Algebra - PSAT Math
Card 0 of 5075
(√(8) / -x ) < 2. Which of the following values could be x?
(√(8) / -x ) < 2. Which of the following values could be x?
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The equation simplifies to x > -1.41. -1 is the answer.
The equation simplifies to x > -1.41. -1 is the answer.
If f(x) = 5x – 10, then what is the value of 5(f(10)) – 10?
If f(x) = 5x – 10, then what is the value of 5(f(10)) – 10?
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The first step is to find what f(10) equals, so f(10)=5(10) – 10 = 40. Then substitute 40 into the second equation: 5(40) – 10 = 200 – 10 = 190.
190 is the correct answer
The first step is to find what f(10) equals, so f(10)=5(10) – 10 = 40. Then substitute 40 into the second equation: 5(40) – 10 = 200 – 10 = 190.
190 is the correct answer
f(x) = 0.1x + 7
g(x) = 1000x + 4
What is g(f(100))?
f(x) = 0.1x + 7
g(x) = 1000x + 4
What is g(f(100))?
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First find the value of f(100) = 0.1(100) + 7 = 10 + 7 = 17
Then find g(17) = 1000(17) + 4 = 17000 + 4 = 17004.
First find the value of f(100) = 0.1(100) + 7 = 10 + 7 = 17
Then find g(17) = 1000(17) + 4 = 17000 + 4 = 17004.
Half of one hundred divided by five and multiplied by one-tenth is .
Half of one hundred divided by five and multiplied by one-tenth is .
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Let's take this step by step. "Half of one hundred" is 100/2 = 50. Then "half of one hundred divided by five" is 50/5 = 10. "Multiplied by one-tenth" really is the same as dividing by ten, so the last step gives us 10/10 = 1.
Let's take this step by step. "Half of one hundred" is 100/2 = 50. Then "half of one hundred divided by five" is 50/5 = 10. "Multiplied by one-tenth" really is the same as dividing by ten, so the last step gives us 10/10 = 1.
Let x&y be defined as (x – y)xy . What is the value of –1_&_2?
Let x&y be defined as (x – y)xy . What is the value of –1_&_2?
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We are told that x&y = (x – y)xy .
–1&2 = (–1 – 2)(–1)(2) = (–3)–2
To simplify this, we can make use of the property of exponents which states that a– b = 1/(ab ).
(–3)–2 = 1/(–3)2 = 1/9
The answer is 1/9.
We are told that x&y = (x – y)xy .
–1&2 = (–1 – 2)(–1)(2) = (–3)–2
To simplify this, we can make use of the property of exponents which states that a– b = 1/(ab ).
(–3)–2 = 1/(–3)2 = 1/9
The answer is 1/9.
The rate of a gym membership costs p dollars the first month and m dollars per month every month thereafter. Which of the following represents the total cost of the gym membership for n months, if n is a positive integer?
The rate of a gym membership costs p dollars the first month and m dollars per month every month thereafter. Which of the following represents the total cost of the gym membership for n months, if n is a positive integer?
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The one-time first-month cost is p, and the monthly cost is m, which gets multipled by every month but the first (of which there are n -1). The total cost is the first-month cost of p, plus the monthly cost for (i.e. times) n -1 months, which makes the total cost equal to p + m (n -1).
The one-time first-month cost is p, and the monthly cost is m, which gets multipled by every month but the first (of which there are n -1). The total cost is the first-month cost of p, plus the monthly cost for (i.e. times) n -1 months, which makes the total cost equal to p + m (n -1).
- If f(x) = (x + 4)/(x – 4) for all integers except x = 4, which of the following has the lowest value?
- If f(x) = (x + 4)/(x – 4) for all integers except x = 4, which of the following has the lowest value?
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Plug each value for x into the above equation and solve for f(x). f(1) provides the lowest value –5/3
Plug each value for x into the above equation and solve for f(x). f(1) provides the lowest value –5/3
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If n and p are positive and 100_n_3_p_-1 = 25_n_, what is n-2 in terms of p ?
If n and p are positive and 100_n_3_p_-1 = 25_n_, what is n-2 in terms of p ?
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To solve this problem, we look for an operation to perform on both sides that will leave n-2 by itself on one side. Dividing both sides by 25_n_-3 would leave n-2 by itself on the right side of the equqation, as shown below:
100n3p–1/25n–3 = 25n/25n–3
Remember that when dividing terms with the same base, we subtract the exponents, so the equation can be written as 100n0p–1/25 = n–2
Finally, we simplify to find 4_p–_1 = _n–_2.
To solve this problem, we look for an operation to perform on both sides that will leave n-2 by itself on one side. Dividing both sides by 25_n_-3 would leave n-2 by itself on the right side of the equqation, as shown below:
100n3p–1/25n–3 = 25n/25n–3
Remember that when dividing terms with the same base, we subtract the exponents, so the equation can be written as 100n0p–1/25 = n–2
Finally, we simplify to find 4_p–_1 = _n–_2.
If 7y = 4x - 12, then x =
If 7y = 4x - 12, then x =
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Adding 12 to both sides and dividing by 4 yields (7y+12)/4.
Adding 12 to both sides and dividing by 4 yields (7y+12)/4.
Which of the statements describes the solution set for **–**7(x + 3) = **–**7x + 20 ?
Which of the statements describes the solution set for **–**7(x + 3) = **–**7x + 20 ?
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By distribution we obtain **–**7x – 21 = – 7x + 20. This equation is never possibly true.
By distribution we obtain **–**7x – 21 = – 7x + 20. This equation is never possibly true.
54 / 25 =
54 / 25 =
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25 = 5 * 5 = 52. Then 54 / 25 = 54 / 52.
Now we can subtract the exponents because the operation is division. 54 / 52 = 54 – 2 = 52 = 25. The answer is therefore 25.
25 = 5 * 5 = 52. Then 54 / 25 = 54 / 52.
Now we can subtract the exponents because the operation is division. 54 / 52 = 54 – 2 = 52 = 25. The answer is therefore 25.
If a(x) = 2x3 + x, and b(x) = –2x, what is a(b(2))?
If a(x) = 2x3 + x, and b(x) = –2x, what is a(b(2))?
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When functions are set up within other functions like in this problem, the function closest to the given variable is performed first. The value obtained from this function is then plugged in as the variable in the outside function. Since b(x) = –2x, and x = 2, the value we obtain from b(x) is –4. We then plug this value in for x in the a(x) function. So a(x) then becomes 2(–43) + (–4), which equals –132.
When functions are set up within other functions like in this problem, the function closest to the given variable is performed first. The value obtained from this function is then plugged in as the variable in the outside function. Since b(x) = –2x, and x = 2, the value we obtain from b(x) is –4. We then plug this value in for x in the a(x) function. So a(x) then becomes 2(–43) + (–4), which equals –132.
Which of the following lists the above quantities from least to greatest?
Which of the following lists the above quantities from least to greatest?
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If p and q are positive integrers and 27p = 9q, then what is the value of q in terms of p?
If p and q are positive integrers and 27p = 9q, then what is the value of q in terms of p?
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The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 33p = 32q. So then 3p = 2q, and q = (3/2)p is our answer.
The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 33p = 32q. So then 3p = 2q, and q = (3/2)p is our answer.
Simplify 272/3.
Simplify 272/3.
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272/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations.
272/3 = (272)1/3 = 7291/3 OR
272/3 = (271/3)2 = 32
Obviously 32 is much easier. Either 32 or 7291/3 will give us the correct answer of 9, but with 32 it is readily apparent.
272/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations.
272/3 = (272)1/3 = 7291/3 OR
272/3 = (271/3)2 = 32
Obviously 32 is much easier. Either 32 or 7291/3 will give us the correct answer of 9, but with 32 it is readily apparent.
|12x + 3y| < 15
What is the range of values for y, expressed in terms of x?
|12x + 3y| < 15
What is the range of values for y, expressed in terms of x?
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Recall that with absolute values and "less than" inequalities, we have to hold the following:
12x + 3y < 15
AND
12x + 3y > –15
Otherwise written, this is:
–15 < 12x + 3y < 15
In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:
–15 – 12x < 3y < 15 – 12x
Now, we have to divide each element by 3:
(–15 – 12x)/3 < y < (15 – 12x)/3
This simplifies to:
–5 – 4x < y < 5 – 4x
Recall that with absolute values and "less than" inequalities, we have to hold the following:
12x + 3y < 15
AND
12x + 3y > –15
Otherwise written, this is:
–15 < 12x + 3y < 15
In this form, we can solve for y. First, we have to subtract x from all 3 parts of the inequality:
–15 – 12x < 3y < 15 – 12x
Now, we have to divide each element by 3:
(–15 – 12x)/3 < y < (15 – 12x)/3
This simplifies to:
–5 – 4x < y < 5 – 4x
|4x + 14| > 30
What is a possible valid value of x?
|4x + 14| > 30
What is a possible valid value of x?
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This inequality could be rewritten as:
4x + 14 > 30 OR 4x + 14 < –30
Solve each for x:
4x + 14 > 30; 4x > 16; x > 4
4x + 14 < –30; 4x < –44; x < –11
Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.
This inequality could be rewritten as:
4x + 14 > 30 OR 4x + 14 < –30
Solve each for x:
4x + 14 > 30; 4x > 16; x > 4
4x + 14 < –30; 4x < –44; x < –11
Therefore, anything between –11 and 4 (inclusive) will not work. Hence, the answer is 7.
Given the inequality, |2_x_ – 2| > 20,
what is a possible value for x?
Given the inequality, |2_x_ – 2| > 20,
what is a possible value for x?
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For this problem, we must take into account the absolute value.
First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).
First step:
2_x_ – 2 > 20
2_x_ > 22
x > 11
Second step:
2_x_ – 2 < –20
2_x_ < –18
x < –9
Therefore, x > 11 and x < –9.
A possible value for x would be –10 since that is less than –9.
Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.
For this problem, we must take into account the absolute value.
First, we solve for 2_x_ – 2 > 20. But we must also solve for 2_x_ – 2 < –20 (please notice that we negate 20 and we also flip the inequality sign).
First step:
2_x_ – 2 > 20
2_x_ > 22
x > 11
Second step:
2_x_ – 2 < –20
2_x_ < –18
x < –9
Therefore, x > 11 and x < –9.
A possible value for x would be –10 since that is less than –9.
Note: the value 11 would not be a possible value for x because the inequality sign given does not include an equal sign.
x f(x) g(x) 9 4 0 10 6 1 11 9 0 12 13 –1
According to the figure above, what is the value of g(12) – √f(9)?
| x | f(x) | g(x) |
|---|---|---|
| 9 | 4 | 0 |
| 10 | 6 | 1 |
| 11 | 9 | 0 |
| 12 | 13 | –1 |
According to the figure above, what is the value of g(12) – √f(9)?
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For this question, we "plug in" the value of x given, which is inside the parentheses, and follow along the table to see what value the f or g functions output. For g(12), the output value is –1, while for f(9), the output value is 4 (be careful not to reverse these!) Thus, we can plug into the equation given:
(–1) – √4) = –1 – 2 = –3.
For this question, we "plug in" the value of x given, which is inside the parentheses, and follow along the table to see what value the f or g functions output. For g(12), the output value is –1, while for f(9), the output value is 4 (be careful not to reverse these!) Thus, we can plug into the equation given:
(–1) – √4) = –1 – 2 = –3.