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2 years ago

you buy two rolls of wrapping paper. each roll has the same lateral surface area. what is the diameter of roll B?

1 answer:

4 0

I saw the image that should have been posted with this word problem.

The rolls of wrapping paper is cylindrical in shape.

Roll A : length = 30 inches ; diameter = 3 inches
Roll B : length = 24 inches ; diameter = unknown

Lateral surface area = 2 π r h

Lateral Surface area of Roll A = Lateral Surface area of Roll B
2 * 3.14 * 1.5in * 30in = 2 * 3.14 * r * 24 in
282.60 in² = 150.72 in * r
282.60 in² / 150.72 in = r
1.875 in = r

diameter is 2 times the radius = 1.875 in * 2 = 3.75 inches

The diameter of Roll B is 3.75 inches.

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Zach borrows $1910 at 2.6% simple interest per month. When Zach pays the loan back 3 years later, how much interest does Zach pa

mixas84 [53]

Answer:

$1884 I think

Step-by-step explanation:

$1910 - 2.6% = $1884

Sorry if it is wrong!

4 0

2 years ago

Read 2 more answers

Help me I'll make brainiest

aev [14]

<h3>Answer: RS = 16</h3>

========================================

Explanation:

The ratio QR : RS : ST is equal to 1 : 4 : 5

This means we have the following three equations

QR = x
RS = 4x
ST = 5x

For some positive number x.

Note the ratio QR : RS : ST turns into x : 4x : 5x, which reduces to 1 : 4 : 5 when we divide all three parts by x .

Along with those three equations, we'll also use QT = 40 as well.

Now turn to the segment addition postulate. Plug in the equations mentioned earlier, and solve for x .

QT = QR + RS + ST

40 = x+4x+5x

40 = 10x

10x = 40

x = 40/10

x = 4

So we know that

QR = x = 4

RS = 4x = 4*4 = 16

ST = 5x = 5*4 = 20

-----------

As a check,

QR + RS + ST = 4 + 16 + 20 = 40

which is the same as QT = 40

Therefore, we've confirmed that QR + RS + ST = QT is correct and we've confirmed our answers.

7 0

2 years ago

Solve the following system of equations:

Vinil7 [7]

The answer is (12, -1)
Work in pic:

8 0

2 years ago

A coffee shop uses two different-sized bins to store coffee beans. The volume of the smaller bin is x cubic inches and the volum

raketka [301]

Answer:

Total volume of all the bins = xS + yL

Step-by-step explanation:

Given: x cubic inches represent the volume of the smaller bin and y cubic inches represents the volume of the larger bin. The store has S smaller bins and L larger bins.

To find: an expression that represents the total volume of all the bins

Solution:

The volume of the smaller bin = x cubic inches

The volume of the larger bin = y cubic inches

Also, the store has S smaller bins and L larger bins

So,

Total volume of all the bins = xS + yL

5 0

2 years ago

<img src="https://tex.z-dn.net/?f=%20%5Ctiny%5Cint_%7Be%7D%5E%7B%7Be%7D%5E%7B2%7D%7D%20%5Cleft%28%20%5Csqrt%7Bx%20%2B%20%5Cfrac%

dlinn [17]

We have the identity

$\left(\sqrt x + \dfrac1{2^n}\right)^2 = x + \dfrac1{2^{n-1}} \sqrt x + \dfrac1{2^{2n}}$

Take the square root of both sides and rearrange terms on the right to get

$\sqrt x + \dfrac1{2^n} = \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt{x} + \dfrac1{2^{n+1}}\right)}$

Decrementing n gives

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \left(\sqrt{x} + \dfrac1{2^{n}}\right)}$

and substituting the previous expression into this, we have

$\sqrt x + \dfrac1{2^{n-1}} = \sqrt{x + \dfrac1{2^{n-2}} \sqrt{x + \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right) } }$

Continuing in this fashion, after k steps we would have

$\sqrt x + \dfrac1{2^{n-k}} = \sqrt{x + \dfrac1{2^{n-(k+1)}} \sqrt{x + \dfrac1{2^{n-k}} \sqrt{x + \dfrac1{2^{n-(k-1)}} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

After a total of n - 2 steps, we arrive at

$\sqrt x + \dfrac14 = \sqrt{x + \dfrac12 \sqrt{x + \dfrac1{2^2} \sqrt{x + \dfrac1{2^3} \sqrt{\cdots \dfrac1{2^{n-1}} \left(\sqrt x + \dfrac1{2^{n+1}}\right)}}}}$

Then as n goes to infinity, the first nested radical converges to √x + 1/4. Similar reasoning can be used to show the other nested radical converges to √x - 1/4. Then the integral reduces to

$\displaystyle \int_e^{e^2} \left(\sqrt x - \frac14\right) + \left(\sqrt x + \frac14\right) \, dx = 2 \int_e^{e^2} \sqrt x \, dx = \boxed{\frac43 \left(e^3 - e^{\frac32}\right)}$

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2 years ago