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3 years ago

. 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

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

length 10 in/6.35-------> 1.57 in*2.54 cm/in-----> 4 cm
width 5 in/6.35---------> 0.79 in*2.54 cm/in-----> 2 cm
height 5 in/6.35--------> 0.79 in*2.54 cm/in----> 2 cm

3 0

3 years ago