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devlian [24]
3 years ago
6

The total interest on a discounted loan is $40. If Bob received $480 into his bank account on drawdown, what is P0, the loan pri

ncipal? Give your answer in dollars to the nearest dollar. Do not include the dollar symbol or commas in your answer.
Mathematics
1 answer:
Elza [17]3 years ago
4 0

Answer:

440

Step-by-step explanation:

Principal+ Interest= Total amount received

Total amount= $480

Interest= $40

Principal=$480-$40=$440

Therefore, principal=440

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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
\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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