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SAT Math : SAT Mathematics

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Example Questions

Example Question #43 : Evaluating Expressions

represents a positive quantity and represents a negative quantity.

$x^{2} = 60$

$y^{2} = 15$

Evaluate $x^{3}+3x^{2}y+3xy^{2}+ y^{3}$ .

Possible Answers:

$-15\sqrt{15}$

$15\sqrt{15}$

$-405\sqrt{15}$

$405\sqrt{15}$

Correct answer:

$15\sqrt{15}$

Explanation:

$x^{3}+3x^{2}y+3xy^{2}+ y^{3}$ can be recognized as the cube of the binomial x+y , so

$x^{3}+3x^{2}y+3xy^{2}+ y^{3} = (x+y)^{3}$

$x^{2} = 60$ ,

so, taking the square root of both sides,

$x = \pm \sqrt{60}$

By the Product of Radicals rule, simplify:

$x = \pm \sqrt{4} \cdot \sqrt{15}$

$= \pm 2\sqrt{15}$

Since represents a positive quantity, we choose

$x = 2\sqrt{15}$

$y^{2} = 15$ ,

so, taking the square root of both sides,

$y = \pm \sqrt{15}$

Since represents a negative quantity.

$y = - \sqrt{15}$

Substituting:

$x^{3}+3x^{2}y+3xy^{2}+ y^{3}$

$= (x+y)^{3}$

$=\left [ (2 \sqrt{15}) + \left ( -\sqrt{15}\right ) \right ] ^{3}$

$= ( \sqrt{15}) ^{3}$

$= ( \sqrt{15}) ^{2} \cdot \sqrt{15}$

$= 15\sqrt{15}$

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Example Question #2531 : Sat Mathematics

and both represent positive quantities.

$x^{2} = 48$

$y^{2} = 12$

Evaluate $x^{3}+ y^{3}$ .

Possible Answers:

$72\sqrt{3}$

$-216\sqrt{3}$

$216\sqrt{3}$

$-72\sqrt{3}$

Correct answer:

$216\sqrt{3}$

Explanation:

One way to evaluate $x^{3}+ y^{3}$ is to note that, as the sum of cubes, this can be factored as

$x^{3}+ y^{3} = (x+ y )(x^{2}-xy+y^{2})$

$x^{2} = 48$ ,

and is positive, so, using the Product of Radicals Rule,

$x = \sqrt{48} = \sqrt{16} \cdot \sqrt{3} = 4 \sqrt{3}$

$y^{2} = 12$ , and is positive, so, similarly,

$y = \sqrt{1 2}= \sqrt{4} \cdot \sqrt{3} = 2 \sqrt{3}$

$xy = 4 \sqrt{3} \cdot 2 \sqrt{3} = 4 \cdot 2 \cdot \sqrt{3} \cdot \sqrt{3} = 4 \cdot 2 \cdot 3 = 24$

Therefore, substituting for $x, y,xy, x^{2}, y^{2}$ in the factored expression:

$x^{3}+ y^{3}$

$= (x+ y )(x^{2}-xy+y^{2})$

$= ( 4 \sqrt{3} +2 \sqrt{3})(48-24+12)$

$= (6\sqrt{3})(36)$

$= 216\sqrt{3}$

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Example Question #2532 : Sat Mathematics

High and low 2

The above double line graph gives the high and low temperatures, in degrees Celsius, for each day in a given week in Washington City. Temperatures given in terms of the Celsius scale can be converted to degrees Farhrenheit using this formula:

F = 1.8C + 32

where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively.

In degrees Fahrenheit, what was the low temperature for Tuesday, June 9 (nearest whole degree?)

Possible Answers:

$72^{\circ } F$

$79^{\circ } F$

$85^{\circ } F$

$59^{\circ } F$

$66^{\circ } F$

Correct answer:

$66^{\circ } F$

Explanation:

The low temperature for Tuesday was $19^{\circ } C$ . This can be converted to degrees Fahrenheit by setting C = 19 in the formula and evaluating :

F = 1.8C + 32

$= 1.8 \cdot 19 + 32$

= 34.2+ 32

=66.2

This rounds to $66 ^{\circ } F$ .

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Example Question #303 : New Sat

If Sandy is running at a pace of $\frac{12\: \text{meters}}{\text{sec}}$ , find how fast sandy is running in $\frac{\text{meters}}{\text{year}}$ .

Possible Answers:

$\frac{3784320\:\text{meters}}{\text{year}}$

$\frac{37843\:\text{meters}}{\text{year}}$

$\frac{12\:\text{meters}}{\text{year}}$

$\frac{2000\:\text{meters}}{\text{year}}$

$\frac{378432000\:\text{meters}}{\text{year}}$

Correct answer:

$\frac{378432000\:\text{meters}}{\text{year}}$

Explanation:

To convert into $\frac{\text{meters}}{\text{year}}$ , we will do the following conversions

$\frac{12\: \text{meters}}{\text{sec}}\cdot\frac{60\:\text{sec}}{\text{min}}\cdot\frac{60\:\text{min}}{\text{hour}}\cdot\frac{24\:\text{hour}}{\text{day}}\cdot\frac{365\:\text{days}}{\text{year}}=\frac{378432000\:\text{meters}}{\text{year}}$

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Example Question #2531 : Sat Mathematics

$\begin{tabular}{cccc} x & \vline & y\\ \hline 0 & \vline & 2\\ 1& \vline &7 \\ 2 & \vline & 12\\ 3 & \vline & 17 \end{tabular}$

Find the equation of a line that fits the above data.

Possible Answers:

y=5x+2

y=-5x+2

y=5x

y=2x+5

y=5x-2

Correct answer:

y=5x+2

Explanation:

We can use point slope form to determine the equation of a line that fits the data.

Point slope form is y-y_0=m(x-x_0) , where (x_0, y_0) , and is the slope, where $m=\frac{y_2-y_1}{x_2-x_1}$ .

Let y_2=17 , y_1=12 , x_2=3 , and x_1=2 .

$m=\frac{17-12}{3-2}=5$

If we do this for every other point, we will see that they have the same slope of .

y-y_0=5(x-x_0)

Now let x_0=0 , and y_0=2 .

y-2=5(x-0)

y-2=5x

y=5x+2

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Example Question #2532 : Sat Mathematics

The equation for the universal gravitation is $F_g=\frac{Gm_1m_2}{r^2}$ , $m_1=\text{mass of object 1}$ , $m_2=\text{mass of object 2}$ , $r=\text{radius between masses}$ , and is the universal gravitational constant. If $m_1=10 \ kg$ , $m_2=20 \ kg$ , $r=100\ m$ , and $G=6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2}$ , what is the Force equal to? Round to the nearest tenth.

Hint: $\frac{\text{kg\:m}}{s^2}=1 \text{Newton}$

Possible Answers:

$F_g=1.3\times10^{-9}N$

$F_g=1.3\times10^{-12} N$

$F_g=1.3\times10^{-10}N$

$F_g=1.3\times10^{-11}N$

$F_g=1.3\times10^{-8}N$

Correct answer:

$F_g=1.3\times10^{-12} N$

Explanation:

All we need to do is plug in the values into the equation.

$F_g=\frac{Gm_1m_2}{r^2}$

$F_g=\frac{(6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2})10\: kg\: 20\: kg}{(100m)^2}$

$F_g=\frac{(6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2})200\: kg^2}{10000m^2}$

$F_g=\frac{(6.67408 \times 10^{-11} m^3 kg^{-1} s^{-2})1\: kg^2}{50m^2}$

$F_g=1.3\times10^{-12}\frac{\text{kg m}}{s^2}$

$F_g=1.3\times10^{-12}N$

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Example Question #2533 : Sat Mathematics

Given a right triangle $\Delta ABC$ whose $m\angle C=90^\circ$ and $\sin B=$ $\frac{2}{3}$ , find $\cos A$ .

Possible Answers:

$\frac{3}{2}$

$\frac{3}{4}$

$\frac{2}{3}$

$72^\circ$

Correct answer:

$\frac{2}{3}$

Explanation:

To solve for $\cos A$ first identify what is known.

The question states that $\Delta ABC$ is a right triangle whose $m\angle C=90^\circ$ and $\sin B=$ $\frac{2}{3}$ . It is important to recall that any triangle has a sum of interior angles that equals 180 degrees.

Therefore, to calculate $\cos A$ use the complimentary angles identity of trigonometric functions.

$\cos A=\sin \left ( 90-A\right )$

and since B=90-A , then

$\cos A=\sin B$

$\cos A=\frac{2}{3}$

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Example Question #71 : Expressions

$(x+1) ^{2} = t$ . is a positive number.

In terms of , which of the following is equal to $(x-8)^{2}$ ?

Possible Answers:

t - 81

$t - 18\sqrt{ t}+ 81$

t + 81

None of these

$t + 18\sqrt{ t}+ 81$

Correct answer:

$t - 18\sqrt{ t}+ 81$

Explanation:

$(x+1) ^{2} = t$ , so, taking the square root of both sides:

$x+1 = \pm\sqrt{ t}$

is positive, so x+1 is as well; consequently,

$x+1 = \sqrt{ t}$

Subtracting 1 from both sides:

$x+1- 1 = \sqrt{ t} - 1$

$x = \sqrt{ t} - 1$

Subtracting 8 from both sides:

$x-8 = \sqrt{ t} - 1 -8$

$x-8 = \sqrt{ t}- 9$

Squaring both sides, and applying the binomial square pattern to the right expression:

$x-8 =( \sqrt{ t}- 9)^{2}$

$= (\sqrt{ t})^{2} - 2\cdot 9 \cdot (\sqrt{ t})+ 9^{2}$

$=t - 18\sqrt{ t}+ 81$

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Example Question #1 : Simplifying Expressions

If x + y = 4, what is the value of x + y – 6?

Possible Answers:

–2

Correct answer:

–2

Explanation:

Substitute 4 for x + y in the expression given.

4 minus 6 equals –2.

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Example Question #1 : Simplifying Expressions

If 6 less than the product of 9 and a number is equal to 48, what is the number?

Possible Answers:

Correct answer:

Explanation:

Write an equation for the written expression: 9x – 6 = 48. When we solve for x we get x = 6.

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SAT Math : SAT Mathematics

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #43 : Evaluating Expressions

Example Question #2531 : Sat Mathematics

Example Question #2532 : Sat Mathematics

Example Question #303 : New Sat

Example Question #2531 : Sat Mathematics

Example Question #2532 : Sat Mathematics

Example Question #2533 : Sat Mathematics

Example Question #71 : Expressions

Example Question #1 : Simplifying Expressions

Example Question #1 : Simplifying Expressions

All SAT Math Resources

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