Tüm alıştırma soruları

612 soru

Soru 41Soru

If 3(k+4)=5k83(k + 4) = 5k - 8, what is the value of kk?

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Cevap: 10

Cevap

10
Distributing the 33 on the left side yields 3k+12=5k83k + 12 = 5k - 8. Subtracting 3k3k from both sides yields 12=2k812 = 2k - 8. Adding 88 to both sides yields 20=2k20 = 2k. Dividing by 22 gives the solution k=10k = 10.

Adım Adım Çözüm

1
Distribute 33 to the terms inside the parentheses
3k+12=5k83k + 12 = 5k - 8
To simplify the left side of the equation
2
Subtract 3k3k from both sides of the equation
12=2k812 = 2k - 8
To group the variable terms on one side
3
Add 88 to both sides of the equation
20=2k20 = 2k
To isolate the variable term
4
Divide both sides by 22
k=10k = 10
To solve for kk

Anahtar Kavram

Solving linear equations with variables on both sides using distributive property
Soru 42Soru

In the equation 34(8x12)+kx=10x9\frac{3}{4}(8x - 12) + kx = 10x - 9, kk is a constant. If the equation has infinitely many solutions, what is the value of kk?

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Cevap: 4

Cevap

The value of kk is 44.
Distributing the fraction on the left side of the equation yields 34(8x)34(12)=6x9\frac{3}{4}(8x) - \frac{3}{4}(12) = 6x - 9. Substituting this back gives 6x9+kx=10x96x - 9 + kx = 10x - 9. Factoring out xx on the left side gives (6+k)x9=10x9(6 + k)x - 9 = 10x - 9. For a linear equation to have infinitely many solutions, the coefficients of xx on both sides must be identical, and the constants must be identical. Since the constants on both sides are already 9-9, we set the coefficients equal: 6+k=106 + k = 10. Subtracting 6 from both sides gives the correct value k=4k = 4.

Adım Adım Çözüm

1
Distribute the fraction 34\frac{3}{4} to the terms inside the parentheses.
6x9+kx=10x96x - 9 + kx = 10x - 9
To simplify the expression and eliminate the parentheses.
2
Factor out xx from the terms on the left side of the equation.
(6+k)x9=10x9(6 + k)x - 9 = 10x - 9
To group the xx terms together to easily compare coefficients.
3
Set the coefficient of xx on the left side equal to the coefficient of xx on the right side.
6+k=106 + k = 10
For the equation to have infinitely many solutions, the coefficients of the variable on both sides must be equal when the constant terms are equal.
4
Solve for kk by subtracting 6 from both sides of the equation.
k=4k = 4
To isolate the constant kk.

Anahtar Kavram

A linear equation in one variable has infinitely many solutions when it can be simplified to an identity of the form Ax+B=Ax+BAx + B = Ax + B, meaning both the coefficients of xx and the constant terms on both sides of the equation are equal.
Soru 43Soru

In the equation below, aa and bb are constants.

13(2a5x)34(xb)=2912x+5\frac{1}{3}(2a - 5x) - \frac{3}{4}(x - b) = -\frac{29}{12}x + 5

If the equation has infinitely many solutions for xx, what is the value of 8a+9b8a + 9b?

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Cevap: 60

Cevap

The value of 8a+9b8a + 9b is 6060.
To find the value of 8a+9b8a + 9b that makes the equation have infinitely many solutions, we first expand and simplify the left side of the equation: 23a53x34x+34b=2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5. Combining the xx terms gives -\frac{29}{12}x + \(\frac{2}{3}a + \frac{3}{4}b\) = -\frac{29}{12}x + 5. Since the coefficients of xx on both sides are equal (2912-\frac{29}{12}), the equation will have infinitely many solutions if the constant terms on both sides are also equal. This requires 23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5. Multiplying this entire equation by the least common multiple of the denominators, which is 12, yields 8a+9b=608a + 9b = 60.

Adım Adım Çözüm

1
Distribute the constants through the parentheses on the left side of the equation.
23a53x34x+34b=2912x+5\frac{2}{3}a - \frac{5}{3}x - \frac{3}{4}x + \frac{3}{4}b = -\frac{29}{12}x + 5
This separates the variable terms from the constant terms so the equation can be simplified.
2
Combine the coefficients of the xx terms on the left side using 12 as the common denominator.
2912x+23a+34b=2912x+5-\frac{29}{12}x + \frac{2}{3}a + \frac{3}{4}b = -\frac{29}{12}x + 5
Simplifying the variable terms allows us to compare the coefficients on both sides of the equation.
3
Equate the constant terms from the left and right sides of the equation.
23a+34b=5\frac{2}{3}a + \frac{3}{4}b = 5
A linear equation has infinitely many solutions when the coefficients of the variable on both sides are equal and the constant terms on both sides are also equal.
4
Multiply both sides of the equation by 12 to eliminate the fractional denominators.
8a+9b=608a + 9b = 60
Multiplying the equation by the common denominator directly evaluates the target expression 8a+9b8a + 9b.

Anahtar Kavram

For a linear equation in one variable to have infinitely many solutions, it must be reducible to an identity of the form cx+d=cx+dcx + d = cx + d, where both the variable coefficients and the constant terms on both sides of the equation are equal.
Soru 44Soru

A factory manufactures solar panels at a constant rate. At the start of a morning shift, the factory has already manufactured 8080 solar panels. The total number of solar panels manufactured tt hours after the shift begins is modeled by a linear function. If the factory has manufactured a total of 200200 solar panels 33 hours after the shift begins, how many hours after the shift begins will the factory have manufactured a total of 440440 solar panels?

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Cevap: 9

Cevap

9
To find the time when the total number of solar panels reaches 440440, we first determine the constant rate of production. The change in solar panels over the first 33 hours is 20080=120200 - 80 = 120 panels. Dividing by the 33 hours gives a constant rate of 4040 panels per hour. The linear equation representing the total panels is N(t)=40t+80N(t) = 40t + 80. Setting N(t)=440N(t) = 440 gives 440=40t+80440 = 40t + 80, which simplifies to 360=40t360 = 40t. Solving for tt yields t=9t = 9.

Adım Adım Çözüm

1
Calculate the constant rate of production (slope) of the linear function.
The rate is 4040 solar panels per hour.
The factory starts with 8080 solar panels and reaches 200200 solar panels in 33 hours. The rate of change is the change in the number of panels divided by the change in time: 2008030=1203=40\frac{200 - 80}{3 - 0} = \frac{120}{3} = 40 panels per hour.
2
Write the linear equation representing the total number of solar panels manufactured, N(t)N(t), after tt hours.
N(t)=40t+80N(t) = 40t + 80
Since the initial quantity is 8080 and the rate of production is 4040 panels per hour, the linear function is N(t)=40t+80N(t) = 40t + 80.
3
Solve for the time tt when the total number of solar panels is 440440.
t=9t = 9 hours
Set N(t)=440N(t) = 440 in the equation: 440=40t+80440 = 40t + 80. Subtract 8080 from both sides to get 360=40t360 = 40t. Divide both sides by 4040 to find t=9t = 9.

Anahtar Kavram

Linear Functions and Graphs
Tahmini Süre:1m 30s
Soru 45Soru

In the equation 12(kx4)23(xk)=56x1\frac{1}{2}(kx - 4) - \frac{2}{3}(x - k) = \frac{5}{6}x - 1, where kk is a constant, the equation has no solution for xx. What is the value of kk?

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Cevap: 3

Cevap

The value of the constant kk for which the equation has no solution is 33.
For a linear equation of the form Ax+B=Cx+DAx + B = Cx + D to have no solution, the coefficients of the variable terms must be equal (A=CA = C) but the constant terms must be unequal (BDB \neq D). Distributing the terms on the left side of the given equation and grouping them yields (12k23)x+(23k2)=56x1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1. Setting the coefficients of xx equal to each other gives 12k23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}. Adding 23\frac{2}{3} to both sides yields 12k=96\frac{1}{2}k = \frac{9}{6}, or 12k=32\frac{1}{2}k = \frac{3}{2}, which simplifies to k=3k = 3. Evaluating the constant terms when k=3k = 3 gives 23(3)2=0\frac{2}{3}(3) - 2 = 0 on the left side and 1-1 on the right side. Since 010 \neq -1, the equation has no solution when k=3k = 3.

Adım Adım Çözüm

1
Distribute the fractional coefficients on the left side of the equation.
12kx223x+23k=56x1\frac{1}{2}kx - 2 - \frac{2}{3}x + \frac{2}{3}k = \frac{5}{6}x - 1
To separate the variable terms from the constants for further algebraic manipulation.
2
Group the terms on the left side into a coefficient for xx and a single constant term.
(12k23)x+(23k2)=56x1\left(\frac{1}{2}k - \frac{2}{3}\right)x + \left(\frac{2}{3}k - 2\right) = \frac{5}{6}x - 1
To represent the equation in the standard linear format Ax+B=Cx+DAx + B = Cx + D.
3
Set the coefficients of xx from both sides equal to each other.
12k23=56\frac{1}{2}k - \frac{2}{3} = \frac{5}{6}
A linear equation has no solution when the variable terms on both sides cancel out, requiring their coefficients to be identical.
4
Solve the linear equation to isolate the constant kk.
k=3k = 3
Add 23\frac{2}{3} to both sides to get 12k=96\frac{1}{2}k = \frac{9}{6}, which simplifies to 12k=32\frac{1}{2}k = \frac{3}{2}. Multiplying both sides by 22 yields k=3k = 3.
5
Verify that the constant terms are not equal when substituting k=3k = 3.
The left-side constant is 00 and the right-side constant is 1-1. Since 010 \neq -1, the condition is satisfied.
To ensure the equation does not simplify to an identity with infinitely many solutions (which occurs when both the variable coefficients and the constant terms are equal).

Anahtar Kavram

Determining the conditions under which a linear equation in one variable has no solution.

Alternatif Yöntem

To avoid working with fractions, multiply every term in the equation by 66 (the least common multiple of 2,3,62, 3, 6) at the start: 3(kx4)4(xk)=5x63(kx - 4) - 4(x - k) = 5x - 6. Expand the parentheses to get 3kx124x+4k=5x63kx - 12 - 4x + 4k = 5x - 6, and group the terms: (3k4)x+(4k12)=5x6(3k - 4)x + (4k - 12) = 5x - 6. For there to be no solution, set the coefficients of xx equal to each other: 3k4=5    3k=9    k=33k - 4 = 5 \implies 3k = 9 \implies k = 3. Check the constant terms with k=3k = 3: 4(3)12=04(3) - 12 = 0, which is unequal to 6-6. This confirms k=3k = 3 is the correct answer.
Tahmini Süre:2m 0s
Soru 46Soru

In the xyxy-plane, the graph of the linear function f(x)=px+qf(x) = px + q, where pp and qq are constants, passes through the point (2,7)(2, 7). Line LL is parallel to the graph of ff and has a yy-intercept that is 33 units below the yy-intercept of the graph of ff. If the xx-intercept of line LL is (6,0)(-6, 0), what is the value of pp?

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Cevap: 0.5

Cevap

The value of pp is 0.50.5 (or the fraction 1/21/2).
The correct answer is 0.50.5 (or 1/21/2). The graph of f(x)=px+qf(x) = px + q passes through (2,7)(2, 7), which means 7=2p+q7 = 2p + q, or q=72pq = 7 - 2p. Line LL is parallel to ff, so its slope is pp, and its yy-intercept is q3q - 3. Thus, the equation of line LL is y=px+q3y = px + q - 3. Since line LL has an xx-intercept at (6,0)(-6, 0), we can substitute x=6x = -6 and y=0y = 0 into its equation, yielding 0=6p+q30 = -6p + q - 3. Substituting q=72pq = 7 - 2p into this equation gives 0=6p+(72p)30 = -6p + (7 - 2p) - 3, which simplifies to 8p+4=0-8p + 4 = 0. Solving for pp gives 8p=48p = 4, or p=0.5p = 0.5.

Adım Adım Çözüm

1
Express the relationship between pp and qq using the given point (2,7)(2, 7) on the graph of ff.
q=72pq = 7 - 2p
The point (2,7)(2, 7) must satisfy the equation f(x)=px+qf(x) = px + q.
2
Formulate the equation of line LL using the parallel slope and the shifted yy-intercept.
y=px+q3y = px + q - 3
Parallel lines have equal slopes, and the yy-intercept of LL is 33 units below the yy-intercept of ff, which is qq.
3
Substitute the xx-intercept (6,0)(-6, 0) into the equation of line LL.
6p+q3=0-6p + q - 3 = 0
The xx-intercept is a point on the line where y=0y = 0.
4
Substitute the expression for qq from Step 1 into the equation from Step 3 and solve for pp.
p=0.5p = 0.5 (or 12\frac{1}{2})
Solving the linear equation 6p+(72p)3=0-6p + (7 - 2p) - 3 = 0 simplifies to 8p+4=0-8p + 4 = 0, giving p=0.5p = 0.5.

Anahtar Kavram

Understanding linear functions, their graphs, slopes of parallel lines, and intercepts.

Alternatif Yöntem

Instead of solving for qq first, you can use the point-slope form. Line LL passes through (6,0)(-6, 0) and has slope pp, so its equation is y=p(x+6)y = p(x + 6), or y=px+6py = px + 6p. The yy-intercept of LL is 6p6p. The yy-intercept of the graph of ff is qq. We are given that the yy-intercept of LL is 33 units below the yy-intercept of ff, so 6p=q36p = q - 3. Since ff passes through (2,7)(2, 7), we have 7=2p+q7 = 2p + q, which means q=72pq = 7 - 2p. Substituting this into 6p=q36p = q - 3 gives 6p=(72p)36p = (7 - 2p) - 3, or 6p=42p6p = 4 - 2p. Adding 2p2p to both sides gives 8p=48p = 4, which results in p=0.5p = 0.5.
Tahmini Süre:2m 30s
Soru 47Soru

A rental company charges a flat fee of 3535 dollars plus 1212 dollars per hour to rent a bicycle. If a customer was charged a total of 9595 dollars, for how many hours did the customer rent the bicycle?

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Cevap: 5

Cevap

5
To find the number of hours the bicycle was rented, we can set up the equation 12h+35=9512h + 35 = 95, where hh is the number of hours. Subtracting 3535 from both sides gives 12h=6012h = 60. Dividing both sides by 1212 gives h=5h = 5. Thus, the customer rented the bicycle for 55 hours.

Adım Adım Çözüm

1
Set up a linear equation representing the total cost.
12h+35=9512h + 35 = 95, where hh represents the number of hours rented.
The total cost consists of a flat fee of 3535 dollars plus an hourly charge of 1212 dollars multiplied by the number of hours hh.
2
Subtract the flat fee from both sides of the equation to isolate the variable term.
12h=6012h = 60
Subtracting 3535 from both sides simplifies the equation to find the total hourly charge portion of the cost.
3
Divide by the hourly rate to solve for the number of hours hh.
h=5h = 5
Dividing both sides by 1212 isolates hh to find the number of rental hours.

Anahtar Kavram

Linear Equations in One Variable
Soru 48Soru

In the xyxy-plane, the graph of the linear function ff has a slope of 3-3 and passes through the point (4,18)(4, 18). What is the yy-coordinate of the yy-intercept of the graph of ff?

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Cevap: 30

Cevap

The correct answer is 30.
The equation of a linear function can be written in slope-intercept form as f(x)=mx+bf(x) = mx + b, where mm is the slope and bb is the yy-coordinate of the yy-intercept. Substituting the given slope m=3m = -3 yields f(x)=3x+bf(x) = -3x + b. Since the point (4,18)(4, 18) lies on the graph of ff, we substitute x=4x = 4 and f(x)=18f(x) = 18 into the equation to get 18=3(4)+b18 = -3(4) + b, which simplifies to 18=12+b18 = -12 + b. Adding 1212 to both sides of the equation yields b=30b = 30. Thus, the yy-coordinate of the yy-intercept of the graph of ff is 3030.

Adım Adım Çözüm

1
Write the general slope-intercept form of a linear equation.
f(x)=3x+bf(x) = -3x + b
The slope of the line is given as 3-3, so we substitute m=3m = -3 into f(x)=mx+bf(x) = mx + b.
2
Substitute the point (4,18)(4, 18) into the equation.
18=3(4)+b18 = -3(4) + b
Since the graph of ff passes through (4,18)(4, 18), these coordinates must satisfy the function equation.
3
Solve for the yy-intercept bb.
b=30b = 30
Simplify to 18=12+b18 = -12 + b, and then add 1212 to both sides to isolate bb.

Anahtar Kavram

Finding the equation of a line using its slope and a point.
Soru 49Soru

In the equation below, mm is a constant.

35(5x10)2(xm)=18\frac{3}{5}(5x - 10) - 2(x - m) = 18

If the solution to the equation is x=12x = 12, what is the value of mm?

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Cevap: 6

Cevap

The correct value of mm is 6.
Substituting x=12x = 12 into the equation yields 35(5(12)10)2(12m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18, which simplifies to 3024+2m=1830 - 24 + 2m = 18. Combining constant terms gives 6+2m=186 + 2m = 18. Subtracting 6 from both sides gives 2m=122m = 12, and dividing by 2 yields m=6m = 6.

Adım Adım Çözüm

1
Substitute x=12x = 12 into the given equation.
35(5(12)10)2(12m)=18\frac{3}{5}(5(12) - 10) - 2(12 - m) = 18
Since x=12x = 12 is the solution, it must satisfy the equation.
2
Simplify the expression inside the first set of parentheses.
5(12)10=6010=505(12) - 10 = 60 - 10 = 50
Evaluate the terms inside the parentheses first.
3
Multiply the simplified term by the fraction 35\frac{3}{5}.
35(50)=30\frac{3}{5}(50) = 30
Multiply 50 by 3 and divide by 5.
4
Substitute 30 back into the equation and distribute 2-2 to the terms inside the second set of parentheses.
3024+2m=1830 - 24 + 2m = 18
Distributing 2-2 to 12m12 - m yields 24+2m-24 + 2m due to sign rules.
5
Combine the constant terms on the left side of the equation.
6+2m=186 + 2m = 18
3024=630 - 24 = 6.
6
Subtract 6 from both sides to isolate the term with mm.
2m=122m = 12
Isolate the variable term on one side of the equation.
7
Divide both sides by 2 to solve for mm.
m=6m = 6
Divide 12 by 2 to find the final value.

Anahtar Kavram

Solving a linear equation in one variable by substitution and isolation.
Soru 50Soru

In the xyxy-plane, line kk has the equation y=2x+10y = -2x + 10. Line mm is parallel to line kk and passes through the point (6,8)(6, 8). Line ll is perpendicular to line kk and intersects the xx-axis at the point (a,0)(a, 0), where a>10a > 10. If the region bounded by lines kk, mm, ll, and the yy-axis has an area of 110110, what is the value of aa?

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Cevap: 25

Cevap

25
To find the value of aa, we determine the equations of the lines mm and ll based on their geometric relationships to line kk. Line mm is parallel to line kk (y=2x+10y = -2x + 10), so its slope is 2-2. Using the point (6,8)(6, 8), its equation is y=2x+20y = -2x + 20. Line ll is perpendicular to line kk, so its slope is 12\frac{1}{2}. It intersects the xx-axis at (a,0)(a, 0), giving the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a. The bounded region formed by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis is a trapezoid. Calculating the area of this trapezoid by dividing it into a parallelogram and a triangle yields the area formula 60+2a60 + 2a. Setting this equal to the given area of 110110 yields 60+2a=11060 + 2a = 110, which solves to a=25a = 25.

Adım Adım Çözüm

1
Determine the equation of line mm using the parallel slope and the given point.
Line mm has the equation y=2x+20y = -2x + 20.
Parallel lines have equal slopes. Since line kk has a slope of 2-2, line mm also has a slope of 2-2. Substituting the point (6,8)(6, 8) into the point-slope form gives y8=2(x6)y - 8 = -2(x - 6).
2
Determine the equation of line ll using the perpendicular slope and its xx-intercept.
Line ll has the equation y=12x12ay = \frac{1}{2}x - \frac{1}{2}a.
Perpendicular lines have negative reciprocal slopes. The negative reciprocal of 2-2 is 12\frac{1}{2}. Using the point (a,0)(a, 0) in the point-slope form gives y0=12(xa)y - 0 = \frac{1}{2}(x - a).
3
Calculate the vertices of the bounded region by finding the intersection points of the boundary lines.
The vertices of the bounded region are (0,20)(0, 20), (0,10)(0, 10), (4+0.2a,20.4a)(4 + 0.2a, 2 - 0.4a), and (8+0.2a,40.4a)(8 + 0.2a, 4 - 0.4a).
The region is bounded by the parallel lines kk and mm, the perpendicular line ll, and the yy-axis (x=0x = 0).
4
Find the area of the region as an algebraic expression in terms of aa.
The area is equal to 60+2a60 + 2a.
The region can be divided into a parallelogram with a vertical base of 1010 and width 4+0.2a4 + 0.2a, and a right triangle with a vertical base of 1010 and width 44. The sum of their areas is 10(4+0.2a)+12(10)(4)=40+2a+20=60+2a10(4 + 0.2a) + \frac{1}{2}(10)(4) = 40 + 2a + 20 = 60 + 2a.
5
Set the area expression equal to the given area of 110110 and solve for aa.
a=25a = 25
Setting the area equal to 110110 yields 60+2a=11060 + 2a = 110, which simplifies to 2a=502a = 50, or a=25a = 25.

Anahtar Kavram

Linear functions, parallel and perpendicular lines, finding line equations, and coordinate geometry area.

Alternatif Yöntem

The area can also be calculated using the geometric properties of a trapezoid. The height of the trapezoid is the perpendicular distance between the parallel lines kk and mm, which is 2010(2)2+12=25\frac{|20 - 10|}{\sqrt{(-2)^2 + 1^2}} = 2\sqrt{5}. The bases of the trapezoid are the segments of lines kk and mm from the yy-axis to their intersection points with line ll. The length of the base on line kk is 959\sqrt{5} and the length of the base on line mm is 13513\sqrt{5} (when a=25a = 25). Using the formula for the area of a trapezoid, Area=95+1352×25=115×25=110\text{Area} = \frac{9\sqrt{5} + 13\sqrt{5}}{2} \times 2\sqrt{5} = 11\sqrt{5} \times 2\sqrt{5} = 110.
Tahmini Süre:3m 0s
Soru 51Soru

A water tank contains 24 gallons of water. Water is being drained from the tank at a constant rate. After 8 minutes, the tank contains 14 gallons of water. If the volume of water in the tank, in gallons, is a linear function of the time in minutes, how many minutes will it take for the tank to contain exactly 4 gallons of water?

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Cevap: 16

Cevap

16
The volume of water in the tank decreases linearly from an initial value of 2424 gallons at t=0t = 0 to 1414 gallons at t=8t = 8. The constant rate of change (slope) is calculated by dividing the change in volume by the change in time: 142480=1.25\frac{14 - 24}{8 - 0} = -1.25 gallons per minute. Using the slope-intercept form, the volume VV at time tt is given by V=1.25t+24V = -1.25t + 24. Setting the volume V=4V = 4 gives the equation 4=1.25t+244 = -1.25t + 24. Solving for tt yields 20=1.25t-20 = -1.25t, which simplifies to t=16t = 16.

Adım Adım Çözüm

1
Determine the initial state and the state after 8 minutes as coordinate points.
The initial state is (0,24)(0, 24) and the state after 88 minutes is (8,14)(8, 14).
These coordinates represent the relationship between time and volume of water in the tank.
2
Calculate the slope (constant rate of change) of the linear function.
Slope m=142480=1.25m = \frac{14 - 24}{8 - 0} = -1.25
The slope represents the constant rate at which water is being drained from the tank.
3
Formulate the linear equation.
V(t)=1.25t+24V(t) = -1.25t + 24
Using the slope-intercept form V(t)=mt+bV(t) = mt + b, where b=24b = 24 is the vertical intercept representing the initial volume.
4
Solve for the time tt when the volume of water is 44 gallons.
4=1.25t+24    20=1.25t    t=164 = -1.25t + 24 \implies -20 = -1.25t \implies t = 16
To find the time at which the volume decreases to exactly 44 gallons.

Anahtar Kavram

Linear Functions and Graphs
Soru 52Soru

If 25(3x4)13(2x+5)=15x+1115\frac{2}{5}(3x - 4) - \frac{1}{3}(2x + 5) = \frac{1}{5}x + \frac{11}{15}, what is the value of 2x72x - 7?

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Cevap: 17

Cevap

The correct answer is 17.
To find the value of 2x72x - 7, first solve the linear equation for xx. Distributing the coefficients on the left side of the equation gives 65x8523x53=15x+1115\frac{6}{5}x - \frac{8}{5} - \frac{2}{3}x - \frac{5}{3} = \frac{1}{5}x + \frac{11}{15}. Combining the variable terms and constants on the left side results in 815x4915=315x+1115\frac{8}{15}x - \frac{49}{15} = \frac{3}{15}x + \frac{11}{15}. Subtracting 315x\frac{3}{15}x and adding 4915\frac{49}{15} to both sides yields 515x=6015\frac{5}{15}x = \frac{60}{15}, which simplifies to 13x=4\frac{1}{3}x = 4, or x=12x = 12. Finally, substituting 12 into the expression 2x72x - 7 gives 2(12)7=172(12) - 7 = 17.

Adım Adım Çözüm

1
Distribute the coefficients to the terms inside the parentheses.
65x8523x53=15x+1115\frac{6}{5}x - \frac{8}{5} - \frac{2}{3}x - \frac{5}{3} = \frac{1}{5}x + \frac{11}{15}
To eliminate parentheses and allow grouping of like terms.
2
Combine the variable terms and the constant terms on the left side using a common denominator of 15.
815x4915=315x+1115\frac{8}{15}x - \frac{49}{15} = \frac{3}{15}x + \frac{11}{15}
To simplify the linear equation into a standard two-sided form.
3
Subtract the variable term from the right side and add the constant term from the left side.
515x=6015\frac{5}{15}x = \frac{60}{15}, which simplifies to x=12x = 12
To isolate the variable xx on one side of the equation.
4
Evaluate the expression 2x72x - 7 using the value of xx.
2(12)7=172(12) - 7 = 17
To solve for the final requested quantity.

Anahtar Kavram

Linear Equations in One Variable
Soru 53Soru

The table below shows some values of the linear function ff.

xxf(x)f(x)
221111
441717
662323

What is the yy-coordinate of the yy-intercept of the graph of y=f(x)y = f(x) in the xyxy-plane?

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Cevap: 5

Cevap

The correct answer is 5, representing the y-coordinate of the y-intercept of the graph of f.
To find the yy-intercept of the linear function, we first determine its slope using the points (2,11)(2, 11) and (4,17)(4, 17) from the table. The slope mm is 171142=62=3\frac{17 - 11}{4 - 2} = \frac{6}{2} = 3. Next, we use the slope-intercept equation f(x)=mx+bf(x) = mx + b. Substituting m=3m = 3 and the point (2,11)(2, 11) gives 11=3(2)+b11 = 3(2) + b, which simplifies to 11=6+b11 = 6 + b. Solving for bb yields 55. Therefore, the yy-coordinate of the yy-intercept of the graph of ff is 55.

Adım Adım Çözüm

1
Calculate the slope (mm) of the linear function using two coordinate pairs from the table.
The slope is m=3m = 3.
A linear function has a constant slope, which can be found using the formula m=f(x2)f(x1)x2x1m = \frac{f(x_2) - f(x_1)}{x_2 - x_1}.
2
Substitute the slope and one of the points into the slope-intercept equation f(x)=mx+bf(x) = mx + b to solve for the yy-intercept bb.
The yy-intercept bb is 55.
The yy-coordinate of the yy-intercept of the graph of y=f(x)y = f(x) is the value of bb in the equation f(x)=mx+bf(x) = mx + b.

Anahtar Kavram

Finding the y-intercept of a linear function from a table of values.
Tahmini Süre:45s
Soru 54Soru

In the xyxy-plane, the graph of the linear function ff is perpendicular to the line with equation 3x+4y=483x + 4y = 48. The graph of ff intersects the xx-axis at the point (p,0)(p, 0) and the yy-axis at the point (0,q)(0, q), where q>0q > 0. If the distance between the points (p,0)(p, 0) and (0,q)(0, q) is 1515, what is the value of qq?

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Cevap: 12

Cevap

12
First, find the slope of the given line by rewriting 3x+4y=483x + 4y = 48 in slope-intercept form: y=34x+12y = -\frac{3}{4}x + 12. The slope is 34-\frac{3}{4}. The graph of the linear function ff is perpendicular to this line, so its slope is the negative reciprocal, 43\frac{4}{3}. With a yy-intercept of (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives the xx-intercept (p,0)=(34q,0)(p, 0) = (-\frac{3}{4}q, 0). The distance between these intercepts is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q (since q>0q > 0). Given that the distance is 1515, we solve 54q=15\frac{5}{4}q = 15 to find q=12q = 12.

Adım Adım Çözüm

1
Find the slope of the line 3x+4y=483x + 4y = 48.
The slope of the line is 34-\frac{3}{4}.
To find the slope of the perpendicular line ff, we first need the slope of the given line. Rewriting 3x+4y=483x + 4y = 48 in slope-intercept form gives y=34x+12y = -\frac{3}{4}x + 12.
2
Determine the slope of ff.
The slope of ff is 43\frac{4}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Find the xx-intercept of ff in terms of qq.
The xx-intercept of ff is (34q,0)(-\frac{3}{4}q, 0).
Since the yy-intercept of ff is (0,q)(0, q), the equation of ff is y=43x+qy = \frac{4}{3}x + q. Setting y=0y = 0 gives 0=43x+q0 = \frac{4}{3}x + q, which simplifies to x=34qx = -\frac{3}{4}q.
4
Use the distance formula between the intercepts to solve for qq.
The value of qq is 1212.
The distance between (34q,0)(-\frac{3}{4}q, 0) and (0,q)(0, q) is (34q)2+q2=2516q2=54q\sqrt{(-\frac{3}{4}q)^2 + q^2} = \sqrt{\frac{25}{16}q^2} = \frac{5}{4}q since q>0q > 0. Setting this distance to 1515 gives 54q=15\frac{5}{4}q = 15, which yields q=12q = 12.

Anahtar Kavram

Properties of perpendicular lines, finding intercepts, and utilizing the distance formula in coordinate geometry.
Soru 55Soru

If 3(2n5)=4n+93(2n - 5) = 4n + 9, what is the value of nn?

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Cevap: 12

Cevap

12
To solve the equation 3(2n5)=4n+93(2n - 5) = 4n + 9, we first distribute the 3 to the terms inside the parentheses to get 6n15=4n+96n - 15 = 4n + 9. Next, we subtract 4n4n from both sides to group the variable terms, giving 2n15=92n - 15 = 9. We then add 15 to both sides to isolate the variable term, resulting in 2n=242n = 24. Finally, dividing both sides by 2 gives the solution n=12n = 12.

Adım Adım Çözüm

1
Distribute 3 to the terms inside the parentheses.
6n15=4n+96n - 15 = 4n + 9
To simplify the left side of the equation and remove parentheses.
2
Subtract 4n4n from both sides of the equation.
2n15=92n - 15 = 9
To group the variable terms on one side of the equation.
3
Add 15 to both sides of the equation.
2n=242n = 24
To isolate the term with the variable nn.
4
Divide both sides of the equation by 2.
n=12n = 12
To solve for nn.

Anahtar Kavram

Solving a linear equation in one variable by applying the distributive property and isolating the variable.
Soru 56Soru

Consider the system of equations below.

x=2yx = 2y
3xy=103x - y = 10

If (x,y)(x, y) is the solution to the system of equations above, what is the value of xx?

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Cevap: 4

Cevap

The value of xx is 44.
The system of equations can be solved by substituting x=2yx = 2y from the first equation into the second equation, which gives 3(2y)y=103(2y) - y = 10. Simplifying this equation yields 6yy=106y - y = 10, or 5y=105y = 10. Dividing by 55 gives y=2y = 2. Substituting y=2y = 2 back into the first equation yields x=2(2)=4x = 2(2) = 4. Therefore, the value of xx is 44.

Adım Adım Çözüm

1
Substitute the expression for xx from the first equation into the second equation.
3(2y)y=103(2y) - y = 10
Since the first equation gives xx in terms of yy, substituting it into the second equation reduces the system to a single linear equation in one variable.
2
Solve the resulting equation for yy.
y=2y = 2
Simplifying 3(2y)y=103(2y) - y = 10 yields 6yy=106y - y = 10, which simplifies further to 5y=105y = 10. Dividing both sides by 55 gives y=2y = 2.
3
Substitute the value of yy back into the first equation to solve for xx.
x=4x = 4
Using x=2yx = 2y and substituting y=2y = 2 gives x=2(2)=4x = 2(2) = 4.

Anahtar Kavram

Solving systems of linear equations using the substitution method.
Soru 57Soru
In the equation below, aa and bb are positive constants.
23(32xa)34(8bx)=52(x3)12\frac{2}{3} \left( \frac{3}{2}x - a \right) - \frac{3}{4} \left( 8 - bx \right) = \frac{5}{2}(x - 3) - \frac{1}{2}
If the equation has infinitely many solutions, what is the value of aba^b?
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Cevap: 9

Cevap

The correct answer is 9.
The correct answer is 9. Expanding the left side of the equation yields (1+34b)x(6+23a)\left(1 + \frac{3}{4}b\right)x - \left(6 + \frac{2}{3}a\right), and simplifying the right side yields 52x8\frac{5}{2}x - 8. For the linear equation to have infinitely many solutions, the coefficient of xx on the left, 1+34b1 + \frac{3}{4}b, must equal the coefficient of xx on the right, 52\frac{5}{2}, which gives b=2b = 2. Similarly, the constant term on the left, (6+23a)-\left(6 + \frac{2}{3}a\right), must equal the constant term on the right, 8-8, which simplifies to 6+23a=86 + \frac{2}{3}a = 8 and gives a=3a = 3. Evaluating aba^b with these values yields 32=93^2 = 9.

Adım Adım Çözüm

1
Expand both sides of the equation to collect like terms.
(1+34b)x(6+23a)=52x8\left( 1 + \frac{3}{4}b \right)x - \left( 6 + \frac{2}{3}a \right) = \frac{5}{2}x - 8
Expanding allows us to compare the coefficient of xx and the constant term on each side of the equation.
2
Equate the coefficients of xx on both sides of the equation.
1+34b=52    b=21 + \frac{3}{4}b = \frac{5}{2} \implies b = 2
For a linear equation to have infinitely many solutions, the coefficient of xx must be identical on both sides.
3
Equate the constant terms on both sides of the equation.
(6+23a)=8    a=3-\left( 6 + \frac{2}{3}a \right) = -8 \implies a = 3
For a linear equation to have infinitely many solutions, the constant terms must also be identical on both sides.
4
Calculate the value of aba^b using the solved values of aa and bb.
32=93^2 = 9
The question asks for the value of the expression aba^b where a=3a = 3 and b=2b = 2.

Anahtar Kavram

A linear equation in one variable of the form Ax+B=Cx+DAx + B = Cx + D has infinitely many solutions if and only if A=CA = C and B=DB = D.

Alternatif Yöntem

Since the equation must hold for all values of xx if it has infinitely many solutions, you can substitute convenient values for xx to solve for aa and bb directly. Substituting x=0x = 0 simplifies the equation to 23a6=8-\frac{2}{3}a - 6 = -8, which quickly yields a=3a = 3. Then, substituting x=2x = 2 and a=3a = 3 simplifies the equation to 34(82b)=3-\frac{3}{4}(8 - 2b) = -3, which yields b=2b = 2. Calculating aba^b gives 32=93^2 = 9.
Tahmini Süre:3m 0s
Soru 58Soru

If 53(3x6)12(4x+8)=2\frac{5}{3}(3x - 6) - \frac{1}{2}(4x + 8) = 2, what is the value of 3x53x - 5?

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Cevap: 11

Cevap

11
Distributing the fraction 53\frac{5}{3} to (3x6)(3x - 6) yields 5x105x - 10. Distributing the term 12-\frac{1}{2} to (4x+8)(4x + 8) yields 2x4-2x - 4. Combining these results gives (5x2x)+(104)=2(5x - 2x) + (-10 - 4) = 2, which simplifies to 3x14=23x - 14 = 2. Adding 14 to both sides yields 3x=163x = 16. Finally, subtracting 5 from both sides of this equation gives 3x5=113x - 5 = 11.

Adım Adım Çözüm

1
Distribute 53\frac{5}{3} and 12-\frac{1}{2} to the terms within their respective parentheses.
5x102x4=25x - 10 - 2x - 4 = 2
This removes the parentheses and allows us to group terms.
2
Combine the variable terms and the constant terms on the left side of the equation.
3x14=23x - 14 = 2
Simplifying the expression on the left side makes it easier to solve.
3
Add 14 to both sides of the equation to isolate the term 3x3x.
3x=163x = 16
Since the question asks for the value of 3x53x - 5, isolating 3x3x allows direct evaluation.
4
Subtract 5 from both sides of the equation 3x=163x = 16.
3x5=113x - 5 = 11
This directly yields the required value without needing to compute the fractional value of xx first.

Anahtar Kavram

Solving linear equations in one variable by distribution, combining like terms, and evaluating expressions.
Tahmini Süre:1m 30s
Soru 59Soru

In the xyxy-plane, the graph of a linear function ff has a positive yy-intercept and a positive xx-intercept. The area of the triangular region in the first quadrant bounded by the graph of ff and the coordinate axes is 3636. If the graph of ff passes through the point (4,4)(4, 4) and has a slope less than 1-1, what is the yy-coordinate of the yy-intercept of the graph of ff?

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Cevap: 12

Cevap

The yy-coordinate of the yy-intercept of the graph of ff is 1212.
The correct answer is 1212 because the system of equations derived from the area constraint (qr=72qr = 72) and the point constraint (q+r=18q+r=18) yields two possible values for the yy-intercept: 66 or 1212. The condition that the slope must be less than 1-1 means that the line must be steeper than a slope of 1-1, which requires the yy-intercept to be larger than the xx-intercept (r>qr > q). Thus, the yy-intercept is 1212.

Adım Adım Çözüm

1
Express the equation of the line using intercept form.
xq+yr=1\frac{x}{q} + \frac{y}{r} = 1, where q>0q > 0 is the xx-intercept and r>0r > 0 is the yy-intercept.
Since the line intersects the positive axes, this form directly relates the intercepts to the coordinates of points on the line.
2
Use the area of the triangle to find a relationship between qq and rr.
qr=72qr = 72
The area of the right triangle formed by the axes and the intercepts is given by Area=12×base×height=12qr=36\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}qr = 36.
3
Substitute the point (4,4)(4, 4) into the equation of the line and simplify using the area relationship.
q+r=18q + r = 18
Plugging in x=4x = 4 and y=4y = 4 yields 4q+4r=1    4(q+r)=qr\frac{4}{q} + \frac{4}{r} = 1 \implies 4(q + r) = qr. Substituting qr=72qr = 72 gives 4(q+r)=72    q+r=184(q + r) = 72 \implies q + r = 18.
4
Solve the system of equations q+r=18q + r = 18 and qr=72qr = 72.
(q,r)=(6,12)(q, r) = (6, 12) or (q,r)=(12,6)(q, r) = (12, 6)
Substituting r=18qr = 18 - q into qr=72qr = 72 yields q(18q)=72    q218q+72=0    (q6)(q12)=0q(18 - q) = 72 \implies q^2 - 18q + 72 = 0 \implies (q - 6)(q - 12) = 0.
5
Apply the slope condition to determine the unique value of rr.
r=12r = 12
The slope of the line is m=rqm = -\frac{r}{q}. If q=12q = 12 and r=6r = 6, then m=0.5m = -0.5, which is not less than 1-1. If q=6q = 6 and r=12r = 12, then m=2m = -2, which is less than 1-1. Thus, r=12r = 12 is the correct yy-intercept.

Anahtar Kavram

Using intercepts and area to determine the equation of a linear function under constraints.
Soru 60Soru

The table below shows some values of xx and the corresponding values of f(x)f(x) for a linear function ff.

xxf(x)f(x)
37
612
917

What is the value of f(15)f(15)?

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Cevap: 27

Cevap

The value of the linear function evaluated at 15 is 27.
To find f(15)f(15) for the linear function, we determine its constant rate of change. Using points (3,7)(3, 7) and (6,12)(6, 12), the change in f(x)f(x) is 127=512 - 7 = 5 for a change in xx of 63=36 - 3 = 3. This gives a slope of 53\frac{5}{3}. Using point-slope form with (3,7)(3, 7) gives the equation f(x)7=53(x3)f(x) - 7 = \frac{5}{3}(x - 3), which simplifies to f(x)=53x+2f(x) = \frac{5}{3}x + 2. Substituting x=15x = 15 yields f(15)=53(15)+2=25+2=27f(15) = \frac{5}{3}(15) + 2 = 25 + 2 = 27. Alternatively, since xx increases by 6 from 9 to 15, which is twice the step size of 3, the value of f(x)f(x) must increase by 2×5=102 \times 5 = 10 from 1717, resulting in 17+10=2717 + 10 = 27.

Adım Adım Çözüm

1
Calculate the slope of the linear function
slope m=53m = \frac{5}{3}
Linear functions have a constant rate of change, which is the slope.
2
Determine the equation of the function
f(x)=53x+2f(x) = \frac{5}{3}x + 2
Using the slope and one point from the table helps define the function for all inputs.
3
Evaluate the function at x=15x = 15
f(15)=27f(15) = 27
Substitute the given input value into the function equation to find the corresponding output value.

Anahtar Kavram

Linear Functions and Graphs
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Tüm alıştırma soruları — SAT | Examkin