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gogolik [260]
3 years ago
7

. Simplify: (-4a+ b–2c) –(3a+ 2b–c)

Mathematics
1 answer:
Naily [24]3 years ago
4 0

Answer:

(-4a + b – 2c) – (3a + 2b –c)  

-4a+b-2c-3a-2b+c  

-7a-b-c

Step-by-step explanation:

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How do I find the slope if a line when y=4x+5
DaniilM [7]

Answer:

slope =4

Step-by-step explanation:

The slope intercept equation of a line is written in the form

y= mx+b

where m is the slope and b is the y intercept.

Since y =4x+5 is written is this form, we can see that 4 is the slope and 5 is the y intercept

3 0
3 years ago
Find the slope of the line through each pair of points (-12,0) (20,0)
AlexFokin [52]

\frac{y2}{x2}  -  \frac{y1}{x1}
That is the formula to find the slope. Hope it will help you.
8 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
Who used less sugar 5/10 or 5/6
abruzzese [7]
It would be 5/6 because the denominator Is smaller than 10

can I be brainliest
7 0
3 years ago
Read 2 more answers
A rectangular prism has a volume of 16 cm3 and a total surface area of 40 cm2. A similar prism has length 10 in., width 5 in., a
Kruka [31]
We know that
[volume of the similar prism]=10*5*5----> 250 in³
volume original prism=16 cm³
 1 in³---------> 16.3871 cm³
X----------> 16
x=16/16.3871------> x=0.9764 in³

[volume of the similar prism]=[scale factor]³*[volume original prism]
[scale factor]³=[volume of the similar prism]/[volume original prism]
[scale factor]³=[250]/[0.9764]------> 256.05
scale factor=∛256.05-------> 6.35

<span>the dimensions of the original prism are
1 in-----> 2.54 cm

</span>length 10 in/6.35-------> 1.57 in*2.54 cm/in-----> 4 cm
<span>width 5 in/6.35---------> 0.79 in*2.54 cm/in-----> 2 cm
</span><span>height 5 in/6.35--------> 0.79 in*2.54 cm/in----> 2 cm</span>
3 0
3 years ago
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