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2 years ago

Please help me answer the middle question please and thank you

Mathematics

2 answers:

oksian1 [2.3K]2 years ago

7 0

.5 to the right side is the correct answer

Vika [28.1K]2 years ago

4 0

5 to the right side is the correct answer

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In a triangle there is three angles..one is 32° other is 116° what is the third one?

frozen [14]

Answer:32 degrees

Step-by-step explanation:a triangle total is 180 degrees. So subtract 116 and 32 by 180 to get 32

8 0

2 years ago

Read 2 more answers

The population of a town increases 14% in 2 years. if the population was 6000 originally, what is the population after the incre

Brilliant_brown [7]

That would be 6000 (1+0.14) = 6840 after two years.

7 0

3 years ago

Determine if the two triangles are congruent. If so, select the appropriate congruence postulate. If not, select the reason why.

morpeh [17]

Answer:

Asa

Step-by-step explanation:

Because if u look left to right on one triangle there is a angle and in between they have a com on line and on the right theres another angle so There for it’s angle side angle

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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

Express the location of the point on the number line as both a fraction and a decimal.

scoundrel [369]

Answer:

Fraction 1/6. Decimal .6 or 1.6

Step-by-step explanation:

3 0

2 years ago