Tüm alıştırma soruları

5556 soru

Soru 2601Soru

For an angle θ\theta in the interval π2<θ<π\frac{\pi}{2} < \theta < \pi, the value of cosθ=35\cos \theta = -\frac{3}{5}. What is the value of sinθ+cosθ\sin \theta + \cos \theta?

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Cevap: 15\frac{1}{5}

Cevap

one-fifth
To evaluate sinθ+cosθ\sin \theta + \cos \theta, we first find the value of sinθ\sin \theta. We substitute cosθ=35\cos \theta = -\frac{3}{5} into the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1, which gives sin2θ+925=1\sin^2 \theta + \frac{9}{25} = 1. Solving for sin2θ\sin^2 \theta yields 1625\frac{16}{25}. Because the angle θ\theta is constrained to the second quadrant (π2<θ<π\frac{\pi}{2} < \theta < \pi), its sine value must be positive, which means sinθ=45\sin \theta = \frac{4}{5}. Adding the values together, we get 45+(35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.

Adım Adım Çözüm

1
Use the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 to find the magnitude of the sine function.
sin2θ=1(35)2=1925=1625\sin^2 \theta = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}
The Pythagorean identity relates sine and cosine for any angle.
2
Determine the correct sign of sinθ\sin \theta based on the given quadrant interval.
Since π2<θ<π\frac{\pi}{2} < \theta < \pi, the angle θ\theta lies in Quadrant II, where the sine function is positive. Thus, sinθ=1625=45\sin \theta = \sqrt{\frac{16}{25}} = \frac{4}{5}.
The trigonometric function values are positive or negative depending on the quadrant on the unit circle.
3
Compute the sum of sinθ\sin \theta and cosθ\cos \theta.
sinθ+cosθ=45+(35)=15\sin \theta + \cos \theta = \frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}
This gives the final value of the requested expression.

Anahtar Kavram

Using the Pythagorean identity sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1 and quadrant rules to calculate trigonometric values.

Alternatif Yöntem

We can sketch a reference right triangle in Quadrant II. Since cosθ=adjacenthypotenuse=35\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = -\frac{3}{5}, we assign the adjacent side a length of 3-3 along the x-axis and the hypotenuse a length of 55. By the Pythagorean theorem, the opposite vertical side is 52(3)2=4\sqrt{5^2 - (-3)^2} = 4. Since the vertical side is in Quadrant II, it is positive. This makes sinθ=oppositehypotenuse=45\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}. Evaluating the sum yields 45+(35)=15\frac{4}{5} + \left(-\frac{3}{5}\right) = \frac{1}{5}.
Tahmini Süre:45s
Soru 2602Soru

A line with a slope of 2.52.5 passes through the point (3,4)(3, 4) in the coordinate plane. What is the y-coordinate of the point on this line whose x-coordinate is 1-1?

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Cevap: -6

Cevap

-6
By applying the slope formula m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1} with the slope equal to 2.52.5, the known point at (3,4)(3, 4), and the target point at (1,y)(-1, y), we obtain the linear equation 2.5=y4132.5 = \frac{y - 4}{-1 - 3}. Simplifying the denominator gives 2.5=y442.5 = \frac{y - 4}{-4}. Multiplying both sides by 4-4 yields 10=y4-10 = y - 4. Finally, adding 44 to both sides gives y=6y = -6.

Adım Adım Çözüm

1
Set up the slope formula relation
2.5=y4132.5 = \frac{y - 4}{-1 - 3}
The slope mm of a line passing through (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is defined as the change in y-values divided by the change in x-values.
2
Solve for the unknown y-coordinate
y=6y = -6
Simplifying the denominator to 4-4, multiplying both sides by 4-4 gives 10=y4-10 = y - 4, and adding 44 to both sides isolates yy.

Anahtar Kavram

Finding a missing coordinate on a line given its slope and another coordinate point.
Soru 2603Soru

Passage

Deep beneath the rugged Chihuahuan Desert of southern New Mexico lies Carlsbad Caverns, a spectacular labyrinth of more than 119 known limestone caves. Formed not by the typical action of surface water carving through rock, but by sulfuric acid eating away at limestone from the bottom up, Carlsbad Caverns represents a geological marvel. The process began millions of years ago when hydrogen sulfide gas, migrating upward from deep oil deposits, mixed with oxygen-rich rainwater that had seeped into the water table. This reaction created sulfuric acid, which dissolved the limestone along fractures and faults, creating massive chambers.

Among these chambers, the most famous is the Big Room, a massive limestone chamber that ranks as one of the largest cave chambers in North America. Measuring 4,000 feet long, 625 feet wide, and 255 feet high at its loftiest point, the Big Room is so spacious that it could comfortably accommodate six football fields. Visitors to the Big Room are treated to an astonishing array of speleothems—secondary mineral deposits formed in caves. These include stalactites, which hang like icicles from the ceiling; stalagmites, which rise from the floor; and delicate soda straws, which are hollow, thin-walled tubes of calcite.

The discovery of Carlsbad Caverns is traditionally attributed to Jim White, a teenage cowboy who first entered the caves in 1898. Riding near the foothills of the Guadalupe Mountains, White noticed a massive dark cloud rising from the desert floor. Drawing closer, he realized the cloud was not smoke, but millions of Mexican free-tailed bats exiting a natural opening in the earth. Armed only with a kerosene lantern and a homemade wire ladder, White spent years exploring the dark passages, marveling at the formations. Although initially met with skepticism by locals who doubted his tales of a vast underground wonderland, White eventually convinced others of the cavern’s significance. His efforts paved the way for Carlsbad Caverns to be designated a national monument in 1923, and later a national park in 1930.

Beyond its geological significance, the cavern ecosystem hosts a diverse array of life. The most notable inhabitants are the Mexican free-tailed bats, which use the cave’s natural entrance as a summer home. From May through October, an estimated 400,000 bats roost in the dark recesses of the cave during the day. At dusk, they emerge in a spectacular swirling column to hunt for insects over the Pecos River Valley. Interestingly, the bats do not roost in the Big Room itself, which remains too cool and damp for their rearing needs; instead, they occupy the Bat Cave, a separate lateral passage near the natural entrance. The bat population provides a vital ecological service, consuming tons of moths and beetles each night, which helps protect local agricultural crops from pests.

Today, park researchers continue to study the caves to better understand the delicate balance between preservation and public access. The introduction of artificial lighting, while necessary for visitor safety, has promoted the growth of algae and disrupted the natural behavior of cave-adapted organisms. Furthermore, the footprints of early explorers and modern tourists alike leave lasting marks on the sensitive cave formations. Preservationists work diligently to monitor humidity levels, airflow, and carbon dioxide concentrations to ensure that this underground treasure remains intact for future generations.

In recent decades, scientific interest in Carlsbad Caverns has expanded beyond traditional geology and ecology into the field of geomicrobiology. Researchers have discovered that the cave walls are home to thriving communities of extremophilic microbes. These microorganisms, which survive in complete darkness without access to sunlight or photosynthesis, derive energy by consuming minerals like manganese, sulfur, and iron found in the cave rock. Some of these unique bacterial strains produce novel chemical compounds that have shown promise in medical research, particularly in the development of new antibiotics. This unexpected scientific frontier highlights the caverns not just as a scenic wonder, but as a living laboratory of evolutionary adaptation.

Question

According to the passage, in which specific area of the caverns do the Mexican free-tailed bats roost during the day?

Cevabı ve açıklamayı göster

Cevap: The Bat Cave

Cevap

The Bat Cave
The correct answer is the Bat Cave because the fourth paragraph explicitly states that the bats do not roost in the Big Room itself; instead, they occupy the Bat Cave, which is a separate lateral passage near the natural entrance.

Adım Adım Çözüm

1
Scan the passage for keywords related to the roosting location of the Mexican free-tailed bats.
The fourth paragraph is identified as the section focusing on the bats, their habits, and their specific locations within the cavern system.
This paragraph explicitly mentions the roosting behaviors and geographical movements of the bat population.
2
Analyze the sentences within the fourth paragraph detailing where the bats spend their time during the day.
The text states: 'Interestingly, the bats do not roost in the Big Room itself... instead, they occupy the Bat Cave, a separate lateral passage near the natural entrance.'
This directly distinguishes between where the bats do not go (the Big Room) and where they actually sleep (the Bat Cave).
3
Match the identified location to the choices provided.
The option containing the Bat Cave matches the explicit details found in the text.
This ensures the selected answer is correct without relying on unsupported inferences.

Anahtar Kavram

Identifying Explicit Details
Tahmini Süre:1m 0s
Soru 2604Soru

An object on a unit circle starts at the point (1,0)(1,0) and rotates counterclockwise through an angle of 990990^\circ. It then rotates clockwise through an angle of 13π4\frac{13\pi}{4} radians. At which of the following coordinate points on the unit circle does the object finish its path?

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Cevap: (22,22)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)

Cevap

The correct coordinate point on the unit circle is (22,22)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).
The correct coordinate point is found by converting the counterclockwise rotation of 990990^\circ to 11π2\frac{11\pi}{2} radians. Subtracting the clockwise rotation of 13π4\frac{13\pi}{4} radians yields 9π4\frac{9\pi}{4} radians. Simplifying this to its coterminal angle in the interval [0,2π)[0, 2\pi) gives π4\frac{\pi}{4} radians. The coordinates of π4\frac{\pi}{4} on the unit circle are (22,22)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).

Adım Adım Çözüm

1
Convert the initial counterclockwise rotation from degrees to radians.
990×π180=11π2990^\circ \times \frac{\pi}{180^\circ} = \frac{11\pi}{2} radians.
To perform calculations with the second rotation which is given in radians, both angle measures should be in the same unit.
2
Subtract the clockwise rotation of 13π4\frac{13\pi}{4} radians from the first rotation.
11π213π4=22π413π4=9π4\frac{11\pi}{2} - \frac{13\pi}{4} = \frac{22\pi}{4} - \frac{13\pi}{4} = \frac{9\pi}{4} radians.
Clockwise rotation corresponds to subtracting the angle from the current position.
3
Find the coterminal angle of 9π4\frac{9\pi}{4} radians within the standard interval [0,2π)[0, 2\pi).
9π42π=π4\frac{9\pi}{4} - 2\pi = \frac{\pi}{4} radians.
Subtracting full rotations (2π2\pi radians) simplifies the angle to its principal equivalent on the unit circle.
4
Determine the coordinates on the unit circle corresponding to the angle π4\frac{\pi}{4}.
(cos(π4),sin(π4))=(22,22)\left(\cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right)\right) = \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right).
On the unit circle, the coordinates of a point at angle θ\theta are (cos(θ),sin(θ))(\cos(\theta), \sin(\theta)).

Anahtar Kavram

Finding the position on the unit circle after multiple rotations by converting degrees to radians, calculating the coterminal angle, and evaluating coordinate values.
Tahmini Süre:2m 0s
Soru 2605Soru

A windshield wiper of length 15 inches15\text{ inches} sweeps through a central angle of 120120^\circ across a windshield. What is the area, in square inches, of the region swept by the wiper?

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Cevap: 75π75\pi

Cevap

75π75\pi square inches
The area of the region swept by the wiper is the area of a circle sector with a radius of 15 inches15\text{ inches} and a central angle of 120120^\circ. The formula for the area of a sector is A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2. Substituting 1515 for the radius and 120120 for the angle gives A=120360×π(15)2=13×225π=75πA = \frac{120}{360} \times \pi (15)^2 = \frac{1}{3} \times 225\pi = 75\pi square inches.

Adım Adım Çözüm

1
Determine the formula for the area of a sector of a circle.
A=θ360×πr2A = \frac{\theta}{360^\circ} \times \pi r^2
The area of a sector is a fraction of the total area of the circle, where the fraction is determined by the central angle θ\theta divided by the total degrees in a circle (360360^\circ).
2
Substitute the given values into the sector area formula, using a radius of 1515 and a central angle of 120120^\circ.
A=120360×π(15)2A = \frac{120^\circ}{360^\circ} \times \pi (15)^2
The windshield wiper length represents the radius r=15 inchesr = 15\text{ inches}, and the sweep angle represents the central angle θ=120\theta = 120^\circ.
3
Simplify the expression to find the final area.
A=13×225π=75πA = \frac{1}{3} \times 225\pi = 75\pi
Reducing the fraction 120360\frac{120}{360} to 13\frac{1}{3} and squaring 1515 to get 225225 yields the area of 75π75\pi square inches.

Anahtar Kavram

The area of a circle sector is found by multiplying the total circle area, πr2\pi r^2, by the ratio of the central angle to the total degree measure of a circle, θ360\frac{\theta}{360^\circ}.
Soru 2606Soru

For any convex quadrilateral ABCDABCD, is the statement that AB2+CD2=BC2+DA2AB^2 + CD^2 = BC^2 + DA^2 if and only if the diagonals ACAC and BDBD are perpendicular true or false?

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Cevap: True

Cevap

The statement is true because the sum of the squares of the lengths of opposite sides in a convex quadrilateral is equal if and only if its diagonals intersect at right angles.
The statement is true because the equality of the sums of the squares of opposite sides is mathematically equivalent to the diagonals being perpendicular in any convex quadrilateral.

Adım Adım Çözüm

1
Define the intersection point of the diagonals ACAC and BDBD as PP and the angle of intersection as θ\theta.
Four triangles are formed: APB\triangle APB, BPC\triangle BPC, CPD\triangle CPD, and DPA\triangle DPA, with angles at PP being θ\theta and 180θ180^\circ - \theta.
This establishes a geometric frame of reference to relate side lengths to diagonal segments.
2
Apply the Law of Cosines to express the square of each side length in terms of the diagonal segments APAP, BPBP, CPCP, and DPDP.
AB2=AP2+BP22(AP)(BP)cosθAB^2 = AP^2 + BP^2 - 2(AP)(BP)\cos\theta, CD2=CP2+DP22(CP)(DP)cosθCD^2 = CP^2 + DP^2 - 2(CP)(DP)\cos\theta, BC2=BP2+CP2+2(BP)(CP)cosθBC^2 = BP^2 + CP^2 + 2(BP)(CP)\cos\theta, and DA2=DP2+AP2+2(DP)(AP)cosθDA^2 = DP^2 + AP^2 + 2(DP)(AP)\cos\theta.
This links the boundary side lengths of the quadrilateral to its internal diagonals.
3
Sum the squares of the opposite sides and compute their difference: (AB2+CD2)(BC2+DA2)(AB^2 + CD^2) - (BC^2 + DA^2).
(AB2+CD2)(BC2+DA2)=2(APBP+CPDP+BPCP+DPAP)cosθ(AB^2 + CD^2) - (BC^2 + DA^2) = -2(AP\cdot BP + CP\cdot DP + BP\cdot CP + DP\cdot AP)\cos\theta.
This algebraic combination isolates the term involving the angle of intersection.
4
Factor the coefficient of 2cosθ-2\cos\theta and simplify the relation.
(AB2+CD2)(BC2+DA2)=2(AP+CP)(BP+DP)cosθ=2(AC)(BD)cosθ(AB^2 + CD^2) - (BC^2 + DA^2) = -2(AP + CP)(BP + DP)\cos\theta = -2(AC)(BD)\cos\theta.
Factoring groups the individual segments into the full lengths of the diagonals ACAC and BDBD.
5
Analyze the condition for the difference to be zero.
AB2+CD2=BC2+DA2    cosθ=0    θ=90AB^2 + CD^2 = BC^2 + DA^2 \iff \cos\theta = 0 \iff \theta = 90^\circ.
Since the lengths ACAC and BDBD must be positive, the difference is zero if and only if the diagonals are perpendicular.

Anahtar Kavram

Orthodiagonal quadrilateral properties and diagonal relations
Soru 2607Soru

A kite has vertices K(2,8)K(2, 8), I(6,3)I(6, 3), T(2,4)T(2, -4), and E(2,3)E(-2, 3) in the standard (x,y)(x, y) coordinate plane. What is the area of kite KITEKITE, in square units?

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Cevap: 48

Cevap

The area of kite KITEKITE is 48 square units.
The area of kite KITEKITE is 48 because the horizontal diagonal IEIE has a length of 8 units, the vertical diagonal KTKT has a length of 12 units, and the area of a kite is calculated as half the product of its diagonal lengths: 12×8×12=48\frac{1}{2} \times 8 \times 12 = 48.

Adım Adım Çözüm

1
Calculate the lengths of the vertical diagonal KTKT and the horizontal diagonal IEIE.
The length of KTKT is 12 and the length of IEIE is 8.
The vertices K(2,8)K(2, 8) and T(2,4)T(2, -4) share the same xx-coordinate, so the diagonal is vertical with length 8(4)=128 - (-4) = 12. The vertices I(6,3)I(6, 3) and E(2,3)E(-2, 3) share the same yy-coordinate, so the diagonal is horizontal with length 6(2)=86 - (-2) = 8.
2
Apply the area formula for a kite: Area=12d1d2\text{Area} = \frac{1}{2} d_1 d_2.
Area = 48
Since the diagonals of a kite are perpendicular, the area is half the product of the lengths of the diagonals: 12×12×8=48\frac{1}{2} \times 12 \times 8 = 48.

Anahtar Kavram

Finding the area of a geometric figure on the coordinate plane by using diagonal lengths.
Soru 2608Soru

In the standard (x,y)(x, y) coordinate plane, a triangle has vertices at D(1,2)D(1, -2), E(4,7)E(4, 7), and F(6,k)F(6, k). If the line containing the altitude from vertex FF to side DEDE has a yy-intercept of 88, what is the value of kk?

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Cevap: 6

Cevap

The value of kk is 66.
The slope of side DEDE is calculated as 7(2)41=3\frac{7 - (-2)}{4 - 1} = 3. An altitude is perpendicular to the side it intersects, so the slope of the altitude is the negative reciprocal of 33, which is 13-\frac{1}{3}. The line containing this altitude has a yy-intercept of 88, giving the equation y=13x+8y = -\frac{1}{3}x + 8. Since the vertex F(6,k)F(6, k) lies on this line, substituting x=6x = 6 and y=ky = k into the equation yields k=13(6)+8=2+8=6k = -\frac{1}{3}(6) + 8 = -2 + 8 = 6. This matches the correct value of 66.

Adım Adım Çözüm

1
Calculate the slope of side DEDE using the coordinates of D(1,2)D(1, -2) and E(4,7)E(4, 7).
The slope of DEDE is 33.
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Substituting the points gives mDE=7(2)41=93=3m_{DE} = \frac{7 - (-2)}{4 - 1} = \frac{9}{3} = 3.
2
Determine the slope of the altitude line, which is perpendicular to side DEDE.
The slope of the altitude is 13-\frac{1}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other. The negative reciprocal of 33 is 13-\frac{1}{3}.
3
Write the equation of the line containing the altitude using its slope and the given yy-intercept of 88.
The equation of the altitude line is y=13x+8y = -\frac{1}{3}x + 8.
The slope-intercept form of a linear equation is y=mx+by = mx + b, where mm is the slope and bb is the yy-intercept.
4
Substitute the coordinates of point F(6,k)F(6, k) into the equation of the altitude to solve for kk.
k=6k = 6.
Since vertex FF lies on the line containing the altitude, its coordinates must satisfy the equation: k=13(6)+8=2+8=6k = -\frac{1}{3}(6) + 8 = -2 + 8 = 6.

Anahtar Kavram

Finding the equation of a line perpendicular to a given line segment and using it to find a missing coordinate of a point on that line.
Soru 2609Soru

In the trapezoid ABCDABCD, bases ABAB and CDCD are parallel, and side ADAD is perpendicular to base ABAB. The length of ABAB is 1010 inches, the length of CDCD is 44 inches, and the measure of angle BB is 6060^\circ. What is the perimeter, in inches, of the trapezoid?

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Cevap: 26+6326 + 6\sqrt{3}

Cevap

The perimeter of the trapezoid is 26+6326 + 6\sqrt{3} inches.
Drawing an altitude from vertex CC to base ABAB splits the trapezoid into a rectangle AECDAECD and a right triangle CEB\triangle CEB. Since AE=CD=4AE = CD = 4, the leg EB=104=6EB = 10 - 4 = 6. The right triangle CEB\triangle CEB is a 30609030^\circ-60^\circ-90^\circ triangle where EBEB is the leg opposite the 3030^\circ angle. The hypotenuse BC=2×6=12BC = 2 \times 6 = 12 and the height CE=AD=63CE = AD = 6\sqrt{3}. Summing all four outer sides of the trapezoid (10+12+4+6310 + 12 + 4 + 6\sqrt{3}) yields the correct perimeter of 26+6326 + 6\sqrt{3} inches.

Adım Adım Çözüm

1
Draw an altitude from vertex CC perpendicular to base ABAB at point EE.
A rectangle AECDAECD and a right triangle CEB\triangle CEB are formed, with AE=CD=4AE = CD = 4 inches and CE=ADCE = AD.
To break down the trapezoid into a rectangle and a right triangle so we can find the unknown side lengths.
2
Find the length of segment EBEB.
EB=ABAE=104=6EB = AB - AE = 10 - 4 = 6 inches.
To find the length of the base of the right triangle CEB\triangle CEB.
3
Use the properties of a 30609030^\circ-60^\circ-90^\circ right triangle to determine the lengths of sides CECE (which is ADAD) and BCBC.
Since B=60\angle B = 60^\circ is opposite to CECE, and EB=6EB = 6 is the shorter leg adjacent to 6060^\circ, the hypotenuse is BC=2×6=12BC = 2 \times 6 = 12 inches and the longer leg is CE=AD=63CE = AD = 6\sqrt{3} inches.
To compute the remaining unknown outer side lengths of the trapezoid.
4
Sum the four outer sides of the trapezoid to find the perimeter.
Perimeter = AB+BC+CD+DA=10+12+4+63=26+63AB + BC + CD + DA = 10 + 12 + 4 + 6\sqrt{3} = 26 + 6\sqrt{3} inches.
The perimeter is the total boundary length of the shape.

Anahtar Kavram

The perimeter of a right trapezoid can be found by drawing an altitude to create a rectangle and a 30609030^\circ-60^\circ-90^\circ special right triangle, then determining the missing side lengths using special right triangle ratios.
Tahmini Süre:1m 30s
Soru 2610Soru

An irregular convex heptagon (7-sided polygon) has three interior angles that each measure 140140^\circ. The remaining four interior angles have measures in the ratio 2:3:3:42:3:3:4. What is the measure of the largest interior angle of this heptagon?

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Cevap: 160160^\circ

Cevap

160160^\circ
The total sum of the interior angles of a convex heptagon is (72)×180=900(7-2) \times 180^\circ = 900^\circ. Subtracting the three angles that each measure 140140^\circ (3×140=4203 \times 140^\circ = 420^\circ) leaves 480480^\circ for the remaining four angles. Using the ratio 2:3:3:42:3:3:4, we represent the angles as 2x2x, 3x3x, 3x3x, and 4x4x, giving the equation 2x+3x+3x+4x=4802x + 3x + 3x + 4x = 480^\circ, which simplifies to 12x=48012x = 480^\circ and yields x=40x = 40^\circ. The largest of these four angles is 4x=4(40)=1604x = 4(40^\circ) = 160^\circ. Since 160160^\circ is larger than 140140^\circ, the largest interior angle of the heptagon is 160160^\circ.

Adım Adım Çözüm

1
Calculate the sum of the interior angles of a heptagon.
900900^\circ
The sum of the interior angles of a convex polygon with nn sides is given by (n2)×180(n-2) \times 180^\circ. For a heptagon (n=7n=7), the sum is (72)×180=5×180=900(7-2) \times 180^\circ = 5 \times 180^\circ = 900^\circ.
2
Find the sum of the three known interior angles.
420420^\circ
Since three interior angles each measure 140140^\circ, their sum is 3×140=4203 \times 140^\circ = 420^\circ.
3
Calculate the sum of the remaining four interior angles.
480480^\circ
Subtracting the sum of the three known angles from the total interior sum gives 900420=480900^\circ - 420^\circ = 480^\circ.
4
Set up and solve an equation for the remaining four angles using the given ratio.
x=40x = 40^\circ
Let the measures of the remaining four angles be 2x2x, 3x3x, 3x3x, and 4x4x. Their sum is 2x+3x+3x+4x=12x2x + 3x + 3x + 4x = 12x. Setting this equal to the remaining sum gives 12x=48012x = 480^\circ, which simplifies to x=40x = 40^\circ.
5
Determine the measures of the remaining angles and identify the largest angle.
160160^\circ
The remaining angles measure 2(40)=802(40^\circ) = 80^\circ, 3(40)=1203(40^\circ) = 120^\circ, 3(40)=1203(40^\circ) = 120^\circ, and 4(40)=1604(40^\circ) = 160^\circ. The largest of these is 160160^\circ. Comparing this with the three 140140^\circ angles, the largest interior angle of the entire heptagon is 160160^\circ.

Anahtar Kavram

Interior angle sum of polygons and ratio division
Soru 2611Soru

For an angle θ\theta such that π<θ<3π2\pi < \theta < \frac{3\pi}{2}, the value of cosθ=0.8\cos\theta = -0.8. What is the value of 10sinθ10\sin\theta?

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Cevap: -6

Cevap

The value of 10sinθ10\sin\theta is 6-6.
The correct answer is 6-6. Substituting cosθ=0.8\cos\theta = -0.8 into the fundamental Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 yields sin2θ+0.64=1\sin^2\theta + 0.64 = 1, which simplifies to sin2θ=0.36\sin^2\theta = 0.36. Since the angle θ\theta lies in the interval π<θ<3π2\pi < \theta < \frac{3\pi}{2} (Quadrant III), its sine value must be negative. Thus, sinθ=0.6\sin\theta = -0.6. Multiplying this value by 10 gives 10sinθ=610\sin\theta = -6.

Adım Adım Çözüm

1
Use the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 with the given cosine value cosθ=0.8\cos\theta = -0.8.
sin2θ+(0.8)2=1sin2θ+0.64=1sin2θ=0.36\sin^2\theta + (-0.8)^2 = 1 \Rightarrow \sin^2\theta + 0.64 = 1 \Rightarrow \sin^2\theta = 0.36
The Pythagorean identity relates the sine and cosine values of any angle.
2
Take the square root of both sides, selecting the correct sign based on the quadrant constraint π<θ<3π2\pi < \theta < \frac{3\pi}{2}.
Since the angle lies in Quadrant III, the sine function must be negative. Thus, sinθ=0.36=0.6\sin\theta = -\sqrt{0.36} = -0.6.
In the third quadrant, y-coordinates on the unit circle are negative, meaning sinθ\sin\theta must be negative.
3
Multiply sinθ\sin\theta by 10 to get the final requested value.
10sinθ=10×(0.6)=610\sin\theta = 10 \times (-0.6) = -6
This calculation yields the final answer requested by the problem.

Anahtar Kavram

Applying the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 and choosing the correct sign based on the angle's quadrant.
Soru 2612Soru

In the standard (x,y)(x, y) coordinate plane, if a line passes through the point (1,3)(1, 3) and has an undefined slope, then the point (1,5)(1, -5) lies on this line.

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Cevap: True

Cevap

The statement is true because a line with an undefined slope is a vertical line, meaning all points on the line share the same x-coordinate. Since the line passes through (1,3)(1, 3), the x-coordinate of all points on the line is 11, which includes the point (1,5)(1, -5).
A line with an undefined slope is a vertical line. Since it passes through (1,3)(1, 3), the x-coordinate of every point on this line must be 11. The point (1,5)(1, -5) has an x-coordinate of 11, so it lies on this vertical line, making the statement true.

Adım Adım Çözüm

1
Identify the geometric properties of a line with an undefined slope.
A line with an undefined slope is a vertical line.
The slope formula is m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. The slope is undefined when the denominator, representing the change in xx, is equal to zero. This occurs only for vertical lines.
2
Determine the equation of the vertical line that passes through the point (1,3)(1, 3).
The equation of the line is x=1x = 1.
Since the line is vertical, all points on the line share the same x-coordinate as the given point (1,3)(1, 3).
3
Verify if the point (1,5)(1, -5) satisfies the equation of the line.
The x-coordinate of the point (1,5)(1, -5) is 11, which satisfies the equation x=1x = 1.
A point lies on a line if its coordinates satisfy the equation of the line.

Anahtar Kavram

Slope of a Line
Soru 2613Soru

An angle θ\theta lies in the second quadrant and satisfies sinθ=513\sin\theta = \frac{5}{13}. What is the value of 12tanθ12\tan\theta?

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Cevap: -5

Cevap

The value of 12tanθ12\tan\theta is -5.
Applying the Pythagorean identity sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1 with sinθ=513\sin\theta = \frac{5}{13} yields cos2θ=144169\cos^2\theta = \frac{144}{169}. Since the angle θ\theta lies in the second quadrant, its cosine is negative, meaning cosθ=1213\cos\theta = -\frac{12}{13}. Using the quotient identity tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}, we find tanθ=512\tan\theta = -\frac{5}{12}. Multiplying by 12 gives the correct value of -5.

Adım Adım Çözüm

1
Use the Pythagorean identity to find the magnitude of cosθ\cos\theta.
cos2θ=144169\cos^2\theta = \frac{144}{169}
The Pythagorean identity states that sin2θ+cos2θ=1\sin^2\theta + \cos^2\theta = 1. Substituting sinθ=513\sin\theta = \frac{5}{13} gives cos2θ=1(513)2\cos^2\theta = 1 - \left(\frac{5}{13}\right)^2.
2
Determine the value of cosθ\cos\theta by applying the quadrant sign rule.
cosθ=1213\cos\theta = -\frac{12}{13}
Since θ\theta is in the second quadrant, the cosine of θ\theta must be negative.
3
Calculate tanθ\tan\theta using the quotient identity.
tanθ=512\tan\theta = -\frac{5}{12}
The quotient identity is tanθ=sinθcosθ\tan\theta = \frac{\sin\theta}{\cos\theta}.
4
Multiply tanθ\tan\theta by 12 to find the required expression's value.
-5
Multiplying the value of tanθ\tan\theta (which is 512-\frac{5}{12}) by 12 yields 5-5.

Anahtar Kavram

Fundamental Trigonometric Identities
Tahmini Süre:1m 0s
Soru 2614Soru

A line, L1L_1, contains the points (2,5)(2, 5) and (6,3)(6, -3) in a coordinate plane. Another line, L2L_2, is perpendicular to L1L_1 and is defined by the equation ax+2y=7ax + 2y = 7. What is the value of aa?

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Cevap: -1

Cevap

The value of aa is 1-1.
The slope of line L1L_1 is 2-2. A line perpendicular to it must have a slope that is the negative reciprocal, which is 12\frac{1}{2}. Rewriting ax+2y=7ax + 2y = 7 in slope-intercept form gives y=a2x+72y = -\frac{a}{2}x + \frac{7}{2}, where the slope is a2-\frac{a}{2}. Setting this slope equal to 12\frac{1}{2} gives a2=12-\frac{a}{2} = \frac{1}{2}, which simplifies to a=1a = -1.

Adım Adım Çözüm

1
Find the slope of line L1L_1 using the points (2,5)(2, 5) and (6,3)(6, -3).
m1=3562=84=2m_1 = \frac{-3 - 5}{6 - 2} = \frac{-8}{4} = -2.
The slope formula is the change in y divided by the change in x.
2
Determine the perpendicular slope for L2L_2.
m2=1m1=12m_2 = -\frac{1}{m_1} = \frac{1}{2}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Rewrite the equation of L2L_2, ax+2y=7ax + 2y = 7, in slope-intercept form to find its slope expression.
2y=ax+7y=a2x+722y = -ax + 7 \Rightarrow y = -\frac{a}{2}x + \frac{7}{2}. The slope is a2-\frac{a}{2}.
Slope-intercept form y=mx+by = mx + b allows direct identification of the slope coefficient.
4
Equate the slope of L2L_2 to the perpendicular slope and solve for aa.
a2=12a=1-\frac{a}{2} = \frac{1}{2} \Rightarrow a = -1.
Solving the equation gives the value of aa required for the lines to be perpendicular.

Anahtar Kavram

Perpendicular lines have slopes that are negative reciprocals of each other.
Soru 2615Soru

The tropical rainforests of Central and South America are home to some of the most complex ecological relationships on Earth. Among these, the partnership between leafcutter ants of the genera Atta and Acromyrmex and a specialized cultivar of fungus (primarily of the family Lepiotaceae) stands out as a marvel of evolutionary cooperation. For millions of years, these tiny insects have practiced a form of agriculture that predates human farming by eons. The ants do not feed on the leaves they harvest; instead, they use the vegetation as a substrate to grow the fungus, which serves as the primary food source for their larvae. This mutual dependency is so absolute that neither partner can survive in the wild without the other.

A leafcutter colony is a highly structured society divided into distinct physical castes, each specialized for specific tasks. The largest ants, the soldiers, possess massive mandibles designed to defend the nest from predators. Below them are the foragers, who travel along cleared trails to locate and cut leaves from the forest canopy. Once the leaves are brought back to the underground nest, smaller workers—the gardeners—take over. These ants chop the leaves into tiny fragments, chew them into a pulp, and mix them with fecal droplets and salivary secretions. This prepared mixture is then added to the subterranean fungal gardens. Finally, the smallest caste of ants, the minims, meticulously tend to the fungus, weeding out parasitic spores and harvesting the nutrient-rich swellings produced by the fungus, known as gongylidia, to feed the colony. Each caste's behavior is genetically preprogrammed, allowing the colony to operate with the efficiency of a single superorganism.

Crucial to the survival of the fungus-ant symbiosis is the management of pathogens. The humid underground chambers that house the fungal gardens are also ideal breeding grounds for Escovopsis, a parasitic microfungus that can rapidly decimate the garden. To combat this threat, the ants have developed a remarkable defense mechanism. They carry a filamentous bacterium, Pseudonocardia, on their cuticles. This bacterium produces highly specific antibiotics that target and inhibit the growth of Escovopsis without harming the cultivar fungus. The ants actively cultivate these bacteria, which are visible as a white, powdery coating on their chests, providing a mobile chemical defense system. This three-way symbiosis illustrates that the agricultural success of leafcutter ants relies not just on cultivation, but also on sophisticated, biological pest control.

Research conducted in the early 2000s shed light on the specificity of this relationship. Scientists discovered that the cultivar fungus is almost entirely dependent on the ants for survival, as it has lost the ability to produce asexual spores and rarely reproduces sexually in the wild. If the colony dies, the fungal garden inevitably perishes shortly thereafter. Conversely, the ants cannot digest the cellulose in the leaves they harvest, relying entirely on the enzymes produced by the fungus to break down the plant material into digestible sugars and proteins. This obligate mutualism highlights the delicate balance of rainforest ecosystems, where the survival of a massive, multi-million-member ant colony hangs on the health of a single, fragile fungal partner. By studying these interactions, ecologists hope to better understand the evolutionary pressures that shape complex symbioses and how they might respond to environmental pressures in the future.

Based on the passage, the statement that leafcutter ants consume the leaves they harvest as their primary food source is:

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Cevap: False

Cevap

The statement is false because the passage explicitly states that the leafcutter ants do not feed on the leaves they harvest.
The passage explicitly states in the first paragraph that the ants do not feed on the leaves they harvest, but rather use them to grow a fungus that serves as food.

Adım Adım Çözüm

1
Locate the details in the text concerning the diet of the leafcutter ants and what they do with the leaves.
In the first paragraph, the passage states: 'The ants do not feed on the leaves they harvest; instead, they use the vegetation as a substrate to grow the fungus...'
Finding the specific sentence that addresses the leafcutter ants' diet is necessary to confirm or deny the statement.
2
Evaluate the statement using the retrieved factual information.
The statement claims that leafcutter ants consume the leaves they harvest as their primary food source, which directly contradicts the text statement that they do not feed on the leaves.
Comparing the statement directly to the explicit detail confirms the statement is false.

Anahtar Kavram

Identifying explicitly stated details in a Reading passage.
Soru 2616Soru

A circle is inscribed inside a sector of a larger circle. The larger circle has a radius of 18 inches18\text{ inches} and the sector has a central angle of 6060^\circ, as shown in the figure. What is the area, in square inches, of the region that is inside the sector but outside the inscribed circle?

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Cevap: 18π18\pi

Cevap

The correct area is 18π18\pi square inches.
The correct answer is 18π18\pi square inches. First, the sector area is calculated as 60360×π×182=54π\frac{60}{360} \times \pi \times 18^2 = 54\pi. Second, using the right triangle formed by the sector's center, the inscribed circle's center, and the point of tangency, we set up sin(30)=r18r\sin(30^\circ) = \frac{r}{18-r}. Solving for rr gives r=6r = 6, so the area of the inscribed circle is π×62=36π\pi \times 6^2 = 36\pi. The difference between the two areas is 54π36π=18π54\pi - 36\pi = 18\pi.

Adım Adım Çözüm

1
Calculate the area of the 6060^\circ sector of the larger circle.
Sector Area = 54π54\pi square inches
The sector has a radius of R=18R = 18 and a central angle of 6060^\circ. The sector's area is a fraction of the total circle's area: Areasector=60360×πR2=16×π×182=324π6=54π\text{Area}_{\text{sector}} = \frac{60}{360} \times \pi R^2 = \frac{1}{6} \times \pi \times 18^2 = \frac{324\pi}{6} = 54\pi.
2
Determine the radius rr of the inscribed circle using trigonometry.
Inscribed Radius r=6r = 6 inches
The center of the inscribed circle, II, lies on the angle bisector of the sector. The line segment from the center of the sector OO to II bisects the 6060^\circ angle, forming a 3030^\circ angle. The distance from OO to the outer boundary of the sector is R=18R = 18, and the distance from II to the boundary is rr, so the hypotenuse OI=18rOI = 18 - r. Drawing a perpendicular from II to one of the straight edges of the sector creates a right triangle with opposite side rr (the radius) and hypotenuse 18r18 - r. Applying the sine ratio: sin(30)=r18r\sin(30^\circ) = \frac{r}{18 - r}. Since sin(30)=0.5\sin(30^\circ) = 0.5, we solve 0.5=r18r18r=2r3r=18r=60.5 = \frac{r}{18 - r} \Rightarrow 18 - r = 2r \Rightarrow 3r = 18 \Rightarrow r = 6.
3
Calculate the area of the inscribed circle.
Inscribed Circle Area = 36π36\pi square inches
The area of the inscribed circle with radius r=6r = 6 is Areacircle=πr2=π×62=36π\text{Area}_{\text{circle}} = \pi r^2 = \pi \times 6^2 = 36\pi.
4
Subtract the area of the inscribed circle from the area of the sector.
Remaining Area = 18π18\pi square inches
The area of the region inside the sector but outside the circle is found by subtracting the inscribed circle area from the sector area: 54π36π=18π54\pi - 36\pi = 18\pi.

Anahtar Kavram

Using trigonometry on angle bisectors to determine the radius of a circle inscribed inside a sector, and applying sector and circle area formulas.
Tahmini Süre:3m 0s
Soru 2617Soru

A convex polygon has nn sides. The sum of the measures of n1n-1 of its interior angles is 20302030^\circ. What is the measure, in degrees, of the remaining interior angle?

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Cevap: 130

Cevap

The measure of the remaining interior angle is 130 degrees.
The sum of the interior angles of a convex polygon must be a multiple of 180180^\circ. The multiple of 180180^\circ immediately greater than 20302030^\circ is 21602160^\circ (which is 12×18012 \times 180^\circ). The difference between the total sum and the sum of the n1n-1 angles is 21602030=1302160^\circ - 2030^\circ = 130^\circ. Since 130130^\circ is less than 180180^\circ, this is a valid interior angle for a convex polygon.

Adım Adım Çözüm

1
Write the formula for the total sum of the interior angles of a convex polygon.
The sum of the interior angles of a polygon with nn sides is (n2)×180(n-2) \times 180^\circ.
This formula relates the number of sides to the total sum of the interior angles.
2
Apply the properties of convex polygons to set up an inequality for the total sum.
Since the remaining interior angle must be greater than 00^\circ and less than 180180^\circ, the total sum of all nn angles must satisfy 2030<(n2)×180<2030+1802030^\circ < (n-2) \times 180^\circ < 2030^\circ + 180^\circ, which simplifies to 2030<(n2)×180<22102030^\circ < (n-2) \times 180^\circ < 2210^\circ.
By definition, every interior angle of a convex polygon is strictly less than 180180^\circ.
3
Solve the inequality to find the integer value of n2n-2.
Dividing the entire inequality by 180180^\circ gives 11.28<n2<12.2811.28 < n-2 < 12.28. Since the number of sides nn must be an integer, n2n-2 must be the integer 1212.
A polygon must have a whole number of sides.
4
Calculate the total sum of all interior angles and find the remaining angle.
The total sum is 12×180=216012 \times 180^\circ = 2160^\circ. Subtracting the sum of the other n1n-1 angles gives the remaining angle: 21602030=1302160^\circ - 2030^\circ = 130^\circ.
The difference between the total sum of all interior angles and the sum of the n1n-1 angles is the measure of the final angle.

Anahtar Kavram

The sum of the interior angles of a convex polygon with nn sides is (n2)×180(n-2) \times 180^\circ, where each interior angle is strictly less than 180180^\circ.
Soru 2618Soru

The Glass Palace of 1851

In the spring of 1851, London became the undisputed center of the industrial world as it hosted the Great Exhibition of the Works of Industry of All Nations. Conceived by Prince Albert, the husband of Queen Victoria, the exhibition aimed to showcase the technological progress of the United Kingdom and its empire, alongside innovations from other nations. At the heart of this massive international event was the Crystal Palace, a temporary exhibition hall that became an architectural icon.

The road to building the exhibition hall was fraught with administrative challenges. The Building Committee had initially rejected over two hundred proposals from established architects, deeming the brick and stone designs too costly, heavy, and slow to construct within the tight timeline. Joseph Paxton, who was primarily known as a gardener and greenhouse designer rather than a formally trained architect, submitted a late proposal. Paxton envisioned a revolutionary structure made almost entirely of cast-iron columns and sheet glass, which would be light, cheap, and rapid to assemble.

Paxton’s innovative design was inspired by his work with giant water lilies (Victoria amazonica) at Chatsworth House, where he served as the head gardener. He observed that the lilies’ massive, floating leaves, which could support the weight of a child, were structurally stabilized by an intricate network of radial ribs and cross-girders beneath the leaf surface. By emulating this organic configuration, Paxton designed a modular framing system of hollow iron columns and lightweight girders that could support vast spans of glass. This structural approach eliminated the need for heavy internal masonry.

The use of prefabricated modular parts was crucial to the project's success. Standardized components were manufactured off-site in Birmingham and transported by rail directly to the Hyde Park construction site. This allowed the building to be assembled in just nine months, beginning in late 1850 and finishing in time for the grand opening on May 1, 1851.

The physical dimensions of the Crystal Palace were unprecedented. The main building was 1,848 feet long and 408 feet wide, covering approximately 19 acres in Hyde Park. These dimensions were chosen to align with the year of the exhibition. One of the most famous and challenging features of the building was the barrel-vaulted transept, which rose to a height of 108 feet. This high, arched ceiling was not part of Paxton's original design but was added to enclose several ancient elm trees growing in Hyde Park, thereby saving them from being cut down and placating a vocal public concerned about the environmental impact of the exhibition.

Inside, the Crystal Palace housed more than 100,000 exhibits from around the globe, displaying the latest advancements in machinery, manufacturing, scientific instruments, and fine arts. Over six million visitors—equivalent to nearly a third of the entire population of Britain at the time—traveled to London to marvel at the exhibits. The enormous success of the exhibition generated a substantial financial surplus, which was later used to establish London’s museum district in South Kensington, including the Science Museum and the Victoria and Albert Museum.

After the exhibition concluded in October 1851, the building was not demolished. Instead, it was disassembled piece by piece and rebuilt in a grander form in Sydenham, a suburb in South London. It remained a popular cultural venue, hosting concerts, political rallies, and public events, for over eight decades. However, on the night of November 30, 1936, the Crystal Palace was completely destroyed by a catastrophic fire that could be seen from miles away. Although the physical structure is gone, the Crystal Palace remains a seminal landmark in the history of modern architecture, pioneering the widespread use of prefabricated elements and glass in civil engineering.

According to the passage, the high, arched transept of the Crystal Palace was specifically added to the design in order to do which of the following?

Cevabı ve açıklamayı göster

Cevap: enclose and preserve existing elm trees in the park

Cevap

The correct answer is the option stating that the transept was added to enclose and preserve existing elm trees in the park.
The correct answer is supported by the fifth paragraph, which explicitly states that the barrel-vaulted transept was added to the design to enclose several ancient elm trees growing in Hyde Park, saving them from being cut down.

Adım Adım Çözüm

1
Locate the key term 'transept' in the passage.
The term is found in the fifth paragraph, which describes the physical dimensions of the Crystal Palace.
Finding the specific location of the detail in the text is the first step in literal comprehension.
2
Read the sentences surrounding the mention of the transept to identify why it was added.
The text states: 'This high, arched ceiling was not part of Paxton's original design but was added to enclose several ancient elm trees growing in Hyde Park, thereby saving them from being cut down...'
This provides the explicit reason for the modification of the original design.
3
Match the explicit detail from the text to the correct option.
The option to 'enclose and preserve existing elm trees in the park' matches the text's explanation of saving the trees from being cut down.
This confirms the correct choice based strictly on literal details.

Anahtar Kavram

Identifying Explicit Details
Soru 2619Soru

An angle θ\theta in standard position is coterminal with an angle of 17π4-\frac{17\pi}{4} radians. If 0θ<2π0 \le \theta < 2\pi, what is the value of θ\theta expressed as a decimal multiple of π\pi? (For example, if the angle were 3π2=1.5π\frac{3\pi}{2} = 1.5\pi, the answer would be 1.5.)

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Cevap: 1.75

Cevap

1.75
To find a coterminal angle in the interval [0,2π)[0, 2\pi) for 17π4-\frac{17\pi}{4}, we add multiples of 2π2\pi. Since 17π4=4.25π-\frac{17\pi}{4} = -4.25\pi, we add 6π6\pi (three full rotations) to get 4.25π+6π=1.75π-4.25\pi + 6\pi = 1.75\pi. The multiple of π\pi is therefore 1.75.

Adım Adım Çözüm

1
Convert the coefficient of the given angle from a fraction to a decimal.
174=4.25-\frac{17}{4} = -4.25
Converting the fraction to a decimal makes it easier to work with the addition of full rotations.
2
Add multiples of 2π2\pi (which corresponds to adding 2 to the coefficient of π\pi) to find a coterminal angle in the interval [0,2π)[0, 2\pi).
4.25+2=2.25-4.25 + 2 = -2.25; 2.25+2=0.25-2.25 + 2 = -0.25; 0.25+2=1.75-0.25 + 2 = 1.75
Adding 2π2\pi representing full counterclockwise rotations on the unit circle results in a coterminal angle. We repeat this process until the coefficient lies in the interval [0,2)[0, 2).
3
Identify the coefficient of π\pi for the coterminal angle.
1.75
The question asks for the angle as a decimal multiple of π\pi, which is the coefficient of π\pi in the expression 1.75π1.75\pi.

Anahtar Kavram

Coterminal angles are angles in standard position that share the same terminal side. They can be found by adding or subtracting multiples of 2π2\pi radians.
Tahmini Süre:1m 30s
Soru 2620Soru

An angle in standard position measures θ=7π6\theta = \frac{7\pi}{6} radians. If the terminal side of the angle is rotated counterclockwise by 120120^\circ, what is the radian measure of the final angle?

Cevabı ve açıklamayı göster

Cevap: 11π6\frac{11\pi}{6}

Cevap

The final angle measures 11π6\frac{11\pi}{6} radians.
The correct answer is 11π6\frac{11\pi}{6} radians. To find the final angle, we convert the rotation angle of 120120^\circ into radians: 120×π180=2π3120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} radians. Since the rotation is counterclockwise, we add this value to the initial angle of 7π6\frac{7\pi}{6} radians: 7π6+2π3=7π6+4π6=11π6\frac{7\pi}{6} + \frac{2\pi}{3} = \frac{7\pi}{6} + \frac{4\pi}{6} = \frac{11\pi}{6} radians.

Adım Adım Çözüm

1
Convert the rotation angle from degrees to radians.
120×π180=2π3120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3} radians.
Both angles must be in radians to add them directly.
2
Set up the sum for a counterclockwise rotation.
7π6+2π3\frac{7\pi}{6} + \frac{2\pi}{3}
Counterclockwise rotation corresponds to adding the angle measure.
3
Find the common denominator and calculate the sum.
7π6+4π6=11π6\frac{7\pi}{6} + \frac{4\pi}{6} = \frac{11\pi}{6} radians.
The least common denominator of 66 and 33 is 66, so we multiply the numerator and denominator of 2π3\frac{2\pi}{3} by 22 before adding.

Anahtar Kavram

Unit Circle and Angle Measures (Radians and Degrees)
Tahmini Süre:1m 30s
ÖncekiSayfa 131 / 278Sonraki
Tüm alıştırma soruları — ACT | Examkin