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3 years ago

PLEASE HELP!!!!! Geometry

Mathematics

1 answer:

barxatty [35]3 years ago

3 0

Answer:

53°

Step-by-step explanation:

i will use ∩ to represent arc since there is no arc sign

if ∩EF = 34, then ∡EFC = 34

think of EFC as an angle in the larger circle

since EFC = 34, ABC = 34 because they are opposite angles and opposite angles are always congruent.

ABD = AB + BD = 19 + 34 = 53

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Find the radius of the button.<br><br> radius: __ in.

Marat540 [252]

Answer:

1 3/4 in

Step-by-step explanation:

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2 years ago

Madeline estimated the product of 3.56 and 8.3

ycow [4]

Answer:29.548

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3 years ago

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"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

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

$\sin^{n-4}ax\cos^4ax=\sin^{n-4}ax(1-\sin^2ax)^2=\sin^{n-4}ax-2\sin^{n-2}ax+\sin^nax$
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So putting everything together, we found

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$I(n,a)=I(n-2,a)-\dfrac{n-3}3\bigg(I(n-4,a)-2I(n-2,a)+I(n,a)\bigg)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$
$\dfrac n3I(n,a)=\dfrac{2n-3}3I(n-2,a)-\dfrac{n-3}3I(n-4,a)+\dfrac1{3a}\sin^{n-3}ax\cos^3ax$

$\implies I(n,a)=\dfrac{2n-3}nI(n-2,a)-\dfrac{n-3}nI(n-4,a)+\dfrac1{na}\sin^{n-3}ax\cos^3ax$

7 0

2 years ago

How do you solve this equation? 5p + 10 = 6p + 22

lisov135 [29]

Answer:

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3 years ago

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densk [106]

(a) The equation is given as Q=45e^1.03t

where e=2.718

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