Variables - SAT Math

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Question

Subtract from .

Answer

Step 1: We need to read the question carefully. It says subtract from. When you see the word "from", you read the question right to left.

I am subtracting the left equation from the right equation.

Step 2: We need to write the equation on the right minus the equation of the left.

Step 3: Distribute the minus sign in front of the parentheses:

Step 4: Combine like terms:

Step 5: Put all the terms together, starting with highest degree. The degree of the terms is the exponent. Here, the highest degree is 2 and lowest is zero.

The final equation is

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Question

Define an operation $\heartsuit$ on the set of real numbers as follows:

For all real ,

$m \heartsuit n = \left ( m-4n\right )^{3}$

How else could this operation be defined?

Answer

$\left ( m-4n\right )^{3}$ , as the cube of a binomial, can be rewritten using the following pattern:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot m ^{2} \cdot 4n + 3 \cdot m \cdot (4n) ^{2}- (4n) ^{3}$

Applying the rules of exponents to simplify this:

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot m \cdot 4 ^{2}\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} - 3 \cdot 4 \cdot m ^{2} \cdot n + 3 \cdot 16 \cdot m\cdot n ^{2}- 4^{3} \cdot n ^{3}$

$\left ( m-4n\right )^{3} = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$

Therefore, the correct choice is that, alternatively stated,

$m \heartsuit n = m ^{3} -12 m ^{2} n + 48mn ^{2}- 64n ^{3}$ .

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Question

Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

Answer

Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.

There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.

$(60\frac{min}{hr})(2.5hr)=150min$

If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.

$(y\frac{ft^2}{min})(150min)=150y\ ft^2$

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Question

The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?

Answer

Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:

y = kx, where k is a constant.

Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:

y = _kx_2_z_3, where k is a constant.

The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.

24 = k(1)2(2)3 = k(1)(8) = 8_k_

24 = 8_k_

Divide both sides by 8.

3 = k

k = 3

Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.

y = _kx_2_z_3

y = 3(3)2(1)3 = 3(9)(1) = 27

y = 27

The answer is 27.

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Question

In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?

Answer

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be .

Alternatively, you can evaluate for each consecutive week.

Week 1:

Week 2:

Week 3:

Week 4:

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Question

and are the radius and volume, respectively, of a given sphere.

$r ^{6} = Q$

$W = \sqrt[3]{V}$ .

Which of the following is a true statement?

Answer

The volume of a sphere can be calculated from its radius as follows:

$V = \frac{4}{3} \pi r^{3}$

Therefore, squaring both sides, we get

$V ^{2}=\left ( \frac{4}{3} \pi r^{3} \right )^{2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{3\cdot 2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W = \sqrt[3]{V}$

$W ^{6 }= \left (\sqrt[3]{V} \right ) ^{6}$

$W ^{6 }= \left ( V ^{\frac{1}{3}}\right ) ^{6} = V ^{\frac{1}{3} \cdot 6} = V^{2}$

Substituting:

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W ^{6 }=\left ( \frac{4}{3} \pi \right )^{2}Q$

$\frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }= \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}} \cdot \left ( \frac{4}{3} \pi \right )^{2}Q$

$Q = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }$

If we let the constant of variation be $K = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}$ , we see that

$Q = K W ^{6 }$ ,

and varies directly as $W^{6}$ , the sixth power of .

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Question

and are the diameter and circumference, respectively, of the same circle.

$d = 12L^{2}$

$C = t^{2}$

Which of the following is a true statement? (Assume all quantities are positive)

Answer

If and are the diameter and circumference, respectively, of the same circle, then

$C = \pi d$ .

By substitution,

$t^{2} = \pi \cdot 12 L^{2}$

$t^{2} = 12 \pi L^{2}$

Taking the square root of both sides:

$\sqrt{t^{2} }= \sqrt{12 \pi L^{2}}$

$t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}$

$t = \sqrt{12 \pi} \cdot L$

Taking $K= \sqrt{12 \pi}$ as the constant of variation, we get

,

meaning that varies directly as .

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Question

is the radius of the base of a cone; is its height; is its volume.

$n = 4r = h^{2}$ ; $m = V ^{2}$ .

Which of the following is a true statement?

Answer

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

$V = \frac{1}{3} \pi r^{2}h$

, so $r = \frac{n}{4}$ .

$n = h^{2}$ , so $h = \sqrt{n}$ .

$V = \frac{1}{3} \pi r^{2}h$ , so by substitution,

$V = \frac{1}{3} \pi\left ( \frac{n}{4} \right ) ^{2} \cdot \sqrt{n}$

$V = \frac{1}{3} \pi\left ( \frac{n^{2}}{16} \right ) \cdot \sqrt{n}$

$V = \frac{1}{48} \pi \cdot n^{2}} \sqrt{n}$

Square both sides:

$V ^{2 }=\left ( \frac{1}{48} \pi \cdot n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \right )^{2} \cdot \left ( \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{4}} \cdot n$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{5}}$

If we take $K = \left ( \frac{1}{48} \pi \right )^{2}$ as the constant of variation, then

$m=K n^{5}}$ ,

and varies directly as the fifth power of .

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Question

The temperature at the surface of the ocean is $25^{\circ}C$ . At meters below the surface, the ocean temperature is $5^{\circ}C$ . By how much does the temperature decrease for every meters below the ocean's surface?

Answer

This may seem confusing, but is pretty straightforward.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{5-25}{-2500-0}$

$\text{Slope}=\frac{-20}{-2500}=\frac{125}{1}$

Thus, for every 125 meters below the surface, the temperature decreases by one degree.

To find how much it decreases with every 100 meters, we need to do the following:

$\frac{100}{125}=\frac{4}{5}$

Thus, the temperature decreases by $\frac{4}{5}^{\circ}$ every 100 meters.

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Question

Phillip can paint square feet of wall per minute. What area of the wall can he paint in 2.5 hours?

Answer

Every minute Phillip completes another _ square feet of painting. To solve for the total area that he completes, we need to find the number of minutes that he works.

There are 60 minutes in an hour, and he paints for 2.5 hours. Multiply to find the total number of minutes.

$(60\frac{min}{hr})(2.5hr)=150min$

If he completes _ square feet per minute, then we can multiply _ by the total minutes to find the final answer.

$(y\frac{ft^2}{min})(150min)=150y\ ft^2$

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Question

The value of varies directly with the square of __and the cube of . If when and , then what is the value of _ when and ?

Answer

Let's consider the general case when y varies directly with x. If y varies directly with x, then we can express their relationship to one another using the following formula:

y = kx, where k is a constant.

Therefore, if y varies directly as the square of x and the cube of z, we can write the following analagous equation:

y = _kx_2_z_3, where k is a constant.

The problem states that y = 24 when x = 1 and z = 2. We can use this information to solve for k by substituting the known values for y, x, and z.

24 = k(1)2(2)3 = k(1)(8) = 8_k_

24 = 8_k_

Divide both sides by 8.

3 = k

k = 3

Now that we have k, we can find y if we know x and z. The problem asks us to find y when x = 3 and z = 1. We will use our formula for direct variation again, this time substitute values for k, x, and z.

y = _kx_2_z_3

y = 3(3)2(1)3 = 3(9)(1) = 27

y = 27

The answer is 27.

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Question

In a growth period, a population of flies triples every week. If the original population had 3 flies, how big is the population after 4 weeks?

Answer

We know that the initial population is 3, and that every week the population will triple.

The equation to model this growth will be , where is the initial size, is the rate of growth, and is the time.

In this case, the equation will be .

Alternatively, you can evaluate for each consecutive week.

Week 1:

Week 2:

Week 3:

Week 4:

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Question

and are the radius and volume, respectively, of a given sphere.

$r ^{6} = Q$

$W = \sqrt[3]{V}$ .

Which of the following is a true statement?

Answer

The volume of a sphere can be calculated from its radius as follows:

$V = \frac{4}{3} \pi r^{3}$

Therefore, squaring both sides, we get

$V ^{2}=\left ( \frac{4}{3} \pi r^{3} \right )^{2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{3\cdot 2}$

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W = \sqrt[3]{V}$

$W ^{6 }= \left (\sqrt[3]{V} \right ) ^{6}$

$W ^{6 }= \left ( V ^{\frac{1}{3}}\right ) ^{6} = V ^{\frac{1}{3} \cdot 6} = V^{2}$

Substituting:

$V ^{2}=\left ( \frac{4}{3} \pi \right )^{2}r^{6}$

$W ^{6 }=\left ( \frac{4}{3} \pi \right )^{2}Q$

$\frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }= \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}} \cdot \left ( \frac{4}{3} \pi \right )^{2}Q$

$Q = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}W ^{6 }$

If we let the constant of variation be $K = \frac{1}{\left ( \frac{4}{3} \pi \right )^{2}}$ , we see that

$Q = K W ^{6 }$ ,

and varies directly as $W^{6}$ , the sixth power of .

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Question

and are the diameter and circumference, respectively, of the same circle.

$d = 12L^{2}$

$C = t^{2}$

Which of the following is a true statement? (Assume all quantities are positive)

Answer

If and are the diameter and circumference, respectively, of the same circle, then

$C = \pi d$ .

By substitution,

$t^{2} = \pi \cdot 12 L^{2}$

$t^{2} = 12 \pi L^{2}$

Taking the square root of both sides:

$\sqrt{t^{2} }= \sqrt{12 \pi L^{2}}$

$t = \sqrt{12 \pi} \cdot \sqrt{L^{2}}$

$t = \sqrt{12 \pi} \cdot L$

Taking $K= \sqrt{12 \pi}$ as the constant of variation, we get

,

meaning that varies directly as .

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Question

is the radius of the base of a cone; is its height; is its volume.

$n = 4r = h^{2}$ ; $m = V ^{2}$ .

Which of the following is a true statement?

Answer

The volume of a cone can be calculated from the radius of its base , and the height , using the formula

$V = \frac{1}{3} \pi r^{2}h$

, so $r = \frac{n}{4}$ .

$n = h^{2}$ , so $h = \sqrt{n}$ .

$V = \frac{1}{3} \pi r^{2}h$ , so by substitution,

$V = \frac{1}{3} \pi\left ( \frac{n}{4} \right ) ^{2} \cdot \sqrt{n}$

$V = \frac{1}{3} \pi\left ( \frac{n^{2}}{16} \right ) \cdot \sqrt{n}$

$V = \frac{1}{48} \pi \cdot n^{2}} \sqrt{n}$

Square both sides:

$V ^{2 }=\left ( \frac{1}{48} \pi \cdot n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot \left ( n^{2}} \right )^{2} \cdot \left ( \sqrt{n} \right )^{2}$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{4}} \cdot n$

$m=\left ( \frac{1}{48} \pi \right )^{2} \cdot n^{5}}$

If we take $K = \left ( \frac{1}{48} \pi \right )^{2}$ as the constant of variation, then

$m=K n^{5}}$ ,

and varies directly as the fifth power of .

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Question

The temperature at the surface of the ocean is $25^{\circ}C$ . At meters below the surface, the ocean temperature is $5^{\circ}C$ . By how much does the temperature decrease for every meters below the ocean's surface?

Answer

This may seem confusing, but is pretty straightforward.

$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\text{Slope}=\frac{5-25}{-2500-0}$

$\text{Slope}=\frac{-20}{-2500}=\frac{125}{1}$

Thus, for every 125 meters below the surface, the temperature decreases by one degree.

To find how much it decreases with every 100 meters, we need to do the following:

$\frac{100}{125}=\frac{4}{5}$

Thus, the temperature decreases by $\frac{4}{5}^{\circ}$ every 100 meters.

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Question

Factor the following variable

(x2 + 18x + 72)

Answer

You need to find two numbers that multiply to give 72 and add up to give 18

easiest way: write the multiples of 72:

1, 72

2, 36

3, 24

4, 18

6, 12: these add up to 18

(x + 6)(x + 12)

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Question

When is factored, it can be written in the form , where , , , , , and are all integer constants, and .

What is the value of ?

Answer

Let's try to factor x2 – y2 – z2 + 2yz.

Notice that the last three terms are very close to y2 + z2 – 2yz, which, if we rearranged them, would become y2 – 2yz+ z2. We could factor y2 – 2yz+ z2 as (y – z)2, using the general rule that p2 – 2pq + q2 = (p – q)2 .

So we want to rearrange the last three terms. Let's group them together first.

x2 + (–y2 – z2 + 2yz)

If we were to factor out a –1 from the last three terms, we would have the following:

x2 – (y2 + z2 – 2yz)

Now we can replace y2 + z2 – 2yz with (y – z)2.

x2 – (y – z)2

This expression is actually a differences of squares. In general, we can factor p2 – q2 as (p – q)(p + q). In this case, we can substitute x for p and (y – z) for q.

x2 – (y – z)2 = (x – (y – z))(x + (y – z))

Now, let's distribute the negative one in the trinomial x – (y – z)

(x – (y – z))(x + (y – z))

(x – y + z)(x + y – z)

The problem said that factoring x2 – y2 – z2 + 2yz would result in two polynomials in the form (ax + by + cz)(dx + ey + fz), where a, b, c, d, e, and f were all integers, and a > 0.

(x – y + z)(x + y – z) fits this form. This means that a = 1, b = –1, c = 1, d = 1, e = 1, and f = –1. The sum of all of these is 2.

The answer is 2.

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Question

Factor 9_x_2 + 12_x_ + 4.

Answer

Nothing common cancels at the beginning. To factor this, we need to find two numbers that multiply to 9 * 4 = 36 and sum to 12. 6 and 6 work.

So 9_x_2 + 12_x_ + 4 = 9_x_2 + 6_x_ + 6_x_ + 4

Let's look at the first two terms and last two terms separately to begin with. 9_x_2 + 6_x_ can be simplified to 3_x_(3_x_ + 2) and 6_x_ + 4 can be simplified into 2(3_x_ + 2). Putting these together gets us

9_x_2 + 12_x_ + 4

= 9_x_2 + 6_x_ + 6_x_ + 4

= 3_x_(3_x_ + 2) + 2(3_x_ + 2)

= (3_x_ + 2)(3_x_ + 2)

This is as far as we can factor.

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Question

Factor and simplify:

$\frac{64y^{2} - 16}{8y + 4}$

Answer

$64y^{2} - 16$ is a difference of squares.

The difference of squares formula is $a^{2} - b^{2} = (a - b)(a + b)$ .

Therefore, $\frac{64y^{2} - 16}{8y + 4}$ = $\frac{(8y + 4)(8y - 4)}{8y + 4} = 8y - 4$ .

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