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MAXImum [283]
3 years ago
13

Please find the perimeter!

Mathematics
1 answer:
puteri [66]3 years ago
7 0

Answer:

1. half circle: C = (2πr) / 2

a. C = (2π4) / 2 ≈ 12.566

2. triangle:

a. find slant / hypotenuse.

i. 4² + 10² = c²

ii. 16 + 100 = c²

iii. c² = 116

iv. c ≈ 10.770

b. add both slants to find triangular area

i. 10.77 + 10.77 ≈ 21.54

3. add:

a. 12.566 + 21.54 ≈ 34.106 cm

hope this helps :)

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How many pieces 3 and three fifths in. long can be cut from 28 metal rods each 15 in.​ long? Disregard waste.
Murrr4er [49]

Answer: 112 pieces


Step-by-step explanation:

1. Rewrite 3^{\frac{3}{5}} as a decimal. Divide the numerator and the denominator of the fractional part and add it to the integer part:

3+0.6=3.6in

2. You know that each metal rod is 15 inches long. Then, you must divide the lenght of each one by 3.6:

pieces=\frac{15in}{3.6in}4.16

3. The waste is 0.16, therefore, you obtain 4 pieces from a metal rod. The total pieces obtained from 28 metal rods are:

total=4pieces*28=112pieces

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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
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