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3 years ago

I'm completely lost on this question please help

Mathematics

2 answers:

Lunna [17]3 years ago

8 0

First, note that y≠0 and y≠4 because that would make the denomenators 0 which is undefined

we got to get rid of fractions
common denomenator is y(y-4)
multiply both sides by y(y-4)
2y+y-4=y(y-1)
3y-4=y²-y
minus 3y both sides
-4=y²-4y
add 4 both sides
0=y²-4y+4
factor
what 2 numbers multiply to get 4 and add to get -4?
-2 and -2
0=(y-2)(y-2)
set to zero

y-2=0
y=2
no extranious roots

y=2 is the solution

solmaris [256]3 years ago

6 0

Multiply the numerators by "y" which yields:
(2y) / (y-4) +1 = (y^2 -y) / (y-4)
(2y) / (y-4) - (y^2 -y) / (y-4) = -1

(2y) - (y^2 -y) / (y-4) = -1

(2y) - (y^2 -y) = -y +1

-y^2 +2y -y +y = 1

-y^2 +2y -1 = 0

Multiplying by -1

y^2 -2y +1 = 0

(y -1) * (y - 1) = 0

x = 1
x= 1

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2 years ago

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valentina_108 [34]

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$\frac{4}{3}$ $\pi$ $x^{2}$ h = 12 $\pi$ $x^{3}$

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