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Luda [366]
3 years ago
13

Trying to finish help

Mathematics
1 answer:
Margaret [11]3 years ago
8 0
It’s the first answer. The function is y = x*2
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the cost of his monthly cable is $40.50 when you plug in 4

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Hazel saves $5.75 each week how much does she save in 2 weeks
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Your answer would be $11.50 I hope this helps you out! :)
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Read 2 more answers
<img src="https://tex.z-dn.net/?f=y%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2B%20%20%5Cfrac%7Bw%7D%7B3%7D%20" id="TexFormula1" ti
ElenaW [278]

Answer:  \bold{w=\dfrac{6y-3}{2}}

<u>Step-by-step explanation:</u>

Isolate w by performing the following steps

  • Multiply by 6 on both sides to clear the denominator
  • Subtract 3 from both sides
  • Divide both sides by 2

y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}

5 0
3 years ago
Which of the following is an extraneous solution of (45-3x)^1/2=x-9?
Iteru [2.4K]

Answer:

C. x =3

Step-by-step explanation:

Extraneous solution is that root of a transformed equation that doesn't satisfy the equation in it's original form because it was excluded from the domain of the original equation.

Let's solve the equation first

\sqrt{45-3x} = x-9\\Taking\ square\ on\ both\ sides\\{(\sqrt{45-3x})}^2 = {(x-9)}^2\\45-3x = x^2-18x+81\\0 = x^2-18x+81-45+3x\\x^2-15x+36 = 0\\x^2-12x-3x+36 = 0\\x(x-12)-3(x-12) = 0\\(x-3)(x-12)\\x-3 = 0\\=> x =3\\x-12 = 0\\x = 12\\We\ will\ check\ the\ solutions\ one\ by\ one\\So,\\for\ x=3\\\sqrt{45-3(3)} = 3-9\\\sqrt{45-9} = -6\\\sqrt{36}= -6\\6\neq -6\\For x=12\\\sqrt{45-3(12)} = 12-9\\\sqrt{45-36} = 6\\\sqrt{36}= 6\\6=6

Hence, we can conclude that x=3 is an extraneous solution of the equation ..

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3 years ago
The diagram shows the width and area of a rectangle.
Mademuasel [1]

Answer:

The answer can be explained here: theraleighregister.com/the-diagram-shows-the-width-and-area-of-a-rectangl.html

Step-by-step explanation:

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