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LiRa [457]
3 years ago
14

TWO questions! PLEASE HELP ME!

Mathematics
2 answers:
Mariulka [41]3 years ago
8 0

Answer:

The first one is C I don't know the second one though

Step-by-step explanation:

Black_prince [1.1K]3 years ago
4 0
The first one is C and I feel like the answer is D for the second one but I dont know.. sorry
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36×78<br><img src="https://tex.z-dn.net/?f=5551%20%5Ctimes%20696" id="TexFormula1" title="5551 \times 696" alt="5551 \times 696"
KATRIN_1 [288]
If the question is asking what 36x78 is the answer is 2,808
3 0
3 years ago
The difference between 9/5 and 7/5 is
Agata [3.3K]

Answer:

2/5

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
Readme [11.4K]

Answer:

3/11

Step-by-step explanation:

Divide both the numerator and denominator by 2.

\dfrac{6}{22}=\dfrac{2*3}{2*11}=\dfrac{3}{11}

8 0
3 years ago
Which equation are equivalent to 1/5+2/3|2-x|=4/15?
disa [49]

Answer:

• 2/3|2 -x| = 1/15

• |2 -x| = 1/10

• 1/5 +|4/3 -2/3x| = 4/15

Step-by-step explanation:

Starting with the given equation, subtract 1/5 = 3/15.

(2/3)|2 -x| = 4/15 - 3/15

(2/3)|2 -x| = 1/15 . . . . . . . . matches the first choice

Now, multiply by 3/2.

|2 -x| = 3/30

|2 -x| = 1/10 . . . . . . . . . . . . matches the third choice

___

If you decide to distribute the coefficient 2/3 instead, you have

1/5 + |2·2/3 -x·2/3| = 4/15

1/5 + |4/3 -2/3x| = 4/15 . . . . matches the selected choice

5 0
3 years ago
Read 2 more answers
Find the half range Fourier sine series of the function
worty [1.4K]
The full range is -\pi (length 2L=2\pi), so the half range is L=\pi. The half range sine series would then be given by

f(x)=\displaystyle\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L=\sum_{n\ge1}b_n\sin nx

where

b_n=\displaystyle\frac2L\int_0^Lf(x)\sin\dfrac{n\pi x}L\,\mathrm dx=\frac2\pi\int_0^\pi(\pi-x)\sin nx\,\mathrm dx

Essentially, this is the same as finding the Fourier series for the function

\begin{cases}g(x)=\begin{cases}\pi-x&\text{for }0

Integrating by parts yields

b_n=\dfrac2\pi\left(\dfrac\pi n-\dfrac{\sin n\pi}{n^2}\right)=\dfrac2n

So the half range sine series for this function is simply

f(x)=\displaystyle\sum_{n\ge1}\frac{2\sin nx}n
5 0
3 years ago
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