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AleksandrR [38]
3 years ago
10

Write a division problem that the quotient is 5/8

Mathematics
1 answer:
kirill115 [55]3 years ago
3 0
One eighth divided by five equals five eighths.
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3x - 2(2x - 5) = 2(x + 3) - 8
nataly862011 [7]

Answer:

x = 4

Step-by-step explanation:

Hello!

We can solve for x by expanding the parentheses and isolating x.

<h3>Solve for x</h3>
  • 3x - 2(2x - 5) = 2(x + 3)-8
  • 3x - 4x + 10 = 2x + 6 - 8
  • -x + 10 = 2x - 2
  • 10 = 3x - 2
  • 12 = 3x
  • x = 4

The value of x is 4.

3 0
2 years ago
*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> 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.)

6 0
3 years ago
Find the product.
dlinn [17]
The correct answer of the given expression above would be the last option. We just solved this using the distributive property.
So: <span>(a - 3)(a - 5)
a^2 -3a -5a + 15
a^2 -8a + 15
Hope this is the answer that you are looking for. Have a great day!</span>
4 0
2 years ago
Solve. 3(4 - x) = 12 - 3x
poizon [28]
The answer is B, all real number
6 0
3 years ago
Read 2 more answers
At least 7 since 0 is a whole number the lowest n+7 can be is 0+7= 7 and there is seven numbers in between 0 and seven.
Nina [5.8K]
Where’s the question
5 0
3 years ago
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