Ask question

Ask question

All categories

2 years ago

11

Find the volume of the solid figure shown below. The area of the base is 40 square inches, and the height of the figure is 7 inc

hes. 280 in. 3 68 in. 3 47 in. 3 108 in. 3

2 answers:

Blizzard [7]2 years ago

7 0

Answer:

280 in. ^3

Step-by-step explanation:

40 x 7

Mice21 [21]2 years ago

6 0

Answer:

280in

Step-by-step explanation:

You might be interested in

If product is addition and factor is to multiply then what is division

harina [27]

Quotient: The answer to a division problem

8 0

2 years ago

A check is _____.

torisob [31]

Answer:

B

Step-by-step explanation:

A check is an I.O.U. or a promissory note.

3 0

2 years ago

Read 2 more answers

Are shape I and shape Il similar? If so, give the dilation that proves they are similar. If not, explain why the shapes are not

joja [24]

Answer:

They are similar.

Step-by-step explanation:

They are similar because, the top triangle has one more square to the height as well as the width. If you take one square away from the height and width the top triangle is the same size as the bottom triangle.

7 0

2 years ago

Which one pls help me

vampirchik [111]

Answer:

C

Step-by-step explanation:

If one answer is longer then the others, Then that's the correct answer

5 0

2 years ago

Read 2 more answers

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

2 years ago