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SAT Practice Test 2015.
$I=\frac{P}{4\pi{r}^2}$
At a large distance r from a radio antenna, the intensity of the radio signal I is related to the power of the signal P by the formula above.
Which of the following expresses the square of the distance from the radio antenna in terms of the intensity of the radio signal and the power of the signal?
A) $r^2=\frac{IP}{4\pi}$
B) $r^2=\frac{P}{4\pi{I}}$
C) $r^2=\frac{4\pi{I}}{P}$
D) $r^2=\frac{I}{4\pi{P}}$
Answer:
$I=\frac{P}{4\pi{r}^2}$
Multiplying both sides of the equation by $r^2$:
$Ir^2=\frac{P}{4\pi}$
Now dividing both sides of the equation by $I$:
$r^2=\frac{P}{4\pi{I}}$
Answer: B
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SAT Practice Test 2015. This question refers to the same data used in the previous question:
For the same signal emitted by a radio antenna, Observer A measures its intensity to be 16 times the intensity measured by Observer B. The distance of Observer A from the radio antenna is what fraction of the distance of Observer B from the radio antenna?
A) 1/4
B) 1/16
C) 1/64
D) 1/256
Answer:
According to the previous question:
$r^2=\frac{P}{4\pi{I}}$
Thus, for Observer B, we should have: ${r_B}^2=\frac{P}{4\pi{I_B}}$
And for Observer A:
${r_A}^2=\frac{P}{4\pi{I_A}}=\frac{P}{4\pi{16I_B}}$
Therefore:
$\frac{{r_B}^2}{{r_A}^2}=\frac{\frac{P}{4\pi{I_B}}}{\frac{P}{4\pi{16I_B}}}$
$\frac{{r_B}^2}{{r_A}^2}=\frac{4\pi{16I_B}}{4\pi{I_B}}=16$
$\frac{{r_A}^2}{{r_B}^2}=\frac{1}{16}$
$\frac{r_A}{r_B}=\frac{1}{4}$
Answer: A
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SAT Practice Test 2015. The expression $\frac{5x-2}{x+3}$ is equivalent to which of the following?
A) $\frac{5-2}{3}$
B) $5-\frac{2}{3}$
C) $5-\frac{2}{x+3}$
D) $5-\frac{17}{x+3}$
Answer:
$\frac{5x-2}{x+3}=$
$\frac{5x+15-17}{x+3}=$
$\frac{5x+15}{x+3}-\frac{17}{x+3}=$
$5-\frac{17}{x+3}=$
Answer: D
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SAT Practice Test 2015. $R=\frac{F}{N+F}$
A website uses the formula above to calculate a seller’s rating, R, based on the number of favorable reviews, F, and unfavorable reviews, N. Which of the following expresses the number of favorable reviews in terms of the other variables?
A) $F=\frac{RN}{R-1}$
B) $F=\frac{RN}{1-R}$
C) $F=\frac{N}{1-R}$
D) $F=\frac{N}{R-1}$
Answer:
$R=\frac{F}{N+F}$
$R(N+F)=F$
$RN+RF=F$
$RN=F-RF$
$F(1-R)=RN$
$F=\frac{RN}{1-R}$
Answer: B
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SAT Test 2015.
$a=1,052+1.08t$
The speed of a sound wave in air depends on the air temperature. The formula above shows the relationship between a, the speed of a sound wave, in feet per second, and t, the air temperature, in degrees Fahrenheit (°F).
Which of the following expresses the air temperature in terms of the speed of a sound wave?
A) $t=\frac{a-1,052}{1.08}$
B) $t=\frac{a+1,052}{1.08}$
C) $t=\frac{1,052-a}{1.08}$
D) $t=\frac{1.08}{a+1,052}$
Answer:
$a=1,052+1.08t$
$a-1,052=1.08t$
$t=\frac{a-1,052}{1.08}$
Answer: A
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SAT Test 2015. If x > 3, which of the following is equivalent to
$\frac{1}{\frac{1}{x+2}+\frac{1}{x+3}}$
A) $\frac{2x+5}{x^2+5x+6}$
B) $\frac{x^2+5x+6}{2x+5}$
C) $2x+5$
D) $x^2+5x+6$
Answer:
$\frac{1}{\frac{1}{x+2}+\frac{1}{x+3}}=$
$\frac{(x+2)(x+3)}{(x+3)+(x+2)}=$
$\frac{x^2+5x+6}{2x+5}$
Answer: B
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SAT Test 2015. If $\frac{a}{b}=2$, what is the value of $\frac{4b}{a}$?
A) 0
B) 1
C) 2
D) 4
Answer:
$\frac{a}{b}=2$
$\frac{b}{a}=\frac{1}{2}$
$\frac{4b}{a}=\frac{4.1}{2}$
$\frac{4b}{a}=2$
Answer: C
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SAT Test 2015. $(x^2y-3y^2+5xy^2)-(-x^2y+3xy^2-3y^2)$
Which of the following is equivalent to the expression above?
A) $4x^2y^2$
B) $8xy^2-6y^2$
C) $2x^2y+2xy^2$
D) $2x^2y+8xy^2-6y^2$
$(x^2y-3y^2+5xy^2)-(-x^2y+3xy^2-3y^2)=$
$x^2y-3y^2+5xy^2+x^2y-3xy^2+3y^2=$
$x^2y+x^2y+5xy^2-3xy^2-3y^2+3y^2=$
$2x^2y+2xy^2$
Answer: C
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