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2 years ago

Y=-x^2+8 y= =3/4x– 3

Mathematics

1 answer:

Dmitrij [34]2 years ago

8 0

Answer:

Yy8cgucucy8cc8yc8ygf7yy7 gdyrvdysrtu ui

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Which of the following is not a rational number? Help please

Pavlova-9 [17]

A, because negative is not rational

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2 years ago

Find f(5). f(x) = x2 + 2x A)15 B)20 C)30 D)35

inysia [295]

f(5) means to replace x in the equation with 5 then solve.

f(x) = x^2 +2x

f(5) = 5^2 + 2(5)

f(5) = 25 +10

f(5) = 35

The answer is D)35

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2 years ago

Point O is the center of the square ABCD. Point E is the midpoint of line segment CD. The area of square ABCD is 24, what is the

GalinKa [24]

Answer: You divided 24 by 3 then you take the answer

Step-by-step explanation:I hope it help

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2 years ago

Subtraction (must show work) 1. d - 27 = 45 2. h - 114 = 28

Arturiano [62]

Answer:

Step-by-step explanation:

d - 27 = 45

add 27

d = 72

h - 114 = 28

add 114

h = 142

-4 + x = 15

add 4

x = 19

-39 + g = 72

add 39

g = 111

7 0

2 years ago

*Asymptotes* g(x) =2x+1/x-3 Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

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3 years ago