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2 years ago

I'm completely lost on this question please help

Mathematics

2 answers:

Lunna [17]2 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]2 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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Please answer correctly !!!!! Will mark brainliest answer !!!!!!!!!!

AnnZ [28]

Answer:

type 2 in the first box,

13/4 in the second box, and

-9/8 in the third one

Step-by-step explanation:Notice that you are asked to write the following quadratic expression in vertex form, so you need to find the "x" value of the vertex, and then the "y" value of the vertex:

$x_{vertex}= -b/2a$

Which in our case is: -13/4

and the value of the y for the vertex is obtained using the functional expression when x equals -13/4:

$f(-13/4)= -9/8$

Then your expression for this quadratic should be:

$f(x)=2\,(x+\frac{13}{4} )^2+(-\frac{9}{8})$

Then type 2 in the first box, 13/4 in the second box, and -9/8 in the third one

7 0

2 years ago

The perimeter of a standard-sized rectangular rug is 36 is 36 ft. the length is 2 ft longer than the width. find the dimensions

evablogger [386]

The formula for the perimeter of a rectangle is P = 2L + 2W, where L is the length and W is the width. Because we don't know either the length or the width we can't solve the problem...too many unknowns. BUT we do have some information that will help with this problem. We are told that the length is 2 feet longer than the width, so we can use that: L = W+2. Now we can make the substitution into the formula along with the value for the perimeter that was given to us: 36=2(W+2) + 2W, and 36 = 2W + 4 + 2W; 36 = 4W + 4; 32 = 4W and W = 8. Now go back to where you said that the length is 2 feet longer than the width. If the width is 8, then 8+2 = 10 for the length.

4 0

2 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

2 years ago

Graph the two linear equations y=-2, x=5

Natalka [10]

Answer:

(hope this helps can I pls have brainlist (crown)☺️)

Step-by-step explanation:

In pic

5 0

1 year ago

Give an example of undefined term and defined term in geometry

vovangra [49]

These are undefined terms:1.plane2.point3.line.
These are defined terms: 1.ray 2.union of sets 3.space 4.subset 5.set 6.proper subset 7.opposite

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2 years ago