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3 years ago

Find the volume of this figure: cm 7 3 cm Cin 3

Mathematics

2 answers:

kenny6666 [7]3 years ago

7 0

I think it’s 63 because to find the volume you have to multiply all numbers

Vladimir79 [104]3 years ago

6 0

Answer:

4 i guess i dont rlly know

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Please help me immediately

saw5 [17]

Answer:

756

Step-by-step explanation:

In the last question the volume of the tank was 168 m³

so to find the cost multiply 168 and 4.5

168*4.5 =756

it will cost 756

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3 years ago

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olga55 [171]

Same solution.

volume = 1/3(8x6)(5) = 16x5 = 80cm^3

answer: 80cm^3

4 0

3 years ago

911+911+666+21−420−69 what this equal

melamori03 [73]

Answer:

2020

Step-by-step explanation:

6 0

2 years ago

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How much times can 900 go into 8,000,000?<br> Please help its due today!!

lakkis [162]

Answer:

8,888.888

Step-by-step explanation:

7 0

2 years ago

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Use the denition of the derivative to find f 0 (3), where f (x) = 3x+5 / 2x−1

Crazy boy [7]

Answer:

$f'(3)= -\frac{13}{25}$

Step-by-step explanation:

We are asked to find $f'(3)$ of function $f(x)=\frac{3x+5}{2x-1}$ using definition of derivatives.

Limit definition of derivatives:

$f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Let us find $f(3+h)$ and $f(3)$ .

$f(3+h)=\frac{3(3+h)+5}{2(3+h)-1}$

$f(3+h)=\frac{9+3h+5}{6+2h-1}\\\\f(3+h)=\frac{3h+14}{2h+5}$

$f(3)=\frac{3(3)+5}{2(3)-1}$

$f(3)=\frac{9+5}{6-1}\\\\f(3)=\frac{14}{5}$

Substituting these values in limit definition of derivatives, we will get:

$f'(3)= \lim_{h \to 0} \frac{f(3+h)-f(3)}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{3h+14}{2h+5}-\frac{14}{5}}{h}$

Make a common denominator:

$f'(3)= \lim_{h \to 0} \frac{\frac{(3h+14)*5}{(2h+5)*5}-\frac{14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{\frac{5(3h+14)-14(2h+5)}{5(2h+5)}}{h}$

$f'(3)= \lim_{h \to 0} \frac{5(3h+14)-14(2h+5)}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{15h+70-28h-70}{5h(2h+5)}$

$f'(3)= \lim_{h \to 0} \frac{-13h}{5h(2h+5)}$

Cancel out h:

$f'(3)= \lim_{h \to 0} \frac{-13}{5(2h+5)}$

$f'(3)= \frac{-13}{5(2(0)+5)}$

$f'(3)= \frac{-13}{5(5)}$

$f'(3)= -\frac{13}{25}$

Therefore, $f'(3)= -\frac{13}{25}$ .

8 0

3 years ago