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SAT Math : Expressions

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Example Question #61 : Expressions

High and low 2

Above is a graph which gives the high and low temperatures, in degrees Celsius, over a one week period for Washington City. Temperature given in degrees Celsius can be converted to the Fahrenheit scale using the following formula, where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively:

F =1.8 C + 32

In degrees Fahrenheit, what was the highest temperature of the week shown (Nearest whole degree)?

Possible Answers:

$69 ^{\circ }F$

$84^{\circ }F$

$64 ^{\circ }F$

$74 ^{\circ }F$

$79 ^{\circ }F$

Correct answer:

$79 ^{\circ }F$

Explanation:

The highest temperature during the week was $26 ^{\circ } C$ , which, as can be seen in the graph, was reached on Tuesday and Friday. Setting C= 26 in the given formula:

F =1.8 C + 32

$F =1.8 \cdot 26 + 32$

=46.8 + 32

= 78.8

Rounded to the nearest whole number, this is $79 ^{\circ }F$ .

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Example Question #62 : Expressions

High and low 2

Above is a graph which gives the high and low temperatures, in degrees Fahrenheit, over a one week period for Washington City. Temperature given in degrees Fahrenheit can be converted to the Celsius scale using the following formula, where and are the temperature expressed in degrees Celsius and degrees Fahrenheit, respectively:

$C = \frac{5}{9} (F-32)$ .

In degrees Celsius, what was the high temperature on Thursday, June 11 (nearest whole degree)?

Possible Answers:

$0^{\circ } C$

$10^{\circ } C$

$5^{\circ } C$

$-5^{\circ } C$

$-10^{\circ } C$

Correct answer:

$-5^{\circ } C$

Explanation:

The high temperature for Thursday was $23 ^{\circ }$ Fahrenheit. Convert this to the Celsius temperature scale by setting F= 23 in the given formula and evaluating:

$C = \frac{5}{9} (F-32)$

$C = \frac{5}{9} (23-32)$

$C = \frac{5}{9} (-9)$

C = -5

In degrees Celsius, Thursday's high temperature was $-5^{\circ } C$ .

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Example Question #63 : Expressions

represents a positive quantity and represents a negative quantity.

$x^{2} = 99$

$y^{2} = 11$

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

Possible Answers:

$88 \sqrt{11}$

$-704 \sqrt{11}$

$-88 \sqrt{11}$

$704 \sqrt{11}$

None of the other choices gives the correct response.

Correct answer:

$704 \sqrt{11}$

Explanation:

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

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

$x^{2} = 99$ ,

so, taking the square root of both sides,

$x = \pm \sqrt{99}$

By the Product of Radicals rule, simplify:

$x = \pm \sqrt{9} \cdot \sqrt{11}$

$= \pm 3 \sqrt{11}$

Since represents a positive quantity, we choose

$x = 3\sqrt{11}$

$y^{2} = 11$ ,

so, taking the square root of both sides,

$y = \pm \sqrt{11}$

Since represents a negative quantity.

$y = - \sqrt{11}$

Substituting:

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

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

$=\left [ (3 \sqrt{11}) - \left ( -\sqrt{11}\right ) \right ] ^{3}$

$= (4 \sqrt{11}) ^{3}$

By the Product of Powers rule, this is

$4 ^{3} \cdot (\sqrt{11}) ^{3}$

$= 4 ^{3} \cdot (\sqrt{11}) ^{2} \cdot \sqrt{11}$

$=64 \cdot 11 \cdot \sqrt{11}$

$= 704 \sqrt{11}$

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Example Question #61 : 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:

$-405\sqrt{15}$

$405\sqrt{15}$

$15\sqrt{15}$

$-15\sqrt{15}$

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 #65 : Expressions

and both represent positive quantities.

$x^{2} = 48$

$y^{2} = 12$

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

Possible Answers:

$-216\sqrt{3}$

$216\sqrt{3}$

$72\sqrt{3}$

$-72\sqrt{3}$

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 #66 : Expressions

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:

$79^{\circ } F$

$59^{\circ } F$

$85^{\circ } F$

$72^{\circ } F$

$66^{\circ } F$

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 #67 : Expressions

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:

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

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

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

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

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

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=2x+5

y=-5x+2

y=5x+2

y=5x

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 #69 : Expressions

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:

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

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

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

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

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

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 #211 : New Sat

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

Possible Answers:

$\frac{3}{4}$

$\frac{3}{2}$

$\frac{2}{3}$

$72^\circ$

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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SAT Math : Expressions

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #61 : Expressions

Example Question #62 : Expressions

Example Question #63 : Expressions

Example Question #61 : Expressions

Example Question #65 : Expressions

Example Question #66 : Expressions

Example Question #67 : Expressions

Example Question #2531 : Sat Mathematics

Example Question #69 : Expressions

Example Question #211 : New Sat

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