Ask question

Ask question

All categories

2 years ago

12

he ratio of your monthly allowance to your friend’s monthly allowance is 5 : 3. Both of your monthly allowances total $40. How m

uch is each allowance?

2 answers:

Elena-2011 [213]2 years ago

6 0

Answer:

your allowance is $25 and your friends is $15

Step-by-step explanation:

40/8*3=15

40/8*5=25

larisa [96]2 years ago

5 0

Yours is 25 and the friends is 15

You might be interested in

Plz help me with this

Mkey [24]

Answer: B) A = 750(1.04)ⁿ

<u>Step-by-step explanation:</u>

The formula for compounded annually is: A = P(1 + r)ⁿ where

A (amount accrued) = <em>unknown</em>
P (amount invested) = $750
r (interest rate) = 4% -->(0.04)
t (time in years) = <em>unknown</em>

A = 750(1 + 0.04)ⁿ

= 750(1.04)ⁿ

8 0

2 years ago

What is the value of s?

beks73 [17]

Answer:

To what

Step-by-step explanation:

3 0

2 years ago

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

as

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

so

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

2 years ago

Find the prime factorization of 15

Anton [14]

The prime factorization of 15 is 1 * 3 * 5

8 0

3 years ago

Read 2 more answers

What is the solution to -(6m+8)=4(17-m)

ale4655 [162]

<u>Answer:</u>

The solution to the problem is -38.

<u>Step-by-step explanation:</u>

-(6m + 8) = 4(17 - m)
=> -6m - 8 = 68 - 4m
=> -6m + 4m = 68 + 8
=> -2m = 76
=> m = 76/-2
=> m = -38

Hence, <u>the solution to the problem is -38.</u>

Hoped this helped.

$BrainiacUser1357$

5 0

2 years ago