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expeople1 [14]
2 years ago
11

NO LINKS PLEASE JUST THE ANSWER TEHEHHE THANKSSSS :))))

Mathematics
1 answer:
goblinko [34]2 years ago
6 0

Answer:

Since this is a multi-part question, just look at the bolded parts under each letter. I hope this helps a bit ;)

Step-by-step explanation:

A)  

All sides of a square are congruent. "s" represents the square's perimeter:

s=4x.

"r" is the rectangle's perimeter:

r= 4x+26

since the perimeter = 2W + 2L:

2W + 2L= 4x+ 26

and W=x, so:

2x + 2L= 4x + 26

Subtract from both sides:

2L= 2x +26

Divide both sides:

the length of the rectangle "L"= x +13.

B) Plug 5 into the equations:

L= 5+ 13 or 18.

2(18) + 2(5)= r

36+ 10 = r or 46

s= s+ 26

s= 46-26 or 20.

20/4= 5...

It seems (at least to me, feel free to give constructive criticism) that the only logical conclusion is that 5 <u>could</u> be the value of x.

C) You would likely need to use substitution to solve, but unless I am much mistaken, this looks like an infinite-solutions equation.

L=w+13

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blagie [28]

Explicación paso a paso:

a+b=16 +

a-b=-4

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2a=12

a=6

b=10

el menor vale 6

<h2>y porfavor branliest! :)</h2>
8 0
2 years ago
Jorge received a 3.1% increase in his pay at work. If he now makes $18.00 per hour, how much was he making before his raise? (Ro
g100num [7]
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How much has James earned since the beginning of the year in 2014? (use pay stub) *
schepotkina [342]
Can you give more information?
4 0
2 years ago
Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
2 years ago
Negative positive or no association
Eddi Din [679]

Answer:

Positive association

Step-by-step explanation:

As the Temperature goes up the more money you have to spend because you pay for the electricity that produces the heat

positive association is when one varriable increases so does the other.

7 0
2 years ago
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