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Vanyuwa [196]
3 years ago
7

CAN SOMEONE HELP ME WITH 12 MATH QUESTIONS THROUGH IG?

Mathematics
1 answer:
lana66690 [7]3 years ago
6 0

Step-by-step explanation:

which question are u saying

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-4(n - 6) = 12<br><br> Solve the Equation.
Romashka-Z-Leto [24]
-4(n - 6) = 12

n - 6 = -12/4   divide both sides by -4

n - 6 = -3

n = -3 + 6    add both sides by 6

n = 3

hope that helps, God bless!
3 0
3 years ago
Options:<br> 14<br> 36<br> 16<br> 12
Lerok [7]

Answer:


12 million dollars.


Step-by-step explanation:


Let the sales in the first year be x million .

So the sales in the second year were x + 4 million, in the third year were 3(x + 4) and this is 48 million dollars.

So we have the equation 3(x + 4) = 48

3x + 12 = 48

3x = 36

x = 12 million dollars.

8 0
3 years ago
A school wants to put on a dance recital Middle School dancers have short programs that run 4 minutes long High School performer
sp2606 [1]

Answer:  The answer is \textup{M}+\textup{H}\leq 10~~\textup{and}4\textup{M}+7\textup{H}\leq 90.


Step-by-step explanation: Given that a school wants to put on a dance recital. The programs of Middle school dancer will be short of 4 minutes each and programmes of High school performers will be long of 7 minutes each.

Number of middle school dances is "M" and number od high school dances is "H".

Since there cannot be more than 10 dances all together, so fisrt inequality can be written as

\textup{M}+\textup{H}\leq 10.

Also, the entire recital must take maximum 90 minutes, so the second inequality can be written as

4\textup{M}+7\textup{H}\leq 90.

Thus, the pair of inequalities is

\textup{M}+\textup{H}\leq 10,\\\\4\textup{M}+7\textup{H}\leq 90.


6 0
3 years ago
2 Complete the statement An B = {...}
slamgirl [31]
B= What’s the question. Maybe try looking at math-way
4 0
3 years ago
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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