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1 year ago

100 POINTS PLEAS please help

Mathematics

2 answers:

mel-nik [20]1 year ago

7 0

(-2u³+5u-1)+(7u³-2u²+9)
-2u³+5u-1+7u³-2u²+9
5u³-2u²+5u-1+9
5u³-2u²+5u+8

Option A

Kamila [148]1 year ago

6 0

Answer:

$\textsf{A)} \quad 5u^3-2u^2+5u+8$

Step-by-step explanation:

Given expression:

$(-2u^3+5u-1)+(7u^3-2u^2+9)$

Remove parentheses:

$\implies -2u^3+5u-1+7u^3-2u^2+9$

Collect like terms:

$\implies -2u^3+7u^3-2u^2+5u-1+9$

Combine like terms:

$\implies 5u^3-2u^2+5u+8$

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klasskru [66]

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3 years ago

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algol [13]

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<h3>What is the mean of a discrete distribution?</h3>

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1 year ago

1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$