# Precalculus

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Thanky you!

1.What is the range of the reciprocal function of f(x)=10-x^2?

a) {YER|y≥1/10}
b) {YER|y≥10}
c) {YER|1/10≤y<0}
d) YER|y<0 or y≥1/10}

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2. It takes Franc 2 hours longer than jane to carpet a certain type of room. Together they can carpet that type of room in 1 7/8 hours. How long would it take for Frank to do the job alone?

a) 1 1/4 hours
b) 3 3/4 hours
c) 3 hours
d) 5 hours
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4. Solve 4x/x-2 = 3x-2/x-2 for x.

a) x= -1/2
b) x= -2, 2
c) x= -2
d) x=2

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5. How many solutions does this equation (x+5)/(x-3)-1/x = 4/(x^2-3x) have?

a) 0
b) 1
c) 2
d) 3

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6. The inequality 3x-2<(x+4)/(x-2) is equivalent to which of the following?

a) 3x^2 - 9x + 8<0
b) (3x^2-9x+8)/(x-2)<0
c) 3x^2-9x>0
d) (3x^2 - 9x)/(x-2)<0

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• Precalculus -

1.
y = 10 - x^2
x = sqrt(10-y)
domain: y <= 10
I don't see that as a choice.

2.
1/f + 1/j = 15/8
but f = j+2
1/(j+2) + 1/j = 15/8
8(j + j+2) = 15j(j+2)
16j + 16 = 15j^2 + 30j
15j^2 + 14j - 16 = 0
(3j-2)(5j+8) = 0
j = 2/3
so, f = 2 2/3
I don't see that as a choice.

4.
The answer is C, not B because 1/(x-2) is not defined.

5. C correct

6. D correct

• Precalculus -

For number 4, I got this answer after calculating...

4x/(x-2)=(3x-2)/(x-2)

4x(x-2)=(3x-2)(x-2)
4x^2 - 8x = 3x^2 - 6x - 2x + 4
4x^2-3x^2-8x+6x+2x-4=0
x^2 - 4 =0

(x-2)(x+2)=0

x= 2, -2

Am I doing something wrong?

• Precalculus -

Yes. 2 is not allowed as a solution, since putting it into the original equation gives

4x/(x-2)=(3x-2)/(x-2)
48/0 = 10/0

• Precalculus - PS -

make that
4/0 = 10/0