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SAT Practice Test 2015. In the xy-plane, the point (3, 6) lies on the graph of the function $f(x)=3x^2-bx+12$. What is the value of b?
Answer:
If x = 3, the value of f(x) is 6. Substituting 3 for x and 6 for f(x) in the give function:
$f(x)=3x^2-bx+12$
$6=3(3^2)-3b+12$
$6=27-3b+12$
$3b=27+12-6$
$3b=33$
$b=11$
Answer: 11
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SAT Practice Test 2015. $y=x^2-6x+8$
The equation above represents a parabola in the xy-plane. Which of the following equivalent forms of the equation displays the x-intercepts of the parabola as constants or coefficients?
A) $y-8=x^2-6x$
B) $y+1=(x-3)^2$
C) $y=x(x-6)+8$
D) $y=(x-2)(x-4)$
Answer:
The factored form of the equation, $y=(x-2)(x-4)$, shows that y equals 0 if and only if x = 2 or x = 4. Thus, these are the x-intercepts of the parabola.
Answer: D
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SAT Practice Test 2015. $2x(3x+5)+3(3x+5)=ax^2+bx+c$
In the equation above, a, b, and c are constants. If the equation is true for all values of x, what is the value of b ?
Answer:
$2x(3x+5)+3(3x+5)=ax^2+bx+c$
$6x^2+10x+9x+15=ax^2+bx+c$
$6x^2+19x+15=ax^2+bx+c$
Therefore,
a=6;
b=19;
c=15
Answer: 19
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SAT Practice Test 2015. What is the sum of all values of m that satisfy
$2m^2-16m+8=0$?
A) $-8$
B) $-4\sqrt{3}$
C) $4\sqrt{3}$
D) $8$
Answer:
$2m^2-16m+8=0$
$m^2-8m+4=0$
$m_1=(8+\sqrt{64-16})/2$
or
$m_2=(8-\sqrt{64-16})/2$
Therefore,
$m_1+m_2=(8+\sqrt{48})/2+(8-\sqrt{48})/2$
$m_1+m_2=8/2+8/2=8$
Answer: D
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SAT 2015 Test. $\sqrt{2k^2+17}-x=0$
If k > 0 and x = 7 in the equation above, what is the value of k?
A) 2
B) 3
C) 4
D) 5
Answer:
$\sqrt{2k^2+17}-x=0$
It is given that x=7, so
$\sqrt{2k^2+17}-7=0$
$\sqrt{2k^2+17}=7$
$2k^2+17=7^2$
$2k^2=49-17$
$2k^2=32$
$k^2=16$
It is given that k>0, then
$k=4$
Answer: C
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SAT 2015 Test.
$h(x)=\frac{1}{(x-5)^2+4(x-5)+4}$
For what value of x is the function h above undefined?
Answer:
The function will be undefined, if the denominator equals zero:
$(x-5)^2+4(x-5)+4=0$
$((x-5)+2)^2=0$
$(x-3)^2=0$
$x=3$
Answer: 3
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SAT 2015 Test.
Which of the following is an equivalent form of the equation of the graph shown in the xy-plane above, from which the coordinates of vertex A can be identified as constants in the equation?
A) $y=(x+3)(x-5)$
B) $y=(x-3)(x+5)$
C) $y=x(x-2)-15$
D) $y=(x-1)^2-16$
Answer:
Any quadratic function can be written in the form $f(x)=a(x-h)^2+k$, where a, h, and k are constants and (h, k) is the vertex of the parabola.
The given equation is:
$y=x^2-2x-15$
$y=x^2-2x+1-16$
$y=(x^2-2x+1)-16$
$y=(x-1)^2-16$
So h=1 and k=-16.
Answer: D
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SAT 2015 Test. $h=-4.9t^2+25t$
The equation above expresses the approximate height h, in meters, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 meters per second. After approximately how many seconds will the ball hit the ground?
A) 3.5
B) 4.0
C) 4.5
D) 5.0
Answer:
When the ball hits the ground h=0:
$h=-4.9t^2+25t$
$0=-4.9t^2+25t$
$(-4.9t+25)t=0$
One of the solutions to this equation is t=0, the moment the ball is launched.
The other solution is:
$-4.9t+25=0$
$4.9t=25$
$t=5.1$
Answer: D
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SAT 2015 Test. If t > 0 and $t^2-4=0$, what is the value of t?
Answer:
$t^2-4=0$
$t^2=4$
$t=2$ or $t=-2$
It is given that t>0, therefore $t=2$.
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SAT 2015 Test. If $(ax+2)(bx+7)=15x^2+cx+14$ for all values of x, and $a+b=8$, what are the two possible values for c ?
A) 3 and 5
B) 6 and 35
C) 10 and 21
D) 31 and 41
Answer:
$(ax+2)(bx+7)=15x^2+cx+14$
$abx^2+(7a+2b)x+14=15x^2+cx+14$
$abx^2+(7a+2b)x=15x^2+cx$
So $ab=15$ and $7a+2b=c$.
First let's find the possible values of $a$ and $b$, by solving the system of equations: $ab=15$ and $a+b=8$ (given):
$ab=15$
$a(8-a)=15$
$8a-a^2=15$
$a^2-8a+15=0$
$a=(8+\sqrt4)/2=5$
or
$a=(8-\sqrt4)/2=3$
Given that $ab=15$, if $a=5$, $b=3$, and if $a=3$, $b=5$
Now let's use these two possible solutions, and find $c$. We already know that $c=7a+2b$, so:
$c_1=7.(5)+2.(3)=35+6=41$
and
$c_2=7.(3)+2.(5)=21+10=31$
Answer: D
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SAT Test 2015. $g(x)=ax^2+24$
For the function g defined above, a is a constant and g(4)=8. What is the value of g(−4)?
A) 8
B) 0
C) −1
D) −8
Answer:
Given that the only term with x is $x^2$, and that $(4)^2=(-4)^2$, we know, regardless of the value of $a$, that $g(-4)=g(4)=8$.
We could also use the information given that $g(4)=8$, and calculate $a$. With $a$ it would be possible to calculate $g(−4)$. If you do that, you will also find $g(-4)=8$.
Answer: A
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SAT Practice Test. If the expression $4x^2/(2x-1)$, is written in the equivalent form $1/(2x-1)+A$, what is A in terms of x?
A. $2x+1$
B. $2x-1$
C. $4x^2$
D. $4x^2-1$
Answer:
$4x^2/(2x-1)=1/(2x-1)+A$
$4x^2/(2x-1)=1/(2x-1)+A.(2x-1)/(2x-1)$
$4x^2=1+A.(2x-1)$
$4x^2-1=A.(2x-1)$
$(2x+1)(2x-1)=A.(2x-1)$
$2x+1=A$
Answer: A
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SAT Practice Test. What is one possible solution to the equation
$\frac{24}{x+1}-\frac{12}{x-1}=1$?
Answer:
$\frac{24}{x+1}-\frac{12}{x-1}=1$
$\frac{24.(x-1)}{(x+1)(x-1)}-\frac{12.(x+1)}{(x+1)(x-1)}=\frac{(x+1)(x-1)}{(x+1)(x-1)}$
$24.(x-1)-12.(x+1)=(x+1)(x-1)$
$24x-24-12x-12=x^2-1$
$x^2-12x+35=0$
Using the Bhaskara Formula:
$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$
$x=\frac{12^+_-\sqrt{(-12)^2-4.1.35}}{2.1}$
$x=\frac{12^+_-\sqrt{144-140}}{2}$
$x=\frac{12^+_-2}{2}$
$x=7$ or $x=5$
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SAT Practice Test. If $a^2+14a=51$ (a, squared plus 14 a, equals 51) and $a>0$ (a, is greater than zero), what is the value of $a+7$ (a, plus 7)?
Answer:
$a^2+14a=51$
$a^2+14a-51=0$
The Bhaskara Fórmula is: $a=\frac{-B^+_-\sqrt{B^2-4AC}}{2A}$. Then,
$a=\frac{-14^+_-\sqrt{14^2-4.1.(-51)}}{2.1}$
$a=\frac{-14^+_-\sqrt{400}}{2}=\frac{-14^+_-20}{2}$
$a_1=\frac{-14+20}{2}=3$
$a_2=\frac{-14-20}{2}=-17$ (should be rejected, because it is given that $a>0$).
So, $a+7=3+7=10$
Answer: $a+7=10$.
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SAT Practice Test. The graph of $y=(2x-4)(x-4)$ (y equals, parenthesis 2 x minus 4 close parenthesis, parenthesis x minus 4 close parenthesis) is a parabola in the x y plane. In which of the following equivalent equations do the x- and y coordinates of the vertex of the parabola appear as constants or coefficients?
A. $y=2x^2-12x+16$ (y equals 2 x squared minus 12 x plus 16).
B. $y=2x(x-16)+16$ (y equals 2 x, parenthesis x minus 6 close parenthesis, plus 16).
C. $y=2(x-3)^2+(-2)$ (y equals 2, parenthesis, x minus 3, close parenthesis, squared, plus, parenthesis, negative 2, close parenthesis).
D. $y=(x-2)(2x-8)$ (y equals parenthesis x minus 2 close parenthesis, parenthesis 2 x minus 8 close parenthesis).
Answer:
The vertex form of a second degree equation is $y=a(x-h)^2+k$, where the vertex is $(h,k)$.
$(h,k)$ is the vertex, because the term $(x-h)^2$ is also a parabola with vertex in $x=h$, and when $x=h$, $y=a(h-h)^2+k=k$.
So,
$y=(2x-4)(x-4)$
$y=2x^2-8x-4x+16$
$y=2x^2-12x+16$
$y=2(x^2-6x+8)$
$y=2(x^2-6x+9+8-9)$
$y=2(x^2-6x+9-1)$
$y=2(x^2-6x+9)-2$
$y=2(x-3)^2+(-2)$
Answer: C.
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