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Svetlanka [38]
3 years ago
7

Solving the equation:-9h-6+ 12h +40=20

Mathematics
2 answers:
BaLLatris [955]3 years ago
7 0
First, I am going to simplify the equation..

-9h-6+12h+40=20     Original equation.

3h-6+40=20     Simplify.

3h-6+6+40=20+6     Add 6 on each side.

3h+40=26     Simplify.

3h+40-40=26-40     Subtract 40 from each side.

3h=-14     Simplify.

3h÷3=-14÷3     Divide each side by 3.

h=-14/3     Simplify.

h=-4 2/3     Change to a mixed number.

The value of h is -4 2/3.
MakcuM [25]3 years ago
5 0
12h-9h=3h move everything else to other side
3h=20-40+6
3h=-14
divide by 3 
h=-14/3
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P greater than or equal to -4
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Pls help with all 4 questions, and show a step by step explanation (show how you set up the problems)
Ainat [17]
<h2>Answer:</h2>

<u>Question 10:</u>

21.75 ÷ 30 = 0.725

Move the decimal backwards 2 times

72.5

<em>Answer: </em>72.5%

<u>Question 11:</u>

Sales: $768

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Commission: $30.72

768 x 4 = 3,072

3,072 ÷ 100 = 30.72

<em>Answer:</em> $30.72

<u>Question 12:</u>

Selling Price: $39.98

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Sales Tax: $1.80

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1.799

1.8

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<u>Question 13:</u>

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8 0
2 years ago
Given that u = 2.89 and v = 5.7, find the value of G when:<br> a G = 5u + 2y
EleoNora [17]

Answer:

28.85

Step-by-step explanation:

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5 0
3 years ago
Solve using Fourier series.
Olin [163]
With 2L=\pi, the Fourier series expansion of f(x) is

\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L
\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx

where the coefficients are obtained by computing

\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx
\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx

\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx
\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx

\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx
\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx

You should end up with

a_0=0
a_n=0
(both due to the fact that f(x) is odd)
b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)

Now the problem is that this expansion does not match the given one. As a matter of fact, since f(x) is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of f(x), which is to say we're actually considering the function

\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3

and enforcing a period of 2L=2\pi. Now, you should find that

\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)

The value of the sum can then be verified by choosing x=0, which gives

\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)
\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots

as required.
5 0
3 years ago
I need help with this​
VLD [36.1K]

Answer:

im sorry wish i could

Step-by-step explanation:

4 0
2 years ago
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