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3 years ago

One morning it was 29°F in Columbus, Ohio and 27°F in Pittsburgh, Pennsylvania.Was it warmer in Columbus or Pittsburgh?

Mathematics

1 answer:

Lina20 [59]3 years ago

5 0

Columbus, higher the number, hotter it is

You might be interested in

Given that p=9i+12j and q=-6i-8j. Evaluate |p-q|-{|p|-|q|}

krok68 [10]

Answer:

$|p-q|-(|p|-|q|) = 20$

Step-by-step explanation:

First let's find the value of 'p-q':

$p - q = 9i + 12j - (-6i - 8j)\\p - q = 9i + 12j + 6i + 8j\\p - q = 15i + 20j\\$

To find |p-q| (module of 'p-q'), we can use the formula:

$|ai + bj| = \sqrt{a^{2}+b^{2}}$

Where 'a' is the coefficient of 'i' and 'b' is the coefficient of 'j'

So we have:

$|p - q| = |15i + 20j| = \sqrt{15^{2}+20^{2}} = 25$

Now, we need to find the module of p and the module of q: $|p| = |9i + 12j| = \sqrt{9^{2}+12^{2}} = 15$

$|q| = |-6i - 8j| = \sqrt{(-6)^{2}+(-8)^{2}} = 10$

Then, evaluating |p-q|-{|p|-|q|}, we have:

$|p-q|-(|p|-|q|) = 25 - (15 - 10) = 25 - 5 = 20$

7 0

3 years ago

Which property of equality could be used to solve x - (-6) = 18?

oksian1 [2.3K]

Answer:subtraction property

Step-by-step explanation:

Two a negative and a negative makes a positive so it would be x+6=18 so u do the opposite on both sides so it would be subtraction

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

Chandra spends 15 minutes doing 4 math problems.She spends the same amount of of time on each problem.How many minutes does she

Serjik [45]

Answer: 3 minutes and 45 seconds

Step-by-step explanation: first, you divide 15 by 4 and that's your answer. Hope this helps!

6 0

3 years ago

?????????????????????

anyanavicka [17]

Answer:

It would be 1/8

Step-by-step explanation:

32 divided by 4 is 8

8 0

3 years ago