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2 years ago

15

The formula A=6s² gives the suface area A of a cube with a side length s. What is the suface area of a cube that has a side leng

th of 5 inches? Plz answer fast

1 answer:

Salsk061 [2.6K]2 years ago

3 0

Answer:

150 square inches.

Step-by-step explanation:

Since the surface area of a cube is 6 times the square of the side length of the cube, then that means the surface area of a cube that has a side length of 5 has a surface area of 6 x 5 x 5. 5 times 5 = 25, and 25 x 6 = 150

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1km make how many cm

Ugo [173]

Answer:

100000 centimetres

Step-by-step explanation:

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2 years ago

Read 2 more answers

X+5+x11=10 <br> Solve for X and show the work please!

kenny6666 [7]

Answer:

x=5/12

Step-by-step explanation:

Simplify left side by adding 11x+x

12x+5=10

Get x alone on left side by subtracting 5 on both sides

12x=5

Get X alone by dividing both sides by 12

x=5/12

6 0

3 years ago

A retailer buys a coat in bulk at a unit price of $45. The coat is marked up to 28% to sell in the store. What is the price of t

inessss [21]

The answer is $57.60

(The original price) 45+ 12.60 (28% of 45) = 57.60

6 0

2 years ago

X²+8x-84=0 <br><br>Formúla General

Brums [2.3K]

Answer:

$x_{1} =-14\\\\x_{2} =6$

Step-by-step explanation:

$x^{2} +8x-84=0\\\\x=\dfrac{-b \pm \sqrt{b^{2}-4ac } }{2a} \\\\a=1; \: \: b=8; \: \: c=-84\\\\x=\dfrac{-8 \pm \sqrt{8^{2}-4 \cdot 1 \cdot (-84) } }{2 \cdot 1}=\dfrac{-8 \pm \sqrt{400 } }{2}=\dfrac{-8 \pm20}{2} =-4 \pm 10\\\\x_{1} =-14\\\\x_{2} =6$

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

2 years ago