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Studentka2010 [4]
2 years ago
11

Write an expression equivalent to the expressions 12x+16y using the gcfpls help

Mathematics
2 answers:
Marrrta [24]2 years ago
8 0

Answer:4(3x+4y)

Step-by-step explanation: the gcf of 12x+16y is 4. we factor out 4 by dividing both terms by 4.

avanturin [10]2 years ago
5 0

Answer:

4(3x+4y)

Step-by-step explanation:

Hey there!

This would be the correct answer because the GCF of 12 and 16 is 4.

Now that we know that it is 4 we can divide the eqiasion by 4 and  then you get

4(3x+4y)

If you need to ckeck if this answer is corret then you will need to use the distributive property to distribute 4 to 3y and 4y.

The equasion would then look like this

(4(3x)) (4(4y))

Next you would need to solve the equasion to get the origional equasion

12x+16y

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Answer:

A and B

Step-by-step explanation:

we would like to solve the following equation:

\rm\displaystyle 5 - 2x =  \sqrt{ {2x}^{2} + x - 1 }  - x

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\rm\displaystyle 5 - 2x + x =  \sqrt{ {2x}^{2} + x - 1 }

simplify addition:

\rm\displaystyle 5 - x =  \sqrt{ {2x}^{2} + x - 1 }

square both sides:

\rm\displaystyle (5 - x {)}^{2}  =  (\sqrt{ {2x}^{2} + x - 1 }  {)}^{2}

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\rm\displaystyle (5 - x {)}^{2}  =  {2x}^{2} + x - 1

use (a-b)²=a²-2ab+b² to expand the left hand side:

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swap the equation:

\rm\displaystyle   {2x}^{2} + x - 1 =  {x}^{2}  - 10x + 25

isolate the right hand side expression to the left hand side and change every sign:

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rewrite the middle term as 13x-2x:

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factor out x:

\rm\displaystyle   x({x}^{} + 13)- 2x  -  26=  0

factor out -2:

\rm\displaystyle   x({x}^{} + 13)- 2(x   +  13)=  0

group:

\rm\displaystyle   (x- 2)(x   +  13)=  0

by <em>Zero</em><em> </em><em>product</em><em> </em><em>property</em> we obtain:

\displaystyle    \begin{cases}x- 2  = 0\\ x   +  13=  0 \end{cases}

solve for x:

\displaystyle    \begin{cases}x = 2\\ x  =   - 13 \end{cases}

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

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solve the inequality for x:

\rm\displaystyle x    \leqslant  5

since 2 and -13 is less than 5 both solutions are valid for x hence,

\displaystyle    \begin{cases}x _{1} = 2\\  x_{2} =   - 13 \end{cases}

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