Question

(04.01 LC)

Simplify quantity x squared plus 5 x plus 6 end quantity over quantity x plus 2

x2 + 1
x2 − 1
x + 3
x − 3
Question 4(Multiple Choice Worth 5 points)
(04.03 MC)

What are the vertical asymptotes of the function f(x) = the quantity of 2x plus 8, all over x squared plus 5x plus 6?

x = −3 and x = −2
x = −3 and x = 2
x = 1 and x = −2
x = 1 and x = 2
Question 5(Multiple Choice Worth 5 points)
(04.02 LC)

Identify the restrictions on the domain of f(x) = quantity x minus 3 over quantity x plus 5.

x ≠ 5
x ≠ −5
x ≠ 3
x ≠ −3
Question 6(Multiple Choice Worth 5 points)
(04.02 LC)

What are the discontinuity and zero of the function f(x) = quantity x squared plus 6 x plus 8 end quantity over quantity x plus 4?

Discontinuity at (4, 6), zero at (−2, 0)
Discontinuity at (4, 6), zero at (2, 0)
Discontinuity at (−4, −2), zero at (−2, 0)
Discontinuity at (−4, −2), zero at (2, 0)
Question 7 (Essay Worth 10 points)
(04.02 MC)

Show all work to identify the asymptotes and zero of the function f of x equals 3 x over quantity x squared minus 9.



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Question 8 (Essay Worth 10 points)
(04.04 MC)

The aquarium has 6 fewer yellow fish than green fish. 40 percent of the fish are yellow. How many green fish are in the aquarium? Show your work.

Answer 1

Following are the solution to the given question:

Explanation:

In question 1:

`\to (x^2+5x+6)/(x+2)\\\\\to (x^2+(3+2)x+6)/(x+2)\\\\\to (x^2+3x+2x+6)/(x+2)\\\\\to (x(x+3)+2(x+3))/(x+2)\\\\\to ((x+3) (x+2))/(x+2)\\\\\to (x+3)`

In question 2:

`\to (2x+8)/(x^2+5x+6)\\\\`

`\to x^2+5x+6=0\\\\\to (x+3)(x+2)=0\\\\\to x+3 =0 \ \ \ \ \ \ x+2 =0\\\\\to x = -3 \ \ \ \ \ \ x=-2 \\\\`

In question 3:

`\to f(x) = (x-3)/(x+5)\\\\\to f(x) \\eq 0\\\\when\\\to x+5=0\\\\\to x=-4 \ excluded \ value \\\\\to x \\eq 5`

In question 4:

`\to f(x) = (x^2+6x+8)/((x+4))\\\\\to f(x) = (x^2+(4+2)x+8)/((x+4))`

`\to f(x) = (x^2+4x+2x+8)/((x+4))\\\\\to f(x) = (x(x+4)+2(x+4))/((x+4))\\\\\to f(x) = ((x+4)(x+2))/((x+4))\\\\\to f(x) = x= -4 and -2`

In question 5:

`\to f(x) =(3x)/(x^2-9) \\\\ \to f(x) =(3x)/(x^2-3^2) \\\\ \to f(x) =(3x)/((x+3)(x-3))`

In question 6:

total number of fish = x

yellow fish
`= (40)/(100)=0.4X`

green fish
`= (60)/(100)) =0.6X`

It is now the case that the yellow fish number is 6 less than the green fish number.

`\therefore \\\\\to 0.6x-0.4x = 6 \\\\\to 0.2x = 6\\\\\to x= (6)/(0.2) \\\\\to x=30`

green fish
`= 0.6 * 30=18`