Tüm alıştırma soruları

2131 soru

Soru 1561Soru

In a circle centered at point OO, the radius is 1818 units. Points PP and QQ lie on the circle such that the area of sector POQPOQ is 54π54\pi square units. What is the ratio of the length of minor arc PQPQ to the total perimeter of sector POQPOQ?

Cevabı ve açıklamayı göster

Cevap: ππ+6\frac{\pi}{\pi + 6}

Cevap

The ratio of the length of minor arc PQPQ to the total perimeter of sector POQPOQ is ππ+6\frac{\pi}{\pi + 6}.
The full circle area is 324π324\pi, making the sector 54π/324π=1/654\pi / 324\pi = 1/6 of the circle. The arc length is 1/6×36π=6π1/6 \times 36\pi = 6\pi. The sector perimeter is the arc length plus two radii (6π+366\pi + 36). Taking the ratio of arc length to sector perimeter yields 6π/(6π+36)=π/(π+6)6\pi / (6\pi + 36) = \pi / (\pi + 6).

Adım Adım Çözüm

1
Calculate the total area of the circle and determine the fractional size of sector POQPOQ.
Total area = π(18)2=324π\pi(18)^2 = 324\pi. Sector fraction = 54π324π=16\frac{54\pi}{324\pi} = \frac{1}{6}.
Determining the fraction of the circle represented by the sector is required to find the arc length.
2
Calculate the length of minor arc PQPQ and the total perimeter of sector POQPOQ.
Minor arc PQ=16×2π(18)=6πPQ = \frac{1}{6} \times 2\pi(18) = 6\pi. Sector perimeter = 6π+2(18)=6π+366\pi + 2(18) = 6\pi + 36.
The sector perimeter consists of the curved arc length plus the two straight radii OPOP and OQOQ.
3
Form and simplify the ratio of arc length to sector perimeter.
6π6π+36=6π6(π+6)=ππ+6\frac{6\pi}{6\pi + 36} = \frac{6\pi}{6(\pi + 6)} = \frac{\pi}{\pi + 6}.
Factoring out 6 from the numerator and denominator simplifies the expression to its lowest form.

Anahtar Kavram

Arc Length and Sector Perimeter Calculations
Soru 1562Soru
A financial analyst models a company's weekly metrics using three variables—revenue RR, operating cost CC, and advertising expenditure AA, measured in thousands of dollars. The metrics satisfy the following system of linear equations, where kk is a real constant:
4R3C+2A=18R+5C6A=145R+2C4A=k\begin{aligned} 4R - 3C + 2A &= 18 \\ R + 5C - 6A &= 14 \\ 5R + 2C - 4A &= k \end{aligned}
If this system of linear equations is consistent (has at least one solution), what is the value of kk?
Cevabı ve açıklamayı göster

Cevap: 32

Cevap

32
Adding the left-hand sides of the first two equations yields (4R3C+2A)+(R+5C6A)=5R+2C4A(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A, which is identical to the left-hand side of the third equation. For the linear system to have at least one solution (to be consistent), the right-hand side constant must satisfy the exact same linear combination: k=18+14=32k = 18 + 14 = 32.

Adım Adım Çözüm

1
Examine the linear combination of the left-hand sides of the first two equations
(4R3C+2A)+(R+5C6A)=5R+2C4A(4R - 3C + 2A) + (R + 5C - 6A) = 5R + 2C - 4A
Observing that the sum of the coefficients of the first two equations matches the left-hand side of the third equation.
2
Apply the condition for system consistency
Right-hand side of Equation 3 must equal Right-hand side of Equation 1 + Right-hand side of Equation 2
For a system with linearly dependent left-hand sides to be consistent, the same linear combination must hold for the right-hand constants.
3
Calculate the value of kk
k=18+14=32k = 18 + 14 = 32
Adding the constants from the right-hand side of the first two equations gives the consistent value for kk.

Anahtar Kavram

Consistency and Linear Dependence in Systems of Linear Equations
Soru 1563Soru

A hollow metallic spherical shell has an inner radius of 33 centimeters and an outer radius of rr centimeters, where r>3r > 3. If the spherical shell is melted down and completely recast into a solid right circular cylinder with base radius rr centimeters and height 77 centimeters, what is the value of rr?

Cevabı ve açıklamayı göster

Cevap: 66

Cevap

6
The volume of metal in the hollow spherical shell is 43π(r333)=43π(r327)\frac{4}{3}\pi(r^3 - 3^3) = \frac{4}{3}\pi(r^3 - 27). The volume of the recast cylinder is πr2h=7πr2\pi r^2 h = 7\pi r^2. Equating these volumes gives 43(r327)=7r2\frac{4}{3}(r^3 - 27) = 7r^2, which simplifies to 4r321r2108=04r^3 - 21r^2 - 108 = 0. Factoring this cubic equation gives (r6)(4r2+3r+18)=0(r - 6)(4r^2 + 3r + 18) = 0. Since the quadratic term has no real roots, the only real solution is r=6r = 6.

Adım Adım Çözüm

1
Set up the formula for the volume of metal in the hollow spherical shell.
Vshell=43π(r333)=43π(r327)V_{\text{shell}} = \frac{4}{3}\pi \left(r^3 - 3^3\right) = \frac{4}{3}\pi \left(r^3 - 27\right)
The metal occupies only the region between the inner sphere of radius 3 cm and outer sphere of radius r cm.
2
Set up the formula for the volume of the recast solid right circular cylinder.
Vcylinder=πr2h=7πr2V_{\text{cylinder}} = \pi r^2 h = 7\pi r^2
The cylinder has base radius r cm and height 7 cm.
3
Equate the two volumes since no metal is lost during melting and recasting.
43π(r327)=7πr2\frac{4}{3}\pi \left(r^3 - 27\right) = 7\pi r^2
Conservation of volume during recasting.
4
Simplify the equation and solve for r.
4(r327)=21r2    4r321r2108=04(r^3 - 27) = 21r^2 \implies 4r^3 - 21r^2 - 108 = 0
Divide both sides by π\pi and multiply by 3 to clear the fraction.
5
Factor the cubic polynomial 4r321r2108=04r^3 - 21r^2 - 108 = 0.
(r6)(4r2+3r+18)=0    r=6(r - 6)(4r^2 + 3r + 18) = 0 \implies r = 6
Testing r=6r = 6 gives 4(216)21(36)108=864756108=04(216) - 21(36) - 108 = 864 - 756 - 108 = 0. The quadratic factor 4r2+3r+184r^2 + 3r + 18 has a negative discriminant and produces no real roots.

Anahtar Kavram

Volume formulas for hollow spheres and right circular cylinders
Soru 1564Soru
For how many real values of the constant aa does the following system of linear equations in xx, yy, and zz have no solution?
x+yz=3x+(a1)y+3z=5x+4y+(a+1)z=a+2\begin{aligned} x + y - z &= 3 \\ x + (a-1)y + 3z &= 5 \\ x + 4y + (a+1)z &= a + 2 \end{aligned}
Cevabı ve açıklamayı göster

Cevap: Exactly one

Cevap

Exactly one
To find when the system has no solution, we first eliminate xx by subtracting the first equation from the second and third equations. This produces a two-variable system in yy and zz: (a2)y+4z=2(a-2)y + 4z = 2 and 3y+(a+2)z=a13y + (a+2)z = a - 1. The determinant of this system's coefficients is (a2)(a+2)12=a216(a-2)(a+2) - 12 = a^2 - 16. Setting the determinant to zero yields two critical values: a=4a = 4 and a=4a = -4. Testing a=4a = 4 simplifies both reduced equations to y+2z=1y + 2z = 1, which means the system is consistent with infinitely many solutions. Testing a=4a = -4 yields 3y+2z=1-3y + 2z = 1 and 3y+2z=5-3y + 2z = 5, which is impossible (1=51 = 5), making the system inconsistent. Thus, there is exactly one real value of aa (a=4a = -4) for which the system has no solution.

Adım Adım Çözüm

1
Eliminate the variable xx from the second and third equations using the first equation.
Subtracting the first equation x+yz=3x + y - z = 3 from the second equation yields:
(a2)y+4z=2(a-2)y + 4z = 2
Subtracting the first equation from the third equation yields:
3y+(a+2)z=a13y + (a+2)z = a - 1
Reducing the 3×33 \times 3 system to a 2×22 \times 2 system in yy and zz simplifies the analysis of linear dependence and consistency.
2
Determine the values of aa for which the reduced 2×22 \times 2 system lacks a unique solution by setting its coefficient determinant to zero.
The determinant of the coefficient matrix is:
D=(a2)(a+2)(3)(4)=a2412=a216D = (a-2)(a+2) - (3)(4) = a^2 - 4 - 12 = a^2 - 16
Setting D=0D = 0 yields a2=16a^2 = 16, which gives a=4a = 4 or a=4a = -4.
A system of linear equations has either a unique solution (when the determinant is non-zero) or non-unique behavior—either no solution or infinitely many solutions—when the determinant is zero.
3
Test a=4a = 4 in the reduced system.
Substituting a=4a = 4 into the reduced equations gives:
2y+4z=2    y+2z=12y + 4z = 2 \implies y + 2z = 1
3y+6z=3    y+2z=13y + 6z = 3 \implies y + 2z = 1
Since both equations are identical, the system is consistent and has infinitely many solutions.
When equation ratios match completely including constant terms, the equations represent identical hyperplanes, yielding infinitely many solutions.
4
Test a=4a = -4 in the reduced system.
Substituting a=4a = -4 into the reduced equations gives:
6y+4z=2    3y+2z=1-6y + 4z = 2 \implies -3y + 2z = 1
3y2z=5    3y+2z=53y - 2z = -5 \implies -3y + 2z = 5
Comparing these gives 1=51 = 5, which is a contradiction. Thus, for a=4a = -4, the system has no solution.
When parallel equations have equal coefficient ratios but unequal constant ratios, the system is inconsistent.

Anahtar Kavram

Parametric Systems of Linear Equations and Consistency Conditions
Soru 1565Soru

Dataset PP consists of the five numbers 10,20,30,40,10, 20, 30, 40, and 5050. Dataset QQ is created by replacing the minimum value in Dataset PP with 1818 and the maximum value with 4242, leaving the remaining three numbers unchanged. Which of the following statements correctly compares the mean and standard deviation of Dataset QQ to those of Dataset PP?

Cevabı ve açıklamayı göster

Cevap: The mean of Dataset QQ is equal to the mean of Dataset PP, and the standard deviation of Dataset QQ is less than the standard deviation of Dataset PP.

Cevap

The mean of Dataset QQ is equal to the mean of Dataset PP, and the standard deviation of Dataset QQ is less than the standard deviation of Dataset PP.
The mean of both datasets is 30 because the decrease of 8 from 50 balances the increase of 8 to 10. Standard deviation quantifies how far data points deviate from the mean. Because 18 and 42 are closer to the mean of 30 than 10 and 50 are, the spread of Dataset Q around the mean is strictly smaller, making its standard deviation smaller.

Adım Adım Çözüm

1
Calculate the mean of Dataset PP.
Mean of P=10+20+30+40+505=1505=30\text{Mean of } P = \frac{10 + 20 + 30 + 40 + 50}{5} = \frac{150}{5} = 30.
To find the baseline central tendency before the dataset values are modified.
2
Calculate the mean of Dataset QQ.
Mean of Q=18+20+30+40+425=1505=30\text{Mean of } Q = \frac{18 + 20 + 30 + 40 + 42}{5} = \frac{150}{5} = 30.
Replacing 1010 with 1818 (+8) and 5050 with 4242 (-8) results in a net change of zero to the sum, so the mean remains unchanged.
3
Compare the dispersion of Dataset QQ relative to Dataset PP.
In Dataset PP, the squared deviations of the modified points from the mean are (1030)2=400(10 - 30)^2 = 400 and (5030)2=400(50 - 30)^2 = 400. In Dataset QQ, the squared deviations of these points are (1830)2=144(18 - 30)^2 = 144 and (4230)2=144(42 - 30)^2 = 144.
Standard deviation measures the average distance of data points from the mean. Since the outer values in Dataset QQ are closer to the mean than in Dataset PP, the overall dispersion and standard deviation decrease.

Anahtar Kavram

Standard deviation measures the spread of data points around their mean; bringing extreme values closer to the mean reduces the standard deviation.
Soru 1566Soru

Dataset WW consists of 1515 distinct real numbers with mean μ\mu, standard deviation σ>0\sigma > 0, interquartile range II, and range RR. A 16th numerical value equal to the mean μ\mu is added to dataset WW to form a new dataset WW'. Which of the following statements MUST be true regarding dataset WW' compared to dataset WW? Select all that apply.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The mean of dataset WW' is equal to the mean of dataset WW.; The standard deviation of dataset WW' is strictly less than the standard deviation of dataset WW.; The range of dataset WW' is equal to the range of dataset WW.

Cevap

The statements asserting that the mean of dataset WW' equals the mean of dataset WW, the standard deviation of dataset WW' is strictly less than that of dataset WW, and the range of dataset WW' equals the range of dataset WW must all be true.
Adding an element equal to the mean leaves the total sum of squared deviations from the mean unchanged while increasing the sample size by 1. Consequently, the mean remains unchanged, the standard deviation decreases by a factor of 15/16\sqrt{15/16}, and because the mean lies strictly inside the range of distinct values, the minimum and maximum remain unchanged, preserving the range.

Adım Adım Çözüm

1
Analyze the impact on the mean when adding x16=μx_{16} = \mu.
The sum of elements in WW' is 15μ+μ=16μ15\mu + \mu = 16\mu. The new mean is 16μ16=μ\frac{16\mu}{16} = \mu.
Adding a value equal to the mean preserves the mean value.
2
Analyze the impact on the range.
Because all 15 elements are distinct real numbers, min(W)<μ<max(W)\min(W) < \mu < \max(W). Adding μ\mu does not change the minimum or maximum values, so Range(W)=max(W)min(W)=R\text{Range}(W') = \max(W) - \min(W) = R.
The range depends solely on the maximum and minimum elements.
3
Analyze the impact on the standard deviation.
The sum of squared deviations for WW' is i=116(xiμ)2=i=115(xiμ)2+(μμ)2=15σ2\sum_{i=1}^{16} (x_i - \mu)^2 = \sum_{i=1}^{15} (x_i - \mu)^2 + (\mu - \mu)^2 = 15\sigma^2. The new variance is σ2=15σ216\sigma'^2 = \frac{15\sigma^2}{16}, so σ=σ1516<σ\sigma' = \sigma \sqrt{\frac{15}{16}} < \sigma.
Increasing the count nn without increasing the total squared deviation reduces overall dispersion around the mean.

Anahtar Kavram

Effect of adding central summary values on measures of dispersion and central tendency.
Soru 1567Soru

If xx is a real number satisfying the exponential equation 9x+132x+1=1629^{x+1} - 3^{2x+1} = 162, what is the value of 4x4^x?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8
By converting 9x+19^{x+1} to 32x+23^{2x+2} and factoring out 32x3^{2x}, the equation simplifies to 632x=1626 \cdot 3^{2x} = 162. Dividing by 6 gives 32x=273^{2x} = 27, so 2x=32x = 3 and x=32x = \frac{3}{2}. Raising 4 to the power of 32\frac{3}{2} yields (4)3=8(\sqrt{4})^3 = 8.

Adım Adım Çözüm

1
Rewrite terms with a common base of 3.
9x+1=(32)x+1=32(x+1)=32x+29^{x+1} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}
Since 9 is a power of 3 (9=329 = 3^2), applying exponent rules converts the equation to base 3.
2
Factor out the common exponential expression 32x3^{2x}.
32x+232x+1=32x3232x31=32x(93)=632x3^{2x+2} - 3^{2x+1} = 3^{2x} \cdot 3^2 - 3^{2x} \cdot 3^1 = 3^{2x}(9 - 3) = 6 \cdot 3^{2x}
Using product rule of exponents (3a+b=3a3b3^{a+b} = 3^a \cdot 3^b) allows combining like terms.
3
Solve for xx.
632x=162    32x=27    32x=33    2x=3    x=326 \cdot 3^{2x} = 162 \implies 3^{2x} = 27 \implies 3^{2x} = 3^3 \implies 2x = 3 \implies x = \frac{3}{2}
Dividing both sides by 6 yields 32x=273^{2x} = 27, and equating exponents of matching bases gives x=32x = \frac{3}{2}.
4
Evaluate the target expression 4x4^x.
43/2=(41/2)3=23=84^{3/2} = (4^{1/2})^3 = 2^3 = 8
Substituting x=32x = \frac{3}{2} into 4x4^x means taking the square root of 4 and raising it to the third power.

Anahtar Kavram

Solving exponential equations using common bases and exponent properties
Soru 1568Soru

A triangle has side lengths of 88, 1111, and xx, where xx is an integer. If the perimeter of the triangle is a positive integer multiple of 55, which of the following could be the value of xx? Select all such values.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: 66; 1111; 1616

Cevap

The possible values for xx are 66, 1111, and 1616.
According to the Triangle Inequality Theorem, the third side xx must be strictly greater than 118=311 - 8 = 3 and strictly less than 11+8=1911 + 8 = 19. The perimeter of the triangle is 8+11+x=19+x8 + 11 + x = 19 + x. For 19+x19 + x to be a positive multiple of 55, 19+x19 + x can be 2525, 3030, or 3535 within the allowed range for xx, giving x=6x = 6, x=11x = 11, and x=16x = 16.

Adım Adım Çözüm

1
Apply the Triangle Inequality Theorem to determine the valid range for the unknown side xx.
The length of xx must satisfy 118<x<11+811 - 8 < x < 11 + 8, which simplifies to 3<x<193 < x < 19.
The sum of any two side lengths of a non-degenerate triangle must be strictly greater than the third side length.
2
Set up an expression for the perimeter PP of the triangle.
P=8+11+x=19+xP = 8 + 11 + x = 19 + x.
Perimeter is the total sum of all three side lengths.
3
Determine which values of xx within the range 3<x<193 < x < 19 make P=19+xP = 19 + x a multiple of 55.
If x=6x = 6, P=25P = 25 (multiple of 55). If x=11x = 11, P=30P = 30 (multiple of 55). If x=16x = 16, P=35P = 35 (multiple of 55).
Adding 1919 to 66, 1111, and 1616 gives multiples of 55 within the strict inequality bounds.
4
Test boundary values outside the inequality bounds.
x=1x = 1 gives P=20P = 20, but 131 \le 3 (invalid). x=21x = 21 gives P=40P = 40, but 211921 \ge 19 (invalid).
Values outside 3<x<193 < x < 19 cannot form a valid triangle.

Anahtar Kavram

Triangle Inequality Theorem and Perimeter Constraints
Soru 1569Soru

In the xyxy-plane, line LL is given by the equation 3x+2y=183x + 2y = 18. Line NN is perpendicular to line LL and passes through the point (1,4)(-1, 4). If line LL and line NN intersect at the point (p,q)(p, q), what is the value of p+qp + q?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

8
Converting the given equation 3x+2y=183x + 2y = 18 into slope-intercept form yields y=32x+9y = -\frac{3}{2}x + 9, so line LL has slope 32-\frac{3}{2}. A perpendicular line must have a slope equal to the negative reciprocal, which is 23\frac{2}{3}. Using point (1,4)(-1, 4) in the point-slope form gives y4=23(x+1)y - 4 = \frac{2}{3}(x + 1), simplifying to 2x3y=142x - 3y = -14. Solving the system of equations formed by line LL (3x+2y=183x + 2y = 18) and line NN (2x3y=142x - 3y = -14) via elimination yields x=2x = 2 and y=6y = 6. Therefore, the intersection point is (2,6)(2, 6), and p+q=2+6=8p + q = 2 + 6 = 8.

Adım Adım Çözüm

1
Determine the slope of line LL
Slope of line LL is mL=32m_L = -\frac{3}{2}
Convert 3x+2y=183x + 2y = 18 to y=32x+9y = -\frac{3}{2}x + 9 to identify the slope coefficient of xx.
2
Calculate the perpendicular slope for line NN
Slope of line NN is mN=23m_N = \frac{2}{3}
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Derive the equation for line NN
Equation of line NN is 2x3y=142x - 3y = -14
Apply point-slope form y4=23(x+1)y - 4 = \frac{2}{3}(x + 1) and rearrange into standard linear form.
4
Solve the system of equations to find the intersection point (p,q)(p, q)
p=2p = 2 and q=6q = 6, giving point (2,6)(2, 6)
Eliminate variable yy by adding 3×(3x+2y=18)3 \times (3x + 2y = 18) and 2×(2x3y=14)2 \times (2x - 3y = -14) to get 13x=2613x = 26.
5
Sum the coordinates pp and qq
p+q=8p + q = 8
Evaluate 2+6=82 + 6 = 8 as required by the stem.

Anahtar Kavram

Perpendicular line slope relationships and linear system intersection
Tahmini Süre:2m 0s
Soru 1570Soru

An executive chartered a private aircraft to complete a trip between two cities separated by a non-stop distance of 600600 miles. On the return flight along the exact same route, strong headwinds reduced the aircraft's average ground speed by 5050 miles per hour compared to its outbound speed. As a result, the return flight took 22 hours longer than the outbound flight. What was the average speed, in miles per hour, of the aircraft on the outbound flight?

Cevabı ve açıklamayı göster

Cevap: 150

Cevap

150 miles per hour
By setting the outbound speed to vv and return speed to v50v - 50, the relationship between outbound flight time 600v\frac{600}{v} and return flight time 600v50\frac{600}{v - 50} yields the rational equation 600v50600v=2\frac{600}{v - 50} - \frac{600}{v} = 2. Solving this equation gives the quadratic v250v15,000=0v^2 - 50v - 15,000 = 0, which factors as (v150)(v+100)=0(v - 150)(v + 100) = 0. The positive root gives an outbound speed of 150 miles per hour.

Adım Adım Çözüm

1
Define variables for the unknown outbound rate and express travel times for both legs.
Let vv be the outbound speed in miles per hour. Outbound time is 600v\frac{600}{v} hours and return time is 600v50\frac{600}{v - 50} hours.
Distance equals speed multiplied by time (d=vtd = v \cdot t), so time equals distance divided by speed.
2
Formulate the algebraic equation using the given difference in flight durations.
\frac{600}{v - 50} - \frac{600}{v} = 2
The return flight took 22 hours longer than the outbound flight.
3
Clear denominators and simplify into standard quadratic form.
v^2 - 50v - 15,000 = 0
Multiplying both sides by v(v50)v(v - 50) yields 600v600v+30,000=2(v250v)600v - 600v + 30,000 = 2(v^2 - 50v), which simplifies to 2v2100v30,000=02v^2 - 100v - 30,000 = 0 or v250v15,000=0v^2 - 50v - 15,000 = 0.
4
Solve the quadratic equation for vv.
v = 150
Factoring (v150)(v+100)=0(v - 150)(v + 100) = 0 gives solutions v=150v = 150 or v=100v = -100. Physical speed must be positive.

Anahtar Kavram

Distance-Rate-Time Quadratic Algebraic Modeling
Soru 1571Soru

In triangle ABCABC, the ratio of the side lengths AB:BC:ACAB : BC : AC is 3:4:53 : 4 : 5, and the total area of triangle ABCABC is 2424 square units. A line segment DEDE is drawn parallel to side BCBC, where point DD lies on side ABAB and point EE lies on side ACAC. If the perimeter of triangle ADEADE is exactly half the perimeter of triangle ABCABC, what is the area of trapezoid DBCEDBCE in square units?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

The area of trapezoid DBCEDBCE is 1818 square units.
Because line segment DEDE is parallel to side BCBC, triangle ADEADE is similar to triangle ABCABC. Given that the perimeter of triangle ADEADE is half the perimeter of triangle ABCABC, the ratio of their side lengths (the linear scale factor) is 1/21/2. The area ratio of similar triangles is the square of the linear scale factor, which is (1/2)2=1/4(1/2)^2 = 1/4. Thus, the area of triangle ADEADE is 1/4×24=61/4 \times 24 = 6 square units. Subtracting this from the total area gives the area of trapezoid DBCEDBCE: 246=1824 - 6 = 18 square units.

Adım Adım Çözüm

1
Determine the linear scale factor between triangle ADEADE and triangle ABCABC.
Linear scale factor k=Perimeter(ADE)Perimeter(ABC)=12k = \frac{\text{Perimeter}(ADE)}{\text{Perimeter}(ABC)} = \frac{1}{2}.
Since DEBCDE \parallel BC, triangle ADEADE is similar to triangle ABCABC, so the ratio of their perimeters equals the ratio of corresponding side lengths.
2
Calculate the area of triangle ADEADE using the area scale factor k2k^2.
Area(ADE)=k2×Area(ABC)=(12)2×24=14×24=6\text{Area}(ADE) = k^2 \times \text{Area}(ABC) = \left(\frac{1}{2}\right)^2 \times 24 = \frac{1}{4} \times 24 = 6 square units.
The ratio of the areas of two similar figures is the square of their linear scale factor.
3
Subtract the area of triangle ADEADE from the area of triangle ABCABC to find the area of trapezoid DBCEDBCE.
Area(DBCE)=Area(ABC)Area(ADE)=246=18\text{Area}(DBCE) = \text{Area}(ABC) - \text{Area}(ADE) = 24 - 6 = 18 square units.
Trapezoid DBCEDBCE is formed by removing triangle ADEADE from triangle ABCABC.

Anahtar Kavram

Properties of Similar Triangles and Area Scaling
Tahmini Süre:1m 30s
Soru 1572Soru

Trapezoid PQRSPQRS has parallel sides PQPQ and RSRS with lengths of 77 centimeters and 1313 centimeters, respectively. If the perpendicular height between these parallel sides is 44 centimeters, what is the area of trapezoid PQRSPQRS, in square centimeters?

Cevabı ve açıklamayı göster

Cevap: 4040

Cevap

The area of trapezoid PQRSPQRS is 4040 square centimeters.
The area of a trapezoid is found by averaging the lengths of the two parallel bases and multiplying by the perpendicular height: Area=7+132×4=10×4=40\text{Area} = \frac{7 + 13}{2} \times 4 = 10 \times 4 = 40 square centimeters.

Adım Adım Çözüm

1
Identify the formula for the area of a trapezoid
Area=b1+b22×h\text{Area} = \frac{b_1 + b_2}{2} \times h, where b1b_1 and b2b_2 are the lengths of the parallel bases and hh is the height.
The area of any trapezoid is equal to the average of its parallel bases multiplied by its perpendicular height.
2
Substitute the given dimensions into the formula
Area=7+132×4\text{Area} = \frac{7 + 13}{2} \times 4
The given bases are b1=7 cmb_1 = 7\text{ cm} and b2=13 cmb_2 = 13\text{ cm}, and the height is h=4 cmh = 4\text{ cm}.
3
Calculate the average base length and multiply by the height
202×4=10×4=40 cm2\frac{20}{2} \times 4 = 10 \times 4 = 40\text{ cm}^2
Simplifying 7+132\frac{7 + 13}{2} gives 1010, and 10×4=4010 \times 4 = 40.

Anahtar Kavram

Trapezoid Area Formula
Tahmini Süre:45s
Soru 1573Soru

In the xyxy-plane, line pp is defined by the equation y=34x+3y = -\frac{3}{4}x + 3. Which of the following statements about line pp must be true? Select all that apply.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: Line pp passes through the point (4,0)(4, 0).; Line pp is perpendicular to any line with a slope of 43\frac{4}{3}.

Cevap

The statements asserting that line pp passes through the point (4,0)(4, 0) and that line pp is perpendicular to any line with a slope of 43\frac{4}{3} are correct.
Substituting x=4x = 4 into the line equation gives y=0y = 0, showing that (4,0)(4, 0) lies on line pp. Additionally, the slope of line pp is 34-\frac{3}{4}, and the negative reciprocal of 34-\frac{3}{4} is 43\frac{4}{3}, making any line with slope 43\frac{4}{3} perpendicular to line pp.

Adım Adım Çözüm

1
Identify the slope and yy-intercept directly from the equation y=34x+3y = -\frac{3}{4}x + 3.
The slope is m=34m = -\frac{3}{4} and the yy-intercept is (0,3)(0, 3).
The equation is given in standard slope-intercept form y=mx+by = mx + b.
2
Verify point inclusion and perpendicular slope relationship.
Substituting x=4x = 4 yields y=34(4)+3=0y = -\frac{3}{4}(4) + 3 = 0, confirming (4,0)(4, 0) is on the line. The negative reciprocal of 34-\frac{3}{4} is 43\frac{4}{3}, confirming the perpendicular line slope.
A point lies on a line if its coordinates satisfy the equation, and perpendicular lines have slopes that multiply to 1-1.
3
Analyze quadrant coverage.
Line pp connects (0,3)(0, 3) on the positive yy-axis to (4,0)(4, 0) on the positive xx-axis, covering Quadrant I. For x<0x < 0, y>3y > 3 (Quadrant II). For x>4x > 4, y<0y < 0 (Quadrant IV). It never enters Quadrant III where both coordinates are negative.
A line with a positive yy-intercept and negative slope crosses Quadrants I, II, and IV only.

Anahtar Kavram

Line properties in coordinate geometry including slope, intercepts, perpendicularity, and quadrant passage.
Soru 1574Soru

In a circle centered at point OO, the radius is 66 units and central angle POQ\angle POQ measures 120120^\circ. Which of the following statements regarding sector POQPOQ and minor arc PQPQ must be true? Select all such statements.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: The length of minor arc PQPQ is 4π4\pi units.; The area of sector POQPOQ is 12π12\pi square units.

Cevap

The correct statements are that the length of minor arc PQPQ is 4π4\pi units and the area of sector POQPOQ is 12π12\pi square units.
The correct statements accurately apply the arc length and sector area formulas by multiplying the full circumference (12π12\pi) and full area (36π36\pi) by the central angle fraction 120360=13\frac{120^\circ}{360^\circ} = \frac{1}{3}, yielding an arc length of 4π4\pi units and a sector area of 12π12\pi square units.

Adım Adım Çözüm

1
Calculate the central angle fraction of the circle.
The fraction of the circle represented by sector POQPOQ is 120360=13\frac{120^\circ}{360^\circ} = \frac{1}{3}.
Arc length and sector area are proportional to the ratio of the central angle to 360360^\circ.
2
Calculate the arc length of minor arc PQPQ.
\text{Arc length} = \frac{1}{3} \times 2\pi(6) = 4\pi \text{ units}.
Arc length equals the central angle fraction times the total circumference 2πr2\pi r.
3
Calculate the area of sector POQPOQ.
\text{Sector area} = \frac{1}{3} \times \pi(6^2) = 12\pi \text{ square units}.
Sector area equals the central angle fraction times the total circle area πr2\pi r^2.
4
Evaluate the given statements against the calculated values.
Statements asserting an arc length of 4π4\pi units and a sector area of 12π12\pi square units are true. Other statements miscalculate by omitting the fraction or adding incorrect boundary components.
Comparing calculated values confirms the valid choices.

Anahtar Kavram

Arc Length and Sector Area Formulas
Soru 1575Soru

In the xyxy-plane, line kk is defined by the equation y=3x4y = 3x - 4. Line LL is perpendicular to line kk and passes through the point (6,2)(6, 2). What is the yy-intercept of line LL?

Cevabı ve açıklamayı göster

Cevap: 44

Cevap

The yy-intercept of line LL is 44.
Line kk has a slope of 33. Since line LL is perpendicular to line kk, the slope of line LL is 13-\frac{1}{3}. Substituting the point (6,2)(6, 2) into the slope-intercept equation y=mx+by = mx + b gives 2=13(6)+b2 = -\frac{1}{3}(6) + b, which simplifies to 2=2+b2 = -2 + b, so b=4b = 4. Therefore, the yy-intercept is 44.

Adım Adım Çözüm

1
Find the slope of line LL
The slope of line LL is 13-\frac{1}{3}.
Line kk has equation y=3x4y = 3x - 4, so its slope is 33. Perpendicular lines have slopes that are negative reciprocals.
2
Substitute the point (6,2)(6, 2) and slope 13-\frac{1}{3} into the slope-intercept form
2=13(6)+b    2=2+b2 = -\frac{1}{3}(6) + b \implies 2 = -2 + b
The equation of a line is y=mx+by = mx + b, where mm is the slope and bb is the yy-intercept.
3
Solve for bb
b=4b = 4
Adding 22 to both sides yields the yy-intercept.

Anahtar Kavram

Perpendicular Line Slopes and Slope-Intercept Form
Soru 1576Soru

In the xyxy-plane, triangle PQRPQR has vertices at P(0,0)P(0, 0), Q(8,0)Q(8, 0), and R(2,6)R(2, 6). Point SS lies on segment PQPQ such that segment RSRS divides triangle PQRPQR into two regions of equal area. Point TT lies on segment QRQR such that segment STST is parallel to segment PRPR. What is the area of triangle QSTQST?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

6
The area of triangle PQRPQR is calculated using base PQ=8PQ = 8 and height h=6h = 6, yielding 12×8×6=24\frac{1}{2} \times 8 \times 6 = 24. Since segment RSRS divides triangle PQRPQR into two regions of equal area that share the height from vertex RR, point SS must be the midpoint of PQPQ, making QS=4QS = 4. Because segment STST is parallel to segment PRPR, triangle QSTQST is similar to triangle QPRQPR with a side length ratio of QSQP=48=12\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}. The ratio of the areas of similar triangles is the square of the side ratio, (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4}. Multiplying the total area 2424 by 14\frac{1}{4} gives the area of triangle QSTQST as 66.

Adım Adım Çözüm

1
Calculate the area of the main triangle PQRPQR.
Area(PQR)=12×8×6=24\text{Area}(PQR) = \frac{1}{2} \times 8 \times 6 = 24.
The base PQPQ lies along the x-axis with length 80=88 - 0 = 8, and the perpendicular height from vertex R(2,6)R(2,6) to the base is 66.
2
Determine the length of segment QSQS.
QS=4QS = 4.
Line segment RSRS splits PQR\triangle PQR into two smaller triangles, PSR\triangle PSR and QSR\triangle QSR, which share the same altitude from vertex RR. For their areas to be equal, their base lengths PSPS and SQSQ must be equal. Therefore, SS is the midpoint of PQPQ, giving QS=82=4QS = \frac{8}{2} = 4.
3
Establish the similarity relationship and scale factor between QST\triangle QST and QPR\triangle QPR.
QSTQPR\triangle QST \sim \triangle QPR with scale factor k=12k = \frac{1}{2}.
Because segment STST is parallel to segment PRPR, corresponding angles are equal (QST=QPR\angle QST = \angle QPR and QTS=QRP\angle QTS = \angle QRP). Thus, QST\triangle QST is similar to QPR\triangle QPR. The ratio of corresponding side lengths is QSQP=48=12\frac{QS}{QP} = \frac{4}{8} = \frac{1}{2}.
4
Compute the area of triangle QSTQST.
Area(QST)=6\text{Area}(QST) = 6.
The ratio of the areas of similar triangles is equal to the square of their linear scale factor: Area(QST)=(12)2×Area(PQR)=14×24=6\text{Area}(QST) = \left(\frac{1}{2}\right)^2 \times \text{Area}(PQR) = \frac{1}{4} \times 24 = 6.

Anahtar Kavram

Area of triangles, midpoint area partitioning, and area ratio scaling in similar triangles
Tahmini Süre:2m 0s
Soru 1577Soru

Dataset SS consists of 100100 distinct positive numbers. The 25th percentile of dataset SS is 4040, the median is 6060, and the 75th percentile is 8080. A new dataset TT is formed by multiplying every number in dataset SS that is strictly greater than the 80th percentile by 22, while keeping all other numbers unchanged. Which of the following statistics MUST be identical for dataset SS and dataset TT?

Cevabı ve açıklamayı göster

Cevap: The interquartile range

Cevap

The interquartile range
The interquartile range is the difference between the 75th percentile (Q3Q_3) and the 25th percentile (Q1Q_1). Because only values strictly greater than the 80th percentile are scaled, the values defining Q1Q_1 and Q3Q_3 are unaffected. Therefore, Q1=40Q_1 = 40 and Q3=80Q_3 = 80 remain the same in both datasets, making the interquartile range (8040=4080 - 40 = 40) identical.

Adım Adım Çözüm

1
Identify which values in the dataset are modified by the transformation
Only numbers strictly greater than the 80th percentile are multiplied by 2. All numbers at or below the 80th percentile remain unchanged.
The transformation criteria specifies that values below or equal to the 80th percentile threshold undergo no change.
2
Evaluate the effect on the 25th percentile (Q1Q_1) and 75th percentile (Q3Q_3)
Since Q1Q_1 (25th percentile) and Q3Q_3 (75th percentile) are position metrics located at or below the 80th percentile mark, neither Q1Q_1 nor Q3Q_3 changes in value.
Modifying elements only above the 80th percentile leaves the ordering and exact values of elements up to the 80th percentile completely intact.
3
Determine the impact on the Interquartile Range (IQR)
The interquartile range is defined as IQR=Q3Q1IQR = Q_3 - Q_1. Because both Q1Q_1 and Q3Q_3 remain constant, IQRIQR is unchanged.
A metric defined solely by unchanged percentile boundaries must itself remain constant.

Anahtar Kavram

Effect of Upper-Tail Data Transformations on Measures of Position and Dispersion
Tahmini Süre:2m 0s
Soru 1578Soru

A community library purchases two types of books: hardcover books for $24\$24 each and paperback books for $15\$15 each. The library spends a total of $360\$360 on these books and purchases at least one book of each type. Which of the following could be the total number of books purchased? Select all such numbers.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: 18; 21

Cevap

18 and 21
The linear modeling equation 24x+15y=36024x + 15y = 360 simplifies to 8x+5y=1208x + 5y = 120. Since xx and yy must be positive integers, xx must be a multiple of 5. The only valid solutions satisfying x1x \ge 1 and y1y \ge 1 are (x=5,y=16)(x=5, y=16) and (x=10,y=8)(x=10, y=8), which yield total book counts of 21 and 18, respectively.

Adım Adım Çözüm

1
Set up the linear equation from the given word problem context.
24x+15y=36024x + 15y = 360, where x1x \ge 1 is the number of hardcover books and y1y \ge 1 is the number of paperback books, with x,yZ+x, y \in \mathbb{Z}^+.
Total expenditure is the sum of cost per hardcover times number of hardcovers plus cost per paperback times number of paperbacks.
2
Simplify the equation by dividing both sides by the greatest common divisor, 3.
8x+5y=1208x + 5y = 120
Simplifying coefficients reduces arithmetic complexity and isolates integer conditions.
3
Express yy in terms of xx to identify valid integer pairs (x,y)(x, y).
y=1208x5=248x5y = \frac{120 - 8x}{5} = 24 - \frac{8x}{5}
For yy to be an integer, 8x8x must be divisible by 5, meaning xx must be a positive multiple of 5.
4
Test valid positive integer values for xx such that y1y \ge 1.
If x=5x = 5, y=248=16y = 24 - 8 = 16, giving total books x+y=21x + y = 21. If x=10x = 10, y=2416=8y = 24 - 16 = 8, giving total books x+y=18x + y = 18. If x15x \ge 15, y0y \le 0, which is invalid.
These are the only integer solutions satisfying x1x \ge 1 and y1y \ge 1.

Anahtar Kavram

Linear Diophantine Equations in Word Problems
Tahmini Süre:1m 30s
Soru 1579Soru

In ABC\triangle ABC, the measure of A\angle A is 4545^\circ, the measure of C\angle C is 3030^\circ, and the length of side ACAC is 6+236 + 2\sqrt{3}. What is the length of side ABAB?

Cevabı ve açıklamayı göster

Cevap: 262\sqrt{6}

Cevap

The length of side ABAB is 262\sqrt{6}.
The correct answer is 262\sqrt{6}. By drawing altitude BDBD from vertex BB to side ACAC, ABC\triangle ABC is decomposed into right triangle BDABDA (a 45459045^\circ-45^\circ-90^\circ triangle) and right triangle BDCBDC (a 30609030^\circ-60^\circ-90^\circ triangle). Setting altitude BD=xBD = x gives AD=xAD = x and CD=x3CD = x\sqrt{3}. Combining these gives AC=x+x3=x(1+3)AC = x + x\sqrt{3} = x(1 + \sqrt{3}). Equating this to 6+23=23(1+3)6 + 2\sqrt{3} = 2\sqrt{3}(1 + \sqrt{3}) gives x=23x = 2\sqrt{3}. The hypotenuse ABAB of the 45459045^\circ-45^\circ-90^\circ triangle is x2=(23)2=26x\sqrt{2} = (2\sqrt{3})\sqrt{2} = 2\sqrt{6}.

Adım Adım Çözüm

1
Draw altitude BDBD perpendicular to side ACAC with point DD lying on segment ACAC.
ABC\triangle ABC is partitioned into two adjacent right triangles: BDA\triangle BDA and BDC\triangle BDC.
Constructing an interior altitude allows the application of special right triangle ratio rules.
2
Analyze BDC\triangle BDC (30609030^\circ-60^\circ-90^\circ right triangle).
If BD=xBD = x, then CD=x3CD = x\sqrt{3} and BC=2xBC = 2x.
In a 30609030^\circ-60^\circ-90^\circ triangle, sides opposite the angles are in the ratio 1:3:21 : \sqrt{3} : 2.
3
Analyze BDA\triangle BDA (45459045^\circ-45^\circ-90^\circ right triangle).
AD=BD=xAD = BD = x, and hypotenuse AB=x2AB = x\sqrt{2}.
In a 45459045^\circ-45^\circ-90^\circ isosceles right triangle, legs are equal and the hypotenuse is leg×2\text{leg} \times \sqrt{2}.
4
Set up an equation for total side length AC=AD+CDAC = AD + CD.
x+x3=6+23    x(1+3)=23(1+3)    x=23x + x\sqrt{3} = 6 + 2\sqrt{3} \implies x(1 + \sqrt{3}) = 2\sqrt{3}(1 + \sqrt{3}) \implies x = 2\sqrt{3}.
Segment addition postulate combines ADAD and CDCD to match given total length ACAC.
5
Calculate requested side length ABAB.
AB=x2=(23)(2)=26AB = x\sqrt{2} = (2\sqrt{3})(\sqrt{2}) = 2\sqrt{6}.
Substitute x=23x = 2\sqrt{3} into the expression for hypotenuse ABAB.

Anahtar Kavram

Partitioning non-right triangles into 30609030^\circ-60^\circ-90^\circ and 45459045^\circ-45^\circ-90^\circ special right triangles by constructing an altitude.
Soru 1580Soru

Two automated assembly lines, Line A and Line B, produce components at constant individual rates. Under normal operating conditions, Line A operating for 33 hours and Line B operating for 44 hours together produce a combined total of 1,4001,400 units. Under adjusted operating conditions, Line A operates at a rate 20%20\% higher than its normal rate, while Line B operates at a rate 10%10\% lower than its normal rate. Operating together under these adjusted conditions for 55 hours, the two lines produce a total of 2,1002,100 units. What is the normal rate of Line A, in units per hour?

Cevabı ve açıklamayı göster

Cevap: 200200

Cevap

The normal rate of Line A is 200200 units per hour.
Let rAr_A and rBr_B represent the normal production rates in units per hour for Line A and Line B, respectively. From the first condition, 3rA+4rB=14003r_A + 4r_B = 1400. From the second condition, operating for 55 hours at rates 1.20rA1.20r_A and 0.90rB0.90r_B yields 5(1.20rA+0.90rB)=21005(1.20r_A + 0.90r_B) = 2100, which simplifies to 1.20rA+0.90rB=4201.20r_A + 0.90r_B = 420, or 4rA+3rB=14004r_A + 3r_B = 1400. Subtracting 3rA+4rB=14003r_A + 4r_B = 1400 from 4rA+3rB=14004r_A + 3r_B = 1400 gives rArB=0r_A - r_B = 0, meaning rA=rBr_A = r_B. Substituting rB=rAr_B = r_A into 3rA+4rA=14003r_A + 4r_A = 1400 gives 7rA=14007r_A = 1400, so rA=200r_A = 200 units per hour.

Adım Adım Çözüm

1
Define variables and set up the equation for normal operating conditions.
3rA+4rB=14003r_A + 4r_B = 1400
Line A operates for 33 hours at rate rAr_A and Line B operates for 44 hours at rate rBr_B to produce 1,4001,400 units.
2
Set up the equation for adjusted operating conditions.
5(1.20rA+0.90rB)=2100    1.20rA+0.90rB=4205(1.20r_A + 0.90r_B) = 2100 \implies 1.20r_A + 0.90r_B = 420
Line A's rate increases by 20%20\% (1.20rA1.20r_A) and Line B's rate decreases by 10%10\% (0.90rB0.90r_B). Divided by 55 hours, their combined hourly adjusted rate is 420420 units per hour.
3
Multiply the simplified adjusted equation by 1010 to clear decimals.
12rA+9rB=4200    4rA+3rB=140012r_A + 9r_B = 4200 \implies 4r_A + 3r_B = 1400
Dividing all terms by 33 simplifies the linear equation for easier elimination.
4
Solve the system of equations for rAr_A.
rA=200r_A = 200
From Step 1, 4rB=14003rA    rB=3500.75rA4r_B = 1400 - 3r_A \implies r_B = 350 - 0.75r_A. Substituting into 4rA+3(3500.75rA)=14004r_A + 3(350 - 0.75r_A) = 1400 gives 4rA+10502.25rA=1400    1.75rA=350    rA=2004r_A + 1050 - 2.25r_A = 1400 \implies 1.75r_A = 350 \implies r_A = 200.

Anahtar Kavram

Linear Modeling of Combined Work and Rates
ÖncekiSayfa 79 / 107Sonraki
Tüm alıştırma soruları — GRE General Test | Examkin