Ask question

Ask question

All categories

2 years ago

9

Mitchell received a stipend at the beginning of the semester to buy textbooks. During the semester, he spent $430.84 on textbook

s. He had $136.04 of the stipend left at the end of the semester.
x - $430.84 = $136.04

How much was Mitchell's stipend?
A.
$566.88
B.
$702.92
C.
$362.82
D.
$294.80

1 answer:

tatyana61 [14]2 years ago

7 0

Answer:

$566.88

Step-by-step explanation:

$430.84 + $136.04 = $566.88

You might be interested in

A man shares a $1,200,00 inheritance between his 3 sons in ratio 1: 3: 6. Calculate the difference between the largest share and

soldi70 [24.7K]

A man's inheritance money = $1200

Given :

He shared his inheritance between his 3 sons in the ratio 1:3:6

Then, the money the first son got :

$= \tt \frac{1}{10} \: of \: 1200$

$=\tt \frac{1}{10} \times 1200$

$=\tt \frac{1 \times 120}{10}$

$= \tt \frac{1200}{10}$

$\color{plum} =\tt\$ \: 120$

▪︎Thus, the first son inherited $120.

Money the second son got :

$= \tt \frac{3}{10} \: of \: 1200$

$= \tt \frac{3}{10} \times 1200$

$=\tt \frac{3 \times 1200}{10}$

$= \tt \frac{3600}{10}$

$\color{plum} =\tt \$360$

▪︎Thus, the second son inherited $360.

Money the third son got :

$=\tt \frac{6}{10} \: of \: 1200$

$=\tt \frac{6}{10} \times 1200$

$= \tt \frac{6 \times 1200}{10}$

$= \tt \frac{7200}{10}$

$\color{plum} = \tt\$720$

▪︎Thus, the third son inherited $720.

We know that :

The largest share = $720

The smallest share = $120

Difference between the largest and smallest share :

$= \tt720 - 120$

$\color{hotpink}\tt = 600$

<h3> ●=> Therefore :</h3>

▪︎<em>The difference between the largest and smallest share = 600</em>

7 0

3 years ago

Is 64 a perfect cube?

3241004551 [841]

It's a perfect square because 8 times 8 equals 64.

6 0

3 years ago

Read 2 more answers

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

2 years ago

I don't understand this.

sladkih [1.3K]

You mean the word or number

7 0

3 years ago

Find angles b,c and d

Mazyrski [523]

Answer:

B = 73°, c = 107°, d= 107°

Step-by-step explanation:

$b + 107 \degree = 180 \degree..(straight \: line \: \angle s) \\ b = 180 \degree - 107 \degree \\ \huge \red{ \boxed{b = 73 \degree}} \\ \\ \huge \purple{ \boxed{c = 107 \degree }}\\..(vertical \: \angle s) \\ \\ \huge \orange{ \boxed{d = 107 \degree }}\\..(corresponding \: \angle s) \\ \\$

8 0

3 years ago