Tüm alıştırma soruları

2131 soru

Soru 281Soru
If xx and yy are positive real numbers satisfying the system of exponential radical equations
xy=3\sqrt{x\sqrt{y}} = 3
yx=9\sqrt{y\sqrt{x}} = 9
which of the following statements must be true? Select all that apply.

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Cevap: xy=81xy = 81; x1/2+y1/2=10x^{1/2} + y^{1/2} = 10; yxxy=80y^x - x^y = 80

Cevap

The statements asserting that xy=81xy = 81, x1/2+y1/2=10x^{1/2} + y^{1/2} = 10, and yxxy=80y^x - x^y = 80 are true.
Solving the system of radical equations gives the unique positive solution pair x=1x = 1 and y=81y = 81. Substituting these values into the statements shows that xy=181=81xy = 1 \cdot 81 = 81, x1/2+y1/2=1+81=10x^{1/2} + y^{1/2} = \sqrt{1} + \sqrt{81} = 10, and yxxy=811181=80y^x - x^y = 81^1 - 1^{81} = 80 are all mathematically valid.

Adım Adım Çözüm

1
Square both sides of each equation to remove the outer radicals
xy=9    x2y=81x\sqrt{y} = 9 \implies x^2 y = 81 and yx=81    y2x=6561y\sqrt{x} = 81 \implies y^2 x = 6561
Squaring A=B\sqrt{A} = B yields A=B2A = B^2, and rewriting radicals as rational powers gives xy1/2=9x y^{1/2} = 9 and yx1/2=81y x^{1/2} = 81.
2
Express yy in terms of xx from the first equation and substitute into the second
y=81x2    (81x2)2x=6561    6561x3=6561y = \frac{81}{x^2} \implies \left(\frac{81}{x^2}\right)^2 x = 6561 \implies \frac{6561}{x^3} = 6561
Substituting yy eliminates the variable yy, leaving a single equation in terms of xx.
3
Solve for xx and yy
x3=1    x=1x^3 = 1 \implies x = 1, which gives y=8112=81y = \frac{81}{1^2} = 81
Since x>0x > 0 is a positive real number, x=1x = 1 is the unique real solution, yielding y=81y = 81.
4
Evaluate each given statement using x=1x = 1 and y=81y = 81
xy=81xy = 81 (True), x1/2+y1/2=1+9=10x^{1/2} + y^{1/2} = 1 + 9 = 10 (True), xy=927\sqrt{xy} = 9 \neq 27 (False), x=1x = -1 is invalid (False), 811181=8081^1 - 1^{81} = 80 (True)
Direct calculation confirms which individual statements hold.

Anahtar Kavram

Simplifying nested radical equations using fractional exponent rules
Tahmini Süre:2m 0s
Soru 282Soru

If xx is a real number such that 2x+1+2x1=402^{x+1} + 2^{x-1} = 40, what is the value of the expression (x+1)x1(x+1)^{x-1}?

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Cevap: 125

Cevap

125
Factoring out 2x2^x from the given equation yields 2x(2+0.5)=402^x(2 + 0.5) = 40, which gives 2.52x=402.5 \cdot 2^x = 40 and 2x=162^x = 16. This determines that x=4x = 4. Substituting x=4x = 4 into (x+1)x1(x+1)^{x-1} gives (4+1)3=53=125(4+1)^{3} = 5^3 = 125.

Adım Adım Çözüm

1
Rewrite the given exponential terms with a common power of 2.
2x+1=2x212^{x+1} = 2^x \cdot 2^1 and 2x1=2x21=2x22^{x-1} = 2^x \cdot 2^{-1} = \frac{2^x}{2}.
Applying product rule for exponents allows factoring out 2x2^x.
2
Factor out 2x2^x and solve for xx.
2x(2+12)=40    2x52=40    2x=16    x=42^x \left(2 + \frac{1}{2}\right) = 40 \implies 2^x \cdot \frac{5}{2} = 40 \implies 2^x = 16 \implies x = 4.
Combining fractional coefficients isolates the exponential term 2x2^x.
3
Substitute x=4x = 4 into the target expression (x+1)x1(x+1)^{x-1}.
(4+1)41=53=125(4+1)^{4-1} = 5^3 = 125.
Simplifying the base and exponent yields the final numerical value.

Anahtar Kavram

Solving exponential equations using power distribution rules and factoring.
Soru 283Soru

What value of xx satisfies the exponential equation 25x45x+1=12525^x - 4 \cdot 5^{x+1} = 125?

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Cevap: 2

Cevap

The correct answer is 22.
Rewriting 25x25^x as (5x)2(5^x)^2 and 45x+14 \cdot 5^{x+1} as 205x20 \cdot 5^x transforms the equation into (5x)220(5x)125=0(5^x)^2 - 20(5^x) - 125 = 0. Substituting u=5xu = 5^x produces u220u125=0u^2 - 20u - 125 = 0, which factors into (u25)(u+5)=0(u - 25)(u + 5) = 0. Because 5x5^x must be greater than zero for all real values of xx, u=5u = -5 yields no valid real solution. Thus, 5x=25=525^x = 25 = 5^2, giving x=2x = 2.

Adım Adım Çözüm

1
Convert exponential expressions to a common base of 5.
25x=(52)x=(5x)225^x = (5^2)^x = (5^x)^2 and 45x+1=455x=205x4 \cdot 5^{x+1} = 4 \cdot 5 \cdot 5^x = 20 \cdot 5^x.
Applying exponent laws am+n=amana^{m+n} = a^m \cdot a^n and (am)n=amn(a^m)^n = a^{mn} expresses terms in quadratic form with respect to 5x5^x.
2
Formulate and factor the quadratic equation in terms of u=5xu = 5^x.
u220u125=0    (u25)(u+5)=0u^2 - 20u - 125 = 0 \implies (u - 25)(u + 5) = 0, yielding u=25u = 25 or u=5u = -5.
The equation reduces to standard quadratic form, which factors easily.
3
Solve for xx while rejecting non-viable real roots.
5x=25=52    x=25^x = 25 = 5^2 \implies x = 2. 5x=55^x = -5 has no real solution.
An exponential function with a positive base produces strictly positive output values for all real domain inputs.

Anahtar Kavram

Solving exponential equations reducible to quadratic form using exponent laws
Soru 284Soru

At a charity fundraising event, standard tickets were sold for $45\$45 each and VIP tickets were sold for $80\$80 each. The number of standard tickets sold was 1212 more than twice the number of VIP tickets sold. If the total revenue generated from standard tickets exceeded the total revenue from VIP tickets by $1,830\$1,830, how many VIP tickets were sold?

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Cevap: 129

Cevap

129 VIP tickets were sold.
Letting vv represent the number of VIP tickets sold, the number of standard tickets sold is 2v+122v + 12. Expressing the revenue condition yields the linear equation 45(2v+12)80v=183045(2v + 12) - 80v = 1830. Distributing 45 gives 90v+54080v=183090v + 540 - 80v = 1830. Combining like terms results in 10v+540=183010v + 540 = 1830. Subtracting 540 from both sides gives 10v=129010v = 1290, which yields v=129v = 129.

Adım Adım Çözüm

1
Define the unknown variable
Let vv be the number of VIP tickets sold.
The question asks for the number of VIP tickets, making vv a direct choice for the variable.
2
Translate the relationship between ticket quantities into an algebraic expression
Standard tickets sold =2v+12= 2v + 12
'12 more than twice the number of VIP tickets' translates directly to 2v+122v + 12.
3
Formulate total revenue expressions and construct the single-variable linear equation
45(2v+12)80v=183045(2v + 12) - 80v = 1830
Total standard revenue minus total VIP revenue equals the given excess of $1,830\$1,830.
4
Distribute and combine like terms to solve for vv
90v+54080v=1830    10v+540=1830    10v=1290    v=12990v + 540 - 80v = 1830 \implies 10v + 540 = 1830 \implies 10v = 1290 \implies v = 129
Applying standard algebraic operations isolates vv on one side of the equation.

Anahtar Kavram

Formulating and solving a linear equation in one variable from a multi-step word problem context
Tahmini Süre:2m 0s
Soru 285Soru

A reservoir initially contains 4,8004,800 gallons of water. A drainage pump operates at a constant rate of 120120 gallons per minute to empty the reservoir. How many minutes of continuous pumping will it take for the amount of water remaining in the reservoir to be reduced to 1,2001,200 gallons?

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Cevap: 3030

Cevap

30 minutes
To find the time needed for 1,2001,200 gallons to remain, subtract 1,2001,200 from the initial 4,8004,800 gallons to determine that 3,6003,600 gallons of water must be removed. Dividing the 3,6003,600 gallons by the drainage rate of 120120 gallons per minute gives exactly 3030 minutes.

Adım Adım Çözüm

1
Calculate the total volume of water that must be drained from the reservoir.
4,800 gallons1,200 gallons=3,600 gallons4,800 \text{ gallons} - 1,200 \text{ gallons} = 3,600 \text{ gallons}
The question asks for the time until 1,2001,200 gallons remain, so the amount removed is the difference between the starting volume and ending volume.
2
Set up the linear rate equation to solve for elapsed time tt in minutes.
Rate×t=Volume Drained    120t=3,600\text{Rate} \times t = \text{Volume Drained} \implies 120t = 3,600
The rate of drainage is constant at 120120 gallons per minute.
3
Solve for tt.
t=3,600120=30 minutest = \frac{3,600}{120} = 30 \text{ minutes}
Dividing total gallons to drain by the rate gives the required time.

Anahtar Kavram

Linear Modeling and Distance-Rate-Time / Work-Rate Relationships
Tahmini Süre:1m 0s
Soru 286Soru

Based on the structural contrast and elaboration signals in the passage below, fill in the blank with the word that best completes the text.

Aşağıdaki boşlukları doldurun

In early twentieth-century ethnomusicology, field recordings captured on wax cylinders were initially celebrated as objective preservation tools that recorded acoustic reality without distortion. Far from capturing an unmediated performance, however, these early acoustic recordings inadvertently the very traditions they sought to document; indeed, the intrusive presence of recording apparatuses frequently compelled musicians to truncate improvisational structures and temper tonal nuances, thereby fundamentally altering authentic performance practices.
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Cevap

compromised
The contrast transition 'Far from... however' signals a turn away from the idea that field recordings were purely objective and non-distorting. The continuation signal 'indeed' after the semicolon elaborates on this pivot, explaining that the presence of recording equipment forced musicians to alter their performance habits. Consequently, the blank requires a word conveying that the integrity or authenticity of the traditions was weakened or altered (such as 'compromised').

Adım Adım Çözüm

1
Analyze the opening contrast transition
The initial clause describes wax cylinders as celebrated 'objective preservation tools' capturing music 'without distortion'. The structural signal 'Far from..., however' indicates a complete reversal of this initial assumption.
Contrast markers signal that the target word must express an outcome opposite to unmediated preservation.
2
Examine the elaboration signal following the semicolon
The continuation signal 'indeed' introduces explicit evidence: the presence of recording equipment forced performers to 'truncate improvisational structures' and 'temper tonal nuances,' resulting in altered performance practices.
Elaboration markers reinforce and clarify the negative impact established by the structural contrast.
3
Select the vocabulary word that synthesizes both structural clues
The term 'compromised' (or acceptable synonyms such as 'vitiated' or 'altered') accurately reflects how the recording process impaired or undermined the authenticity of the musical traditions.
It reconciles the structural reversal ('however') with the explanatory details of performance modification ('indeed').

Anahtar Kavram

Deciphering Meaning via Structural Clues and Contrast/Continuation Signals
Soru 287Soru

Let mm and nn be negative integers such that m<nm < n and mnm - n is an odd integer. Which of the following expressions must be a positive even integer?

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Cevap: (mn)2+1(m - n)^2 + 1

Cevap

(mn)2+1(m - n)^2 + 1 must be a positive even integer.
Because mnm - n is given as an odd integer and m<nm < n, mnm - n is a negative odd integer. Squaring any negative odd integer gives a positive odd integer. Adding 11 to a positive odd integer yields a positive even integer, which guarantees the result is always positive and even.

Adım Adım Çözüm

1
Analyze the parity and sign of mnm - n
mnm - n is a negative odd integer.
Since m<nm < n, mn<0m - n < 0. The problem specifies that mnm - n is odd.
2
Evaluate the expression (mn)2(m - n)^2
(mn)2(m - n)^2 is a positive odd integer.
Squaring any non-zero real number yields a positive result. Squaring an odd integer always yields an odd integer.
3
Add 1 to (mn)2(m - n)^2
(mn)2+1(m - n)^2 + 1 is an even integer greater than or equal to 2.
Adding 1 to a positive odd integer converts it into a positive even integer.

Anahtar Kavram

Parity rules under arithmetic operations and exponents with signed integers
Tahmini Süre:1m 15s
Soru 288Soru

A logistics warehouse uses two sizes of shipping crates: small crates and large crates. A shipment containing 33 small crates and 55 large crates has a total weight of 4747 pounds. A second shipment containing 66 small crates and 22 large crates has a total weight of 3838 pounds. What is the total weight, in pounds, of a shipment containing 22 small crates and 33 large crates?

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Cevap: 29

Cevap

29 pounds
By setting up the linear system 3s+5l=473s + 5l = 47 and 6s+2l=386s + 2l = 38, using elimination gives s=4s = 4 pounds for a small crate and l=7l = 7 pounds for a large crate. Substituting these values into the target expression 2s+3l2s + 3l yields 2(4)+3(7)=292(4) + 3(7) = 29 pounds.

Adım Adım Çözüm

1
Set up a system of linear equations from the given context.
Let ss be the weight of a small crate and ll be the weight of a large crate. The equations are: (1) 3s+5l=473s + 5l = 47 and (2) 6s+2l=386s + 2l = 38.
Translate the physical constraints of each shipment into algebraic equations.
2
Solve for the variable ll by eliminating ss.
Multiply Equation (1) by 22 to get 6s+10l=946s + 10l = 94. Subtract Equation (2) (6s+2l=386s + 2l = 38) from this new equation: (6s+10l)(6s+2l)=9438    8l=56    l=7(6s + 10l) - (6s + 2l) = 94 - 38 \implies 8l = 56 \implies l = 7.
Align the coefficients of ss so that elimination via subtraction yields a single-variable linear equation for ll.
3
Substitute l=7l = 7 back into Equation (2) to solve for ss.
6s+2(7)=38    6s+14=38    6s=24    s=46s + 2(7) = 38 \implies 6s + 14 = 38 \implies 6s = 24 \implies s = 4.
Determine the individual weight of a small crate.
4
Calculate the requested total weight for 22 small crates and 33 large crates.
2s+3l=2(4)+3(7)=8+21=292s + 3l = 2(4) + 3(7) = 8 + 21 = 29.
Substitute the individual values of ss and ll into the target expression.

Anahtar Kavram

Solving 2x2 Systems of Linear Equations by Elimination and Linear Combination Evaluation
Tahmini Süre:1m 30s
Soru 289Soru

Let xx and yy be non-zero integers that satisfy all of the following conditions:
I. (x)y+1<0(-x)^{y + 1} < 0
II. x2y+xx^2 y + x is odd
III. x<yx < y

Which of the following statements MUST be true? Select all that apply.

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Cevap: yxy^x is an even integer; xyyx^y - y is an odd integer; (x+y)2(x + y)^2 is an odd integer

Cevap

The statements that must be true are 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer'.
From the given conditions, we deduce that xx is a positive odd integer and yy is a positive even integer with x<yx < y. Therefore:
- Raising the positive even integer yy to the positive power xx yields an even integer (yxy^x is even).
- Raising the odd integer xx to the positive power yy yields an odd integer, and subtracting the even integer yy leaves an odd integer (xyyx^y - y is odd).
- The sum of odd xx and even yy is odd, and squaring an odd integer yields an odd integer ((x+y)2(x + y)^2 is odd).

Adım Adım Çözüm

1
Analyze Condition I: (x)y+1<0(-x)^{y + 1} < 0
x>0x > 0 (a positive integer) and yy is an even integer.
For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence, x<0    x>0-x < 0 \implies x > 0, and y+1y + 1 is odd     y\implies y is even.
2
Analyze Condition II: x2y+xx^2 y + x is odd
xx is an odd integer.
Factor the expression as x(xy+1)x(xy + 1). A product of two integers is odd if and only if both factors are odd. Thus, xx must be odd. (Additionally, xy+1xy + 1 must be odd     xy\implies xy is even, which holds since yy is even).
3
Analyze Condition III: x<yx < y
yy is a positive even integer.
Since xx is a positive odd integer (x1x \ge 1) and x<yx < y, yy must also be a positive integer (y2y \ge 2).
4
Evaluate the statements based on the derived properties (xx is positive odd, yy is positive even, x<yx < y)
The statements 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer' are guaranteed to be true.
1) evenpositive odd=even\text{even}^{\text{positive odd}} = \text{even}. 2) oddpositive eveneven=oddeven=odd\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}. 3) (odd+even)2=odd2=odd(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}.

Anahtar Kavram

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.
Tahmini Süre:2m 0s
Soru 290Soru

Two water reservoirs, Reservoir A and Reservoir B, contain a combined total of 360360 liters of water. After 3030 liters of water are transferred from Reservoir A to Reservoir B, the volume of water in Reservoir B is equal to 23\frac{2}{3} of the volume of water remaining in Reservoir A. How many liters of water were originally in Reservoir A?

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Cevap: 246246

Cevap

246 liters were originally in Reservoir A.
Let xx represent the original volume of Reservoir A. Since the combined volume is 360360 liters, Reservoir B initially contains 360x360 - x liters. After transferring 3030 liters from A to B, Reservoir A contains x30x - 30 liters and Reservoir B contains (360x)+30=390x(360 - x) + 30 = 390 - x liters. Setting up the equation 390x=23(x30)390 - x = \frac{2}{3}(x - 30) and multiplying by 3 gives 11703x=2x601170 - 3x = 2x - 60. Combining like terms yields 5x=12305x = 1230, so x=246x = 246 liters. Thus, 246246 is the correct original volume.

Adım Adım Çözüm

1
Define the variable for the unknown quantity.
Let xx be the original volume of water in Reservoir A (in liters).
Choosing a single variable simplifies setting up a linear equation.
2
Express the original volume of Reservoir B in terms of xx.
Original volume in Reservoir B is 360x360 - x.
The total volume across both reservoirs is given as 360360 liters.
3
Write expressions for the volumes in each reservoir after the transfer of 30 liters.
Volume in Reservoir A after transfer: x30x - 30.
Volume in Reservoir B after transfer: (360x)+30=390x(360 - x) + 30 = 390 - x.
Transferring 30 liters removes 30 liters from A and adds 30 liters to B.
4
Set up the linear equation based on the given relationship.
390x=23(x30)390 - x = \frac{2}{3}(x - 30)
The problem specifies that the new volume in Reservoir B is 23\frac{2}{3} of the new volume in Reservoir A.
5
Solve the equation for xx.
3(390x)=2(x30)    11703x=2x60    1230=5x    x=2463(390 - x) = 2(x - 30) \implies 1170 - 3x = 2x - 60 \implies 1230 = 5x \implies x = 246.
Multiply both sides by 3 to eliminate the fraction, then collect like terms.

Anahtar Kavram

Formulating and solving a linear equation in one variable from a real-world conservation/transfer scenario.
Soru 291Soru

If xx is a real number such that 2x5=x+4|2x - 5| = |x + 4|, which of the following could be the value of x2x^2? Select all such values.

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Cevap: 19\frac{1}{9}; 8181

Cevap

The possible values of x2x^2 are 19\frac{1}{9} and 8181.
To solve 2x5=x+4|2x - 5| = |x + 4|, set 2x52x - 5 equal to both x+4x + 4 and (x+4)-(x + 4). Solving 2x5=x+42x - 5 = x + 4 gives x=9x = 9, which leads to x2=81x^2 = 81. Solving 2x5=x42x - 5 = -x - 4 gives 3x=13x = 1, so x=13x = \frac{1}{3}, which leads to x2=19x^2 = \frac{1}{9}. Therefore, the options representing 19\frac{1}{9} and 8181 are correct.

Adım Adım Çözüm

1
Set up equations to remove the absolute value bars
Two linear equations: 2x5=x+42x - 5 = x + 4 or 2x5=(x+4)2x - 5 = -(x + 4).
For real numbers AA and BB, A=B|A| = |B| implies A=BA = B or A=BA = -B.
2
Solve the first equation 2x5=x+42x - 5 = x + 4
x=9x = 9.
Subtract xx and add 55 to both sides.
3
Solve the second equation 2x5=x42x - 5 = -x - 4
3x=1    x=133x = 1 \implies x = \frac{1}{3}.
Add xx and add 55 to both sides.
4
Compute x2x^2 for each possible value of xx
If x=9x = 9, then x2=81x^2 = 81. If x=13x = \frac{1}{3}, then x2=19x^2 = \frac{1}{9}.
The question asks for the values of x2x^2, not xx.

Anahtar Kavram

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.
Soru 292Soru

If xx and yy are real numbers such that x43|x - 4| \le 3 and y+25|y + 2| \le 5, what is the maximum possible value of xy|x - y|?

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Cevap: 14

Cevap

The maximum possible value of xy|x - y| is 14.
Solving x43|x - 4| \le 3 gives the closed interval [1,7][1, 7] for xx. Solving y+25|y + 2| \le 5 gives the closed interval [7,3][-7, 3] for yy. The maximum possible value of xy|x - y| is the maximum distance between a point in [1,7][1, 7] and a point in [7,3][-7, 3], which is 7(7)=147 - (-7) = 14.

Adım Adım Çözüm

1
Determine the range of possible values for xx.
1x71 \le x \le 7
The inequality x43|x - 4| \le 3 represents all numbers within distance 3 of 4 on the number line.
2
Determine the range of possible values for yy.
7y3-7 \le y \le 3
The inequality y+25|y + 2| \le 5 represents all numbers within distance 5 of -2 on the number line.
3
Find the maximum distance between any point xx in [1,7][1, 7] and any point yy in [7,3][-7, 3].
14
The maximum absolute difference xy|x - y| occurs between the upper endpoint of the xx-interval (77) and the lower endpoint of the yy-interval (7-7).

Anahtar Kavram

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables
Soru 293Soru

If nn is a positive integer such that 2n2+2n1+2n+2n+1=4802^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480, what is the value of nn?

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Cevap: 7

Cevap

The value of nn is 7.
Factoring out 2n22^{n-2} converts the sum into 2n2(1+2+4+8)=152n2=4802^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480. Dividing 480 by 15 gives 2n2=322^{n-2} = 32. Since 32=2532 = 2^5, setting n2=5n - 2 = 5 yields n=7n = 7.

Adım Adım Çözüm

1
Factor out the lowest power of 2, which is 2n22^{n-2}, from all terms on the left side of the equation.
2n2(1+2+4+8)=4802^{n-2}(1 + 2 + 4 + 8) = 480
Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.
2
Evaluate the constant factor inside the parentheses and solve for the exponential expression 2n22^{n-2}.
152n2=480    2n2=3215 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32
Summing 1+2+4+81 + 2 + 4 + 8 yields 15. Dividing both sides by 15 isolates the base-2 term.
3
Express 32 as a power with base 2 and equate the exponents.
2n2=25    n2=5    n=72^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7
Since 32=2532 = 2^5 and the bases are identical, the exponents must be equal.

Anahtar Kavram

Factoring sum of exponential terms with common bases
Soru 294Soru

Two concentric circles centered at point OO have radii of 66 units and 1010 units. A central angle θ\theta defines a sector that intersects the region between the concentric circles, creating a region bounded by an outer arc, an inner arc, and two straight line segments. If the area of this bounded region is 16π16\pi square units, what is the perimeter of the region?

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Cevap: 8+8π8 + 8\pi

Cevap

The perimeter of the region is 8+8π8 + 8\pi units.
The region between the concentric circles has an area equal to the fraction of the central angle times the area of the ring: θ360π(10262)=64π(θ360)\frac{\theta}{360^\circ} \cdot \pi (10^2 - 6^2) = 64\pi \left(\frac{\theta}{360^\circ}\right). Setting this equal to 16π16\pi gives θ360=14\frac{\theta}{360^\circ} = \frac{1}{4}. The outer arc length is 14(2π10)=5π\frac{1}{4}(2\pi \cdot 10) = 5\pi, and the inner arc length is 14(2π6)=3π\frac{1}{4}(2\pi \cdot 6) = 3\pi. The region is bounded on the sides by two segments of length 106=410 - 6 = 4 units each. The total perimeter is 5π+3π+4+4=8+8π5\pi + 3\pi + 4 + 4 = 8 + 8\pi.

Adım Adım Çözüm

1
Express the area of the bounded region in terms of the central angle fraction f=θ360f = \frac{\theta}{360^\circ}.
Area=fπ(R2r2)=fπ(10262)=64πf\text{Area} = f \cdot \pi(R^2 - r^2) = f \cdot \pi(10^2 - 6^2) = 64\pi f.
The area of the region between two concentric sector arcs is the difference between the outer sector area and the inner sector area.
2
Solve for the fraction ff using the given area of 16π16\pi.
64πf=16π    f=16π64π=1464\pi f = 16\pi \implies f = \frac{16\pi}{64\pi} = \frac{1}{4}.
Equating the algebraic expression for the region's area to 16π16\pi allows finding the proportion of the circle represented by the central angle.
3
Calculate the lengths of the outer arc, inner arc, and straight boundary segments.
Outer Arc=14(2π10)=5π\text{Outer Arc} = \frac{1}{4}(2\pi \cdot 10) = 5\pi, Inner Arc=14(2π6)=3π\text{Inner Arc} = \frac{1}{4}(2\pi \cdot 6) = 3\pi, Segment Length=106=4\text{Segment Length} = 10 - 6 = 4.
Arc lengths use the formula f2πrf \cdot 2\pi r, and each straight boundary segment is the radial distance between the two circles.
4
Sum all boundary components to find the total perimeter.
Perimeter=5π+3π+4+4=8+8π\text{Perimeter} = 5\pi + 3\pi + 4 + 4 = 8 + 8\pi.
The complete perimeter of the bounded region consists of the outer arc, inner arc, and two radial segments.

Anahtar Kavram

Annular Sector Area and Perimeter
Tahmini Süre:1m 30s
Soru 295Soru

If xx is a real number satisfying the exponential equation 4x+14x1=1204^{x+1} - 4^{x-1} = 120, what is the value of 22x+12^{2x + 1}?

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Cevap: 64

Cevap

The value of 22x+12^{2x + 1} is 64.
Factoring 4x14^{x-1} from 4x+14x14^{x+1} - 4^{x-1} gives 4x1(161)=154x1=1204^{x-1}(16 - 1) = 15 \cdot 4^{x-1} = 120. Dividing by 15 yields 4x1=84^{x-1} = 8. Rewriting with base 2 gives (22)x1=22x2=23(2^2)^{x-1} = 2^{2x-2} = 2^3, so 2x2=32x - 2 = 3, meaning 2x=52x = 5. Substituting 2x=52x = 5 into 22x+12^{2x+1} gives 25+1=26=642^{5+1} = 2^6 = 64.

Adım Adım Çözüm

1
Factor out 4x14^{x-1} from the left side of the equation 4x+14x1=1204^{x+1} - 4^{x-1} = 120.
4x1(421)=1204^{x-1}(4^2 - 1) = 120, which simplifies to 4x1(15)=1204^{x-1}(15) = 120.
Factoring out the lowest power of 4 allows simplification of the terms on the left side.
2
Divide both sides by 15 to isolate 4x14^{x-1}.
4x1=12015=84^{x-1} = \frac{120}{15} = 8.
Isolating the exponential expression is necessary to solve for xx.
3
Express both sides with a common base of 2.
(22)x1=23    22(x1)=23    22x2=23(2^2)^{x-1} = 2^3 \implies 2^{2(x-1)} = 2^3 \implies 2^{2x - 2} = 2^3.
Converting to a common prime base allows equating the exponents.
4
Equate exponents to solve for 2x2x.
2x2=3    2x=52x - 2 = 3 \implies 2x = 5.
Since the bases are equal and positive, their exponents must be equal.
5
Substitute 2x=52x = 5 into the target expression 22x+12^{2x + 1}.
25+1=26=642^{5 + 1} = 2^6 = 64.
Evaluating the exact expression requested in the stem.

Anahtar Kavram

Factoring exponential expressions with variable exponents and converting bases
Tahmini Süre:1m 30s
Soru 296Soru

In triangle ABCABC, the length of side ABAB is 77 and the length of side BCBC is 1515. If the area of triangle ABCABC is 4242, which of the following could be the length of side ACAC? Select all such lengths.

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Cevap: 2372\sqrt{37}; 2020

Cevap

The possible lengths of side ACAC are 2372\sqrt{37} and 2020.
Using the triangle area formula Area=12absinB\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin B, we find sinB=45\sin B = \frac{4}{5}. Because sinB\sin B is positive in both Quadrant I and Quadrant II, angle BB can be either acute or obtuse. If angle BB is acute, cosB=35\cos B = \frac{3}{5}, giving AC=72+1522(7)(15)(0.6)=237AC = \sqrt{7^2 + 15^2 - 2(7)(15)(0.6)} = 2\sqrt{37}. If angle BB is obtuse, cosB=35\cos B = -\frac{3}{5}, giving AC=72+1522(7)(15)(0.6)=20AC = \sqrt{7^2 + 15^2 - 2(7)(15)(-0.6)} = 20. Both values represent valid triangle configurations.

Adım Adım Çözüm

1
Determine the sine of angle BB using the area formula.
Area=12ABBCsinB    42=12715sinB    sinB=84105=45\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B \implies 42 = \frac{1}{2} \cdot 7 \cdot 15 \cdot \sin B \implies \sin B = \frac{84}{105} = \frac{4}{5}.
The area of a triangle with two given sides and an included angle is 12absinθ\frac{1}{2} a b \sin \theta.
2
Find the possible values for cosB\cos B.
Since sinB=45\sin B = \frac{4}{5}, cosB\cos B can be either 35\frac{3}{5} (if angle BB is acute) or 35-\frac{3}{5} (if angle BB is obtuse).
Sine is positive in both the first and second quadrants, permitting both acute and obtuse angles for triangle ABCABC.
3
Calculate the length of side ACAC when angle BB is acute.
AC2=72+1522(7)(15)(35)=49+225126=148    AC=148=237AC^2 = 7^2 + 15^2 - 2(7)(15)\left(\frac{3}{5}\right) = 49 + 225 - 126 = 148 \implies AC = \sqrt{148} = 2\sqrt{37}.
Apply the Law of Cosines: AC2=AB2+BC22(AB)(BC)cosBAC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B with cosB=0.6\cos B = 0.6.
4
Calculate the length of side ACAC when angle BB is obtuse.
AC2=72+1522(7)(15)(35)=49+225+126=400    AC=400=20AC^2 = 7^2 + 15^2 - 2(7)(15)\left(-\frac{3}{5}\right) = 49 + 225 + 126 = 400 \implies AC = \sqrt{400} = 20.
Apply the Law of Cosines with cosB=0.6\cos B = -0.6.

Anahtar Kavram

Triangle Area via Included Angle and Dual Solutions in Non-Right Triangles
Soru 297Soru

In triangle PQRPQR, the measure of angle PQRPQR is 9090^\circ. Point SS lies on segment PRPR such that line segment QSQS is perpendicular to PRPR. If PS=4PS = 4 and SR=9SR = 9, what is the area of triangle PQRPQR?

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Cevap: 3939

Cevap

The area of triangle PQRPQR is 39.
By the geometric mean theorem (right triangle altitude theorem), the altitude QSQS satisfies QS2=PSSR=49=36QS^2 = PS \cdot SR = 4 \cdot 9 = 36, so QS=6QS = 6. The hypotenuse PR=PS+SR=4+9=13PR = PS + SR = 4 + 9 = 13. Substituting base 1313 and height 66 into the triangle area formula 12bh\frac{1}{2}bh gives 12×13×6=39\frac{1}{2} \times 13 \times 6 = 39.

Adım Adım Çözüm

1
Find the length of hypotenuse PRPR
PR=PS+SR=4+9=13PR = PS + SR = 4 + 9 = 13
Point SS lies on segment PRPR, so the total length of the hypotenuse is the sum of its two segments.
2
Calculate altitude QSQS using the geometric mean theorem for right triangles
QS=PS×SR=4×9=36=6QS = \sqrt{PS \times SR} = \sqrt{4 \times 9} = \sqrt{36} = 6
In a right triangle, the altitude to the hypotenuse is the geometric mean of the two segments into which the hypotenuse is divided.
3
Calculate the area of triangle PQRPQR
Area=12×base×height=12×13×6=39\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 13 \times 6 = 39
The area of any triangle is half the product of its base and corresponding height.

Anahtar Kavram

Altitude to the hypotenuse in right triangles and triangle area calculation

Alternatif Yöntem

Alternatively, use similar triangles PQSQRS\triangle PQS \sim \triangle QRS. The ratio of corresponding sides gives PQPS=PRPQ    PQ2=PSPR=413=52\frac{PQ}{PS} = \frac{PR}{PQ} \implies PQ^2 = PS \cdot PR = 4 \cdot 13 = 52, and QR2=SRPR=913=117QR^2 = SR \cdot PR = 9 \cdot 13 = 117. Since PQR\triangle PQR is a right triangle at QQ, its area is 12PQQR=1252117=126084=1278=39\frac{1}{2} \cdot PQ \cdot QR = \frac{1}{2} \sqrt{52 \cdot 117} = \frac{1}{2} \sqrt{6084} = \frac{1}{2} \cdot 78 = 39.
Tahmini Süre:1m 30s
Soru 298Soru

On the real number line, xx is a negative real number such that 52x=11|5 - 2x| = 11, and yy is a positive real number such that 3y+1=13|3y + 1| = 13. What is the distance on the real number line between xx and yy?

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Cevap: 7

Cevap

The distance between xx and yy on the real number line is 7.
Solving 52x=11|5 - 2x| = 11 with x<0x < 0 yields x=3x = -3. Solving 3y+1=13|3y + 1| = 13 with y>0y > 0 yields y=4y = 4. The distance between 3-3 and 44 on the number line is 4(3)=7|4 - (-3)| = 7.

Adım Adım Çözüm

1
Solve for the negative real number xx using the equation 52x=11|5 - 2x| = 11.
x=3x = -3
The equation splits into 52x=115 - 2x = 11 (yielding x=3x = -3) and 52x=115 - 2x = -11 (yielding x=8x = 8). Since xx must be negative, x=3x = -3 is selected.
2
Solve for the positive real number yy using the equation 3y+1=13|3y + 1| = 13.
y=4y = 4
The equation splits into 3y+1=133y + 1 = 13 (yielding y=4y = 4) and 3y+1=133y + 1 = -13 (yielding y=143y = -\frac{14}{3}). Since yy must be positive, y=4y = 4 is selected.
3
Compute the distance between xx and yy on the number line.
7
The distance between two points on the number line is given by yx=4(3)=7|y - x| = |4 - (-3)| = 7.

Anahtar Kavram

Distance on the real number line between two points aa and bb is given by ab|a - b|, solved by evaluating absolute value equations under given sign constraints.

Alternatif Yöntem

Plot the candidate solutions for xx (x=3x = -3 and x=8x = 8) and yy (y=4y = 4 and y=14/3y = -14/3) on a number line, then directly count units between the valid points x=3x = -3 and y=4y = 4.
Tahmini Süre:1m 15s
Soru 299Soru
A manufacturing company's monthly total production cost CC, in dollars, is related to the number of units produced, xx, by the linear equation:
3(C1,200)5=2x+180\frac{3(C - 1,200)}{5} = 2x + 180
If the total production cost in a given month was between $4,000\$4,000 and $5,000\$5,000, inclusive, which of the following could be the number of units produced in that month? Select all such numbers.

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Cevap: 800; 950; 1,000

Cevap

The numbers of units that could be produced are 800, 950, and 1,000.
Solving the linear equation for xx gives x=3C4,50010x = \frac{3C - 4,500}{10}. Substituting the boundary cost values C=4,000C = 4,000 and C=5,000C = 5,000 yields x=750x = 750 and x=1,050x = 1,050, respectively. Therefore, any valid number of units must lie in the range [750,1,050][750, 1,050]. The values 800, 950, and 1,000 fall strictly within this interval.

Adım Adım Çözüm

1
Express x in terms of C by isolating the variable x in the linear equation.
3(C1,200)=5(2x+180)    3C3,600=10x+900    10x=3C4,500    x=3C4,500103(C - 1,200) = 5(2x + 180) \implies 3C - 3,600 = 10x + 900 \implies 10x = 3C - 4,500 \implies x = \frac{3C - 4,500}{10}
Isolating xx allows direct evaluation of the unit limits based on the cost boundary values.
2
Calculate the minimum value of x corresponding to the minimum cost C = 4,000.
xmin=3(4,000)4,50010=12,0004,50010=7,50010=750x_{\min} = \frac{3(4,000) - 4,500}{10} = \frac{12,000 - 4,500}{10} = \frac{7,500}{10} = 750
Determines the lower bound of units produced.
3
Calculate the maximum value of x corresponding to the maximum cost C = 5,000.
xmax=3(5,000)4,50010=15,0004,50010=10,50010=1,050x_{\max} = \frac{3(5,000) - 4,500}{10} = \frac{15,000 - 4,500}{10} = \frac{10,500}{10} = 1,050
Determines the upper bound of units produced.
4
Filter the given choices to find all values within the interval [750, 1,050].
The valid values are 800, 950, and 1,000.
Only values between 750 and 1,050 inclusive satisfy the cost constraint.

Anahtar Kavram

Linear Equations in One Variable
Soru 300Soru

In a school club of 4040 students, the ratio of the number of boys to the number of girls is 3:53 : 5. Which of the following statements must be true? Select all such statements.

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Cevap: The number of boys in the club is 1515.; The ratio of the number of boys to the total number of students is 3:83 : 8.; The number of girls in the club exceeds the number of boys by 1010.

Cevap

The correct statements are those indicating that there are 15 boys in the club, that the ratio of boys to total students is 3:8, and that there are 10 more girls than boys in the club.
The total number of ratio parts is 3+5=83 + 5 = 8, representing the entire group of 4040 students. Dividing 4040 by 88 gives 55 students per part. Therefore, the number of boys is 3×5=153 \times 5 = 15, the number of girls is 5×5=255 \times 5 = 25, and the ratio of boys to total students is 15:40=3:815:40 = 3:8. Subtracting the number of boys from girls (2515=1025 - 15 = 10) shows there are 1010 more girls than boys.

Adım Adım Çözüm

1
Calculate total ratio parts and determine the value of one part.
Total ratio parts = 3+5=83 + 5 = 8. Each part corresponds to 408=5\frac{40}{8} = 5 students.
Converting ratio terms into parts allows calculation of actual quantities from the total group size.
2
Compute the total number of boys and girls.
Number of boys = 3×5=153 \times 5 = 15. Number of girls = 5×5=255 \times 5 = 25.
Multiply each component's ratio share by the value of one part.
3
Evaluate the given statements against calculated quantities.
There are 1515 boys; the ratio of boys to total students is 15:40=3:815:40 = 3:8; and the difference between girls and boys is 2515=1025 - 15 = 10.
Determines which options represent true statements about the group.

Anahtar Kavram

Part-to-Part vs. Part-to-Whole Ratios
ÖncekiSayfa 15 / 107Sonraki
Tüm alıştırma soruları — GRE General Test | Examkin