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erastova [34]
2 years ago
12

Which equation shows a proportional relationship between x and y?

Mathematics
1 answer:
Ad libitum [116K]2 years ago
6 0

Answer:

D. y=4x

Step-by-step explanation:

It asks for a proportional relationship between x and y

And is the only answer

because

When x and y is 1 the answer is

4/1

And when x and y is 2 the answer is

8/2

If you divide both of them are still 4

And this will always be the result no matter how many times you do it

like if you put it as 150

150 times 4 is 600/150

600 divided by 150 is 4

Hope this helps!

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The product 3 and a number x is 2/3. What is the value of x? A. X=7/3 B. x=2/9 C. x=2 D. X= 2 1/3
BabaBlast [244]

Answer:

the answer is C.

Step-by-step explanation:

3(2/3)

=3×2/3

=2×3/3

=2

7 0
3 years ago
I need help with this
andre [41]

Answer:

  f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}

Step-by-step explanation:

Each vertical asymptote corresponds to a zero in the denominator. When the function does not change sign from one side of the asymptote to the other, the factor has even degree. The vertical asymptote at x=-4 corresponds to a denominator factor of (x+4). The one at x=2 corresponds to a denominator factor of (x-2)², because the function does not change sign there.

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Each zero corresponds to a numerator factor that is zero at that point. Again, if the sign doesn't change either side of that zero, then the factor has even multiplicity. The zero at x=1 corresponds to a numerator factor of (x-1)².

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Each "hole" in the function corresponds to numerator and denominator factors that are equal and both zero at that point. The hole at x=-3 corresponds to numerator and denominator factors of (x-3).

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Taken altogether, these factors give us the function ...

  \boxed{f(x)=\dfrac{(x+3)(x-1)^2}{(x+4)(x+3)(x-2)^2}}

8 0
3 years ago
Limit definition for slope of the graph, equation of tangent line point for<br> f(x)=2x^2 at x=(-1)
Tems11 [23]

The slope of the tangent line to f at x=-1 is given by the derivative of f at that point:

f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}

Factorize the numerator:

2x^2-2=2(x^2-1)=2(x-1)(x+1)

We have x approaching -1; in particular, this means x\neq-1, so that

\dfrac{2x^2-2}{x+1}=\dfrac{2(x-1)(x+1)}{x+1}=2(x-1)

Then

f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4

and the tangent line's equation is

y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2

6 0
2 years ago
What are the steps for this equation: -5x-16 = -10x-11?
sineoko [7]

Answer:

x=1

Step-by-step explanation:

-5x-16=-10x-11

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8 0
3 years ago
How did shape EFGH translate to E'F'G'H
agasfer [191]

Answer:

Hi

If u haven't given the pic it would be easy.

Thanks.

^ - ^

3 0
2 years ago
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