Ask question

Ask question

All categories

3 years ago

13

Area of the base of cuboid shaped tank is 8400 cm. When 420 liters of water is put into the tank. find the height of the water i

n tank.

1 answer:

Oksi-84 [34.3K]3 years ago

4 0

Answer:

hope this helps you get a good grades

You might be interested in

Mary Lou has 2 more nickels than pennies, and she has 30 coins all together. Use x for the number of pennies.

qwelly [4]

She has 20 pennies and two nickels

6 0

3 years ago

A school principal is trying to decide how long the breaks should be between periods. He plans to time how long it takes several

Andreyy89

he could use a bar graph to plot data

6 0

3 years ago

Given limit of f(x) = -4 as x approaches c and limit of g(x) =1/5 as x approaches c. What is limit of g(x)/f(x) as x approaches

Dahasolnce [82]

Answer:

$\displaystyle \frac{-1}{20}$

General Formulas and Concepts:

<u>Calculus</u>

Limit Property [Division]: $\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{ \lim_{x \to c} f(x)}{ \lim_{x \to c} g(x)}$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

<em /> $\displaystyle \lim_{x \to c} f(x) = -4$ <em />

<em /> $\displaystyle \lim_{x \to c} g(x) = \frac{1}{5}$ <em />

<em />

<u>Step 2: Solve</u>

Substitute in limits [Limit Property - Division]: $\displaystyle \lim_{x \to c} \frac{g(x)}{f(x)} = \frac{ \frac{1}{5} }{ -4 }$
Simplify: $\displaystyle \lim_{x \to c} \frac{g(x)}{f(x)} = \frac{-1}{20}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

Book: College Calculus 10e

8 0

3 years ago

Read 2 more answers

Question 2 ciencjejcjebxjd

Lelechka [254]

Answer:

x6 7 8 9 8 y 77899

Step-by-step explanation:

4 0

3 years ago

Find the average value of the function on the given interval.<br> f(x)=x^2e^2x; [0,2]

ss7ja [257]

Answer:

$f_{avg}=\frac{1}{8}(5e^4-1)$

Step-by-step explanation:

We are given that a function

$f(x)=x^2e^{2x}$

We have to find the average value of function on the given interval [0,2]

Average value of function on interval [a,b] is given by

$\frac{1}{b-a}\int_{a}^{b}f(x)dx$

Using the formula

$f_{avg}=\frac{1}{2-0}\int_{0}^{2}x^2e^{2x} dx$

By Parts integration formula

$\int(uv)dx=u\int vdx-\int(\frac{du}{dx}\int vdx)dx$

u= $x^2$ and v= $e^{2x}$

Apply by parts integration

$f_{avg}=\frac{1}{2-0}([\frac{x^2e^{2x}}{2}]^{2}_{0}-\int_{0}^{2}(2x\times \frac{e^{2x}}{2}dx)$

$f_{avg}=\frac{1}{2}(2e^4-(\int_{0}^{2}xe^{2x})dx$

$f_{avg}=\frac{1}{2}(2e^4-0-([\frac{xe^{2x}}{2}]^{2}-\frac{1}{4}[e^{2x}]^{2}_{0}))$

$f_{avg}=\frac{1}{2}(2e^4-e^4+\frac{1}{4}(e^4-1))=\frac{1}{2}(e^4+\frac{1}{4}e^4-\frac{1}{4})$

$f_{avg}=\frac{1}{8}(5e^4-1)$

5 0

4 years ago