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3 years ago

PLEASE HELP I GIVE THANKS

Mathematics

1 answer:

oksano4ka [1.4K]3 years ago

6 0

Look at the pictures.

1. D (y=3)
2. y=3
3. y=-2
4. y=4
5. B (52 lb)

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I need help asap someone please help me i dont understand this question so can someone help me

eduard

Answer:

we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

Step-by-step explanation:

Given the expression

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}$

$=\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3}$

$=\frac{2p}{4p^2-1}\times \frac{2p+1}{2p^3}$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}$

$=\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}$

cancel the common factor: 2

$=\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}$

cancel the common factor: p

$=\frac{2p+1}{p^2\left(4p^2-1\right)}$

$=\frac{2p+1}{p^2\left(2p+1\right)\left(2p-1\right)}$

cancel the common factor: 2p+1

$=\frac{1}{p^2\left(2p-1\right)}$

Expanding

$=\frac{1}{2p^3-p^2}$

Thus, we conclude that:

$\frac{2p}{4p^2-1}\div \frac{6p^3}{6p+3}=\frac{1}{2p^3-p^2}$

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3 years ago

Giving Brainliest please help

Free_Kalibri [48]

Answer:

i think its f :) double check tho

Step-by-step explanation:

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B. it is the area of a circle with a<br> diameter of 1. <br> True of false

Dominik [7]

It has to be true because you would need to divide

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IgorC [24]

Answer:

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Step-by-step explanation:

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3 years ago

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otez555 [7]

Answer:for #2 it's answer 1. #3 is " five less than twice a number is 2x-5. Then 5 times the sum of two numbers if 5x+y. The the difference of 5 and a number squared is (5-x)*2

Step-by-step explanation:

Hope this helps

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3 years ago