almost exclusively on stellar parallax to calculate distances. This geometric method, which measures the apparent shift of a nearby star against more distant background stars as Earth orbits the Sun, was highly accurate but severely limited. Due to the limits of early twentieth-century telescopes, parallax could only be used to measure distances to stars within approximately 100 light-years of Earth. Beyond this narrow bubble, the universe was a vast, unmeasurable expanse, and astronomers actively debated whether the Milky Way constituted the entirety of the cosmos.

Leavitt’s period-luminosity relation, often referred to as the cosmic "standard candle," shattered these boundaries. In 1924, astronomer Edwin Hubble located Cepheid variables in the Andromeda Nebula. Using Leavitt’s relationship, Hubble calculated that Andromeda was roughly 900,000 light-years away—far outside the boundaries of the Milky Way. This single calculation proved that Andromeda was not a cloud of gas within our galaxy, but an independent galaxy of its own. Hubble’s subsequent discovery of the expanding universe was built directly upon the foundation of Leavitt’s meticulous work with the glass plates of Harvard.

According to the passage, prior to Henrietta Swan Leavitt's discovery of the period-luminosity relation, what method did astronomers primarily use to calculate astronomical distances?

Cevabı ve açıklamayı göster

Cevap: Stellar parallax

Cevap

Stellar parallax
The correct answer is correct because the passage explicitly states that before Henrietta Swan Leavitt's breakthrough, astronomers relied almost exclusively on stellar parallax to calculate distances.

Adım Adım Çözüm

1
Locate the portion of the text that discusses the methods astronomers used to calculate astronomical distances prior to Henrietta Swan Leavitt's breakthrough.
The fifth paragraph of the passage states: 'Prior to Leavitt’s breakthrough, astronomers relied almost exclusively on stellar parallax to calculate distances.'
Finding the specific sentence in the text that explicitly addresses the timeframe before Leavitt's discovery is the direct way to identify the correct detail.
2
Compare the retrieved detail with the given options to find an exact or paraphrased match.
The option containing 'Stellar parallax' matches the explicit wording from the fifth paragraph of the passage.
Matching the retrieved text with the correct option confirms the correct answer without requiring any extra inferences.

Anahtar Kavram

Identifying Explicit Details
Tahmini Süre:2m 0s
ÖncekiSayfa 135 / 278Sonraki
Tüm alıştırma soruları — ACT | Examkin