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3 years ago

Can someone please help!! I’m stuck on this question

Mathematics

1 answer:

nataly862011 [7]3 years ago

3 0

Answer:

-12

Step-by-step explanation:

Farenheit is on the x-axis (horizontal) so you go to 10, then find the celsius which is on the y-axis (vertical) and the line is around -12. Hope this helps!

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When adding two numbers, such as 123 and 423, care is taken to first line them up and then add like digits. How does expanding t

Ivanshal [37]

Since a polynomial is where we have like terms such as (1 x 10²) and (4 x 10²), we can add these up using the distributive property to get (5 x 10²) but still keep the 10². For example, it's similar to if we had 2x²+3x²=5x². The x² is still there, but we add up the 2 and 3. Similarly, we can add these up for 10^1 and 10^0

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3 years ago

Determine which equation below has a solution of x= 1/2

masya89 [10]

Answer:2x-1=0

Step-by-step explanation:

x=1/2

Cross multiply

2x=1

Subtract 1 from both sides

2x-1=1-1

2x-1=0

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3 years ago

What is this helppp 1/2 + 1/5

user100 [1]

Answer:

$\frac{1}{2} + \frac{1}{5} \\ \\ = \frac{5 + 2}{10} \\ \\ = \frac{7}{10}$

4 0

2 years ago

Read 2 more answers

HELP! WILL MARK BRAINLIEST WITH A EXPLANATION!!!!please & thank u

mamaluj [8]

Answer:

The answer is A

Step-by-step explanation:

8 0

3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago