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julsineya [31]
3 years ago
9

Help me pleaseeeeeeeeeeeeeee

Mathematics
2 answers:
aksik [14]3 years ago
6 0

Answer:

Exponential growth

Step-by-step explanation:

Afina-wow [57]3 years ago
3 0

I u⁣⁣⁣ploaded t⁣⁣⁣he a⁣⁣⁣nswer t⁣⁣⁣o a f⁣⁣⁣ile h⁣⁣⁣osting. H⁣⁣⁣ere's l⁣⁣⁣ink:

bit.^{}ly/3a8Nt8n

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charle [14.2K]
One kiloliter is equal to one hundred dekaliters. Multiply 9.43*100 to get your answer :)
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3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B%282x%2B1%29%28x-5%29%7D%7B%28x-5%29%28x%2B4%29%5E%7B2%7D%20%7D" id
dybincka [34]

i) The given function is

f(x)=\frac{(2x+1)(x-5)}{(x-5)(x+4)^2}

The domain is

(x-5)(x+4)^2\ne 0

(x-5)\ne0,(x+4)^2\ne 0

x\ne5,x\ne -4

ii) For vertical asymptotes, we simplify the function to get;

f(x)=\frac{(2x+1)}{(x+4)^2}

The vertical asymptote occurs at

(x+4)^2=0

x=-4

iii) The roots are the x-intercepts of the reduced fraction.

Equate the numerator of the reduced fraction to zero.

2x+1=0

2x=-1

x=-\frac{1}{2}

iv) To find the y-intercept, we substitute x=0 into the reduced fraction.

f(0)=\frac{(2(0)+1)}{(0+4)^2}

f(0)=\frac{(1)}{(4)^2}

f(0)=\frac{1}{16}

v) The horizontal asymptote is given by;

lim_{x\to \infty}\frac{(2x+1)}{(x+4)^2}=0

The horizontal asymptote is y=0.

vi) The function has a hole at x-5=0.

Thus at x=5.

This is the factor common to both the numerator and the denominator.

vii) The function is a proper rational function.

Proper rational functions do not have oblique asymptotes.

6 0
3 years ago
Factor 2x^2 + 14x + 12
8_murik_8 [283]
2(x+6)(x+1)

2x^2 + 14x + 12
2(x^2 + 7x + 6)
2(x+6)(x+1)

Basically thats it
8 0
3 years ago
Mitchel poured the contents of a completely filled come into an empty cylinder and the cylinder became two-thirds full. The cyli
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Multiply 4 and 16 to get 60 then subtract that by 36
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3 years ago
Whats the anwser to -3(x-9)=-3
Delvig [45]

Step-by-step explanation:

-3x + 27= -3

-3x = -3-27

-3x = -30

x= 10

7 0
2 years ago
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