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beks73 [17]
4 years ago
5

How can division patterns help us divide multiples of 10

Mathematics
1 answer:
KengaRu [80]4 years ago
7 0
Division using multiples of 10 is different than how most of us learned how to divide. <span>The idea of multiple is what number can 10 go into without a remainder. That is easy. Ten ends in a zero. Thus 10 goes into numbers ending in zero. An example is 60. Ten ends in a zero; 60 ends in a zero. It will divide evenly. </span>
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SOMEONE PLEASE JUST ANSWER THIS FOR BRAINLIEST!!!
castortr0y [4]

ANSWER

Step 1

EXPLANATION

The given polynomials are:

( {k}^{2}  + 2k - 13) - ( - 3 { k}^{2}  + 9k + 1)

Annie's expansion should have been,

{k}^{2}  + 2k - 13+ 3 { k}^{2}   -  9k  - 1

But she got:

{k}^{2}  + 2k - 13 + 3 { k}^{2}  + 9k + 1

Therefore Annie made an error in the first step.

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3 years ago
Which choice correctly shows the line<br> y = 2x+3?
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It’s c! Hope this helps

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(02.07)Which relationship is always true for the angles x, y, and z of triangle ABC? A triangle is shown with a leg extending pa
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How far will a cyclist riding at 25 miles per<br> hour travel in 3.5 hours?
Mademuasel [1]

Answer:

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2 years ago
Which statement best reflects the solution(s) of the equation?
Inessa [10]

x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Step-by-step explanation:

We need to solve the equation \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1} and find values of x.

Solving:

Find the LCM of denominators x-1,x and x-1. The LCM is x(x-1)

Multiply the entire equation with x(x-1)

\frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}\\\frac{1}{x-1}*x(x-1)+\frac{2}{x}*x(x-1)=\frac{x}{x-1}*x(x-1)\\Cancelling\,\,out\,\,the\,\,same\,\,terms:\\x+2(x-1)=x^2\\x+2x-2=x^2\\3x-2=x^2\\x^2-3x+2=0

Now, factoring the term:

x^2-2x-x+2=0\\x(x-2)-1(x-2)=0\\(x-1)(x-2)=0\\x-1=0\,\,and\,\, x-2=0\\x=1\,\,and\,\, x=2

The values of x are x=1 and x=2

Checking for extraneous roots:

Extraneous roots: The root that is the solution of the equation but when we put it in the equation the answer turns out not to be right.

If we put x=1 in the equation, \frac{1}{x-1}+\frac{2}{x}=\frac{x}{x-1}  the denominator becomes zero i.e

\frac{1}{1-1}+\frac{2}{1}=\frac{1}{1-1}\\\frac{1}{0}+2=\frac{1}{0}

which is not correct as in fraction anything divided by zero is undefined. So, x=1 is an extraneous solution.

If we put x=2  in the equation,

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

\frac{1}{2-1}+\frac{2}{2}=\frac{2}{2-1}\\\frac{1}{1}+1=\frac{2}{1}\\1+1=2\\2=2

So, x=2 is only solution while x=1 is extraneous solution

Option C is correct.

Keywords: Solving Equations and checking extraneous solution

Learn more about Solving Equations and checking extraneous solution at:

  • brainly.com/question/1626495
  • brainly.com/question/2959656
  • brainly.com/question/2456302

#learnwithBrainly

5 0
4 years ago
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