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nalin [4]
4 years ago
15

If 6 more than the product of a number and - 2 is greater than 10, which of the following could be that number?

Mathematics
1 answer:
Minchanka [31]4 years ago
6 0

Answer:

-2x + 6 > 10

Step-by-step explanation:

Let the unknown number be x

the product of x and -2 is -2x

6 more than -2x is written as

-2x + 6

if the result of the above is greater than 10, then

-2x + 6 > 10

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Sonja [21]
The shaded area is 75%
5 0
3 years ago
Write the equation of the line shown (3,6) (3,-4)
lozanna [386]

Answer:

The equation of line is given as  x=3

Step-by-step explanation:

Point on the line given :

(3,6),(3,-4)

We can find the slope of the line first from the given points.

Slope(m)=\frac{y_2-y_1}{x_2-x_1} = \frac{-4-6}{3-3}=\frac{-10}{0}== Undefined.

The slope of the line is undefined which means that the line is parallel to y-axis.

From the points known to us we can tell that the line passes through x=3

∴ The equation of line is given as  x=3

6 0
3 years ago
A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rat
snow_lady [41]

Your question: A population of insects grows exponentially, as shown in the table. Suppose the increase in population continues at the same rate.

Answer: 1730

That's the answer mate, if you need more help or if you need to know how I got the answer.. just message me.

Good luck!

6 0
3 years ago
Kim’s hair grows a half inch per month. How many months will it take for her hair to grow 12 inches?
Contact [7]

Answer:

24 months

Step-by-step explanation:

Cross-multiply

0.5 inches -------------- 1 month

12 inches ---------------- x months

0.5/12 = 1/x ===> 0.5x = 12 ===> x = 12/0.5  ===> x = 24 months

3 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
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