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Minchanka [31]
3 years ago
14

..........................help me pls?

Mathematics
1 answer:
Ludmilka [50]3 years ago
6 0

Answer:

7x-7=y

Step-by-step explanation:

5x+7y=10

5x-10=10-7y-10

5x-10=-7y

(1/7)5x-10=7y(1/7)

5x/35-10/70=y

7x-7=y

<em>Sana</em><em> </em><em>maka</em><em>tulong</em><em> </em>

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Which of the following would be 1,000 times a base unit?
Zolol [24]
Kilo :) hope this helps


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What is the expanded form of this number?<br><br> 503.208<br> (Trying to help my sister do math!)
Blababa [14]

Answer:

500+3+ .200+.008

Step-by-step explanation:

You add 500 and 3 then after you get your number add .200 lastly add .008.

when expanding numbers you need to know what place value it's in so that you k ow where to put the zeros.

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The bottom of a rectangular swimming pool has a perimeter of 62 meters. Its area is 220 square meters. What are the dimensions o
jek_recluse [69]

Half the perimeter is the length + the width

62/2 = 31

Area = length x width

16 + 15 = 31, 15 x 16 = 240

17 + 14 = 31, 17 x 14 = 238

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3 0
3 years ago
What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?
Ahat [919]

Answer:

\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}

So the quotient is 1x^2+5x+-1 and the remainder is -6.

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form x-c.

Since we are dividing by x+2=x-(-2), this is our setup for the synthetic division:

-2 |     1    7    9    -8

   |         -2   -10     2

   ______________

          1   5    -1       -6

So the quotient is 1x^2+5x+-1 and the remainder is -6.

So \frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}.

We can also do long division.

                 x^2+5x-1

               ____________________

        x+2| x^3+7x^2+9x-8

              -(x^3+2x^2)

              -------------------

                       5x^2+9x-8

                    -( 5x^2+10x)

                      --------------------

                                 -x-8

                               -(-x-2)

                                --------------

                                      -6

So we see here we get the same quotient, x^2+5x-1. and the same remainder, -6.

Now let's check our result that:

\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}.

So I'm going to rewrite the right hand side as a single fraction:

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}.

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}

Now let's focus on multiplying (x+2)(x^2+5x-1).

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

x(x^2)=x^3

x(5x)=5x^2

x(-1)=-x

2(x^2)=2x^2

2(5x)=10x

2(-1)=-2

---------------------------Combine like terms:

x^3+(5x^2+2x^2)+(-x+10x)+-2

x^3+7x^2+9x-2

So let's go back where we were in our check of \frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}:

\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}

\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}

\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}

We have the exact same thing on both sides so we did good.

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