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olchik [2.2K]
3 years ago
12

Solve for the unknown by using the additive inverse. Type the FULL answer in the box, without using any spaces (ex., X=5).

Mathematics
1 answer:
zavuch27 [327]3 years ago
3 0

Answer:

X=16

Step-by-step explanation:

10x-8=9x+8

First you have to simplify.

10x-9x=1x

So now the equation looks like this

x-8=8

Now you have to get x by itself

-8+8=0 and we have to do the same thing on both sides of the equal sign 8+8=16

The equation now looks like this x=16

And that's your answer

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4+6+6+2=18(balls)
Another way 1)4+6=10(balls)-green and blue
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What percent is equivalent to 2/5
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Which statement describes the inverse of m(x) = x^2 – 17x?
DochEvi [55]

Given:

The function is

m(x)=x^2-17x

To find:

The inverse of the given function.

Solution:

We have,

m(x)=x^2-17x

Substitute m(x)=y.

y=x^2-17x

Interchange x and y.

x=y^2-17y

Add square of half of coefficient of y , i.e., \left(\dfrac{-17}{2}\right)^2 on both sides,

x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2

x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2        [\because (a-b)^2=a^2-2ab+b^2]

Taking square root on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}

Add \dfrac{17}{2} on both sides.

\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y

Substitute y=m^{-1}(x).

m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}

We know that, negative term inside the root is not real number. So,

x+\left(\dfrac{17}{2}\right)^2\geq 0

x\geq -\left(\dfrac{17}{2}\right)^2

Therefore, the restricted domain is x\geq -\left(\dfrac{17}{2}\right)^2 and the inverse function is m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}.

Hence, option D is correct.

Note: In all the options square of \dfrac{17}{2} is missing in restricted domain.

7 0
3 years ago
Sophie has 3/4 quart of lemonade. if she divides the lemonade into glasses that hold 1/16 quart, how many glasses can Sophie fil
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So I'll start with the mixed number to improper fraction.
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Now for improper fraction to mixed number
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