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2 years ago

Find a formula for the nth term

Mathematics

1 answer:

Airida [17]2 years ago

7 0

Answer:

$a_{n}$ = - 1.3n - 2.4

Step-by-step explanation:

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference, thus

$a_{n}$ = - 3.7 - 1.3(n - 1) = - 3.7 - 1.3n + 1.3 = - 1.3n - 2.4

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Negative 4,200 divided by 300

Pachacha [2.7K]

Answer:

-14

Step-by-step explanation:

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2 years ago

What are the zeros of the function?

d1i1m1o1n [39]

F(x) factors as
f(x) = x(x+1)(x-3)
so has zeros of x = -1, 0, 3, because these values make the factors be zero.

The appropriate choice is ...
C. -1, 0, and 3

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3 years ago

572000 square miles rounded ro the nearest hundred thousand

baherus [9]

It would be rounding it to 600000 because 600000 is the closest 100000x amount where x is an integer. I hope this helped and if I am right could I have brainliest? :)

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3 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

Mandy is saving money to buy a new car. For the past 5 weeks,she has $291 each week to her account . She also added $313 to her

arlik [135]

Answer: $2215

Step-by-step explanation:

For the past 5 weeks, Mandy adds $291 each week to her account. This gives $291×5= $1455

Money received on birthday= $313

Money started with in the account= $527

Total money in Mandy's account will be= $1455+$313+$527 = $2295

The total amount in Mandy's account is $2215

7 0

2 years ago