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3 years ago

What are the common factors for 88 and 112

2 answers:

7 0

88 - 112 / 2
44 - 56 / 2
22 - 28 / 2
11 - 14 / 2
11 - 7 / 7
11 - 1 / 11
1

2 · 2 · 2 · 2 · 7 · 11 = 1232

Hope it helped!

SCORPION-xisa [38]3 years ago

4 0

Hello!

Common factors are numbers that both numbers can be divided by

The common factors are 2, 4 and 8

The answers are 2, 4 and 8

Hope this helps!

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One hundred football players were asked how many broken bones they have suffered. The probability distribution shows the probabi

777dan777 [17]

The probability for a football player to suffer more than three roken bones will be the the sum of probabilities for 4 broken bones and five broken bones. Looking at the chart provided we can see that the probablity for 4 four broken bones is P(4)=0.10, and for 5 broken bones is P(5)=0.15. Thus the probaility,
P(x>3)=0.10+0.15=0.25.

6 0

4 years ago

Please solve −0.18÷0.03

steposvetlana [31]

Answer: -6

Step-by-step explanation: When dividing a negative number by a positive number your answer will be always negative.

Therefore, -0.18 ÷ 0.03 = -6

8 0

3 years ago

Use the method of cylindrical shells to find the volume v generated by rotating the region bounded by the curves about the given

kondor19780726 [428]

Answer:

The volume is $\frac{16\pi}{e}$

Step-by-step explanation:

* Lets talk about the shell method

- The shell method is to finding the volume by decomposing

a solid of revolution into cylindrical shells

- Consider a region in the plane that is divided into thin vertical

rectangle

- If each vertical rectangle is revolved about the y-axis, we

obtain a cylindrical shell, with the top and bottom removed.

- The resulting volume of the cylindrical shell is the surface area

of the cylinder times the thickness of the cylinder

- The formula for the volume will be: $V=\int\limits^b_a {2\pi xf(x)} \, dx$

where 2πx · f(x) is the surface area of the cylinder shell and dx is its

thickness

* Lets solve the problem

- To find the volume V generated by rotating the region bounded

by the curves y = 4e^x and y = 4e^-x about the y-axis by use

cylindrical shells

- Consider that the height of the cylinder is y = (4e^x - 4e^-x)

- Consider that the radius of the cylinder is x

- The limits are x = 0 and x = 1

- Lets take 2π and 4 as a common factor out the integration

∴ $V=\int\limits^1_0 {2\pi x(4e^{x}-4e^{-x})} \, dx$

∴ $V=2\pi(4)\int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx$

- To integrate $xe^{x}$ and $xe^{-x}$ we will use

integration by parts methods $\int\ {uv'=uv-\int{v}\,u' }\,$

∵ u = x

∴ u' = du/dx = 1 ⇒ differentiation x with respect to x is 1

∵ v' = dv/dx = e^x

- The integration e^x is e^x ÷ differentiation of x (1)

∴ $v=\int\ {e^{x}}\, dx= e^{x}$

∴ $\int\ {xe^{x}} \, dx=xe^{x}-\int\ e^{x}\, dx=xe^{x}-e^{x}$

- Similar we will integrate xe^-x

∵ u = x

∴ u' = du/dx = 1

∵ v' = dv/dx = e^-x

- The integration e^-x is e^x ÷ differentiation of -x (-1)

∴ $v=\int\ {e^{-x}} \, dx=-e^{-x}$

∴ $\int\ {x}e^{-x}\, dx=-xe^{-x}+\int\ {e^{-x}} \, dx=-xe^{-x}-e^{-x}$

∴ V = $8\pi \int\limits^1_0 ({xe^{x}-xe^{-x})} \, dx=8\pi[xe^{x}-e^{x}+xe^{-x}+e^{-x}]$ from 0 to 1

- Lets substitute x = 1 minus x = 0

∴ $V=8\pi[(1)(e^{1})-(e^{1})+(1)(e^{-1})+(e^{-1})-(0)(e^{0})+(e^{0})-(0)(e^{0})-(e^{0})]$

∴ $V=8\pi[e^{1}-e^{1}+e^{-1}+e^{-1}-0+1-0-1]=8\pi[2e^{-1}]=16\pi e^{-1}$

∵ $e^{-1}=\frac{1}{e}$

∴ $V=\frac{16\pi}{e}$

3 0

4 years ago

Determine if the function is odd even or neither calculator

BlackZzzverrR [31]

It is not odd or even. It is neither. It's a calculator.

5 0

3 years ago

I need help please!!! My teacher doesn't teacher and she has asked us to do a problem I don't know how to do it. Can somebody ex

ra1l [238]

Using exponential functions, it is found that:

a) Since the amount of caffeine will be less than 50 mg, the patient will be ready for the blood test by 6 a.m.

b) The patient could have ingest 231 milligrams of caffeine.

A decaying exponential function is modeled by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.
r is the decay rate, as a decimal.

In this problem:

Caffeine metabolize at a rate of 13% per hour, hence $r = 0.13$ .

Then:

$A(t) = A(0)(1 - r)^t$

$A(t) = A(0)(1 - 0.13)^t$

$A(t) = A(0)(0.87)^t$

Item a:

The coffee cup contains 150 milligrams of caffeine, hence $A(0) = 150$ .

At 6 a.m., it is 8 hours after drinking the coffee, hence we have to find A(8).

$A(t) = A(0)(0.87)^t$

$A(8) = 150(0.87)^8$

$A(8) = 49.2$

Since the amount of caffeine will be less than 50 mg, the patient will be ready for the blood test by 6 a.m.

Item b:

This A(0), considering A(11) = 50, hence:

$50 = A(0)(0.87)^{11}$

$A(0) = \frac{50}{(0.87)^{11}}$

$A(0) = 231$

The patient could have ingest 231 milligrams of caffeine.

A similar problem is given at brainly.com/question/25537936

3 0

2 years ago