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Vesnalui [34]
3 years ago
5

HELP ME ASAP! BRAINLIEST UP FOR GRABS

Mathematics
2 answers:
natita [175]3 years ago
8 0

Answer:

-5 ≤ x≤ 3

Step-by-step explanation:

The domain is the values for x

x starts and -5 and includes -5 since the circle is closed

and goes to 3 and  includes 3 since the circle is closed

-5 ≤ x≤ 3

k0ka [10]3 years ago
6 0

Answer:

first option

Step-by-step explanation:

The domain are the values from the x- axis that can be input into the function.

The closed circles at the ends of the graph indicate that x can equal these values.

left side value of x = - 5 and right hand value of x = 3, thus

domain is - 5 ≤ x ≤ 3

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What do you think Thomas Edison meant by the word perspiration? How do those words apply to what you've learned about businesses
EastWind [94]

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<h3>What does Thomas Edison's reference mean?</h3>

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5 0
2 years ago
"find the reduction formula for the integral" sin^n(18x)
dexar [7]
Let

I(n,a)=\displaystyle\int\sin^nax\,\mathrm dx
For demonstration on how to tackle this sort of problem, I'll only work through the case where n is odd. We can write

\displaystyle\int\sin^nax\,\mathrm dx=\int\sin^{n-2}ax\sin^2ax\,\mathrm dx=\int\sin^{n-2}ax(1-\cos^2ax)\,\mathrm dx
\implies I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx

For the remaining integral, we can integrate by parts, taking

u=\sin^{n-3}ax\implies\mathrm du=a(n-3)\sin^{n-4}ax\cos ax\,\mathrm dx\mathrm dv=\sin ax\cos^2ax\,\mathrm dx\implies v=-\dfrac1{3a}\cos^3ax

\implies\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx=-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{a(n-3)}{3a}\int\sin^{n-4}ax\cos^4ax\,\mathrm dx

For this next integral, we rewrite the integrand

\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax
\implies\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx=I(n-4,a)-2I(n-2,a)+I(n,a)

So putting everything together, we found

I(n,a)=I(n-2,a)-\displaystyle\int\sin^{n-2}ax\cos^2ax\,\mathrm dx
I(n,a)=I(n-2,a)-\left(-\dfrac1{3a}\sin^{n-3}ax\cos^3ax+\dfrac{n-3}3\displaystyle\int\sin^{n-4}ax\cos^4ax\,\mathrm dx\right)
I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax
\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax

\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax
7 0
2 years ago
Find a6 for an arithmetic sequence where a1=3x+1 and d=2x+6​
abruzzese [7]

Answer:

a₆ = 13x + 31

Step-by-step explanation:

The nth term of an arithmetic sequence is

a_{n} = a₁ + (n - 1)d

Here a₁ = 3x + 1 and d = 2x + 6 , then

a₆ = 3x + 1 + 5(2x + 6)

    = 3x + 1 + 10x + 30

    = 13x + 31

7 0
3 years ago
Read 2 more answers
What is the value of x
VMariaS [17]
The two angles on top are supplementary angles meaning they should add up to 180
72+3x=180
3x=180-72
3x=108
x=36
Mark me brainliest!

7 0
2 years ago
Why do square numbers not end in 2, 3, 7 or 8. Why is it so that to identify whether a number is a square or not you look at the
Maksim231197 [3]

Answer:

I don't know dear

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3 0
3 years ago
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