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tankabanditka [31]
3 years ago
5

Sara Blake's grandmother gave her $3000.00 to save for college. She put it in a savings account that earns 6% interest per year.

The money was in the account for 8 years. How much simple interest did she earn?
Mathematics
2 answers:
Elina [12.6K]3 years ago
6 0

• P is the principal amount, $3000.00.

• r is the interest rate, 6% per year, or in decimal form, 6/100=0.06.

• t is the time involved, 8 years time periods.

• So, t is 8 year time periods.

To find the simple interest, we multiply 3000 × 0.06 × 8 to get that:

The interest is: $1440.00

Zigmanuir [339]3 years ago
5 0

Answer:

1440

Step-by-step explanation:

to get the interest you have to multiply the principal and rate and time (P x R x T)

I = 3000 x 0.06 x 8     (remember 6% turns into a decimal 0.06)

I = 1140

~batmans wife dun dun dun.....

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A six-sided die with sides labeled 1 through 6 will be rolled once. Each number is equally likely to be rolled.
ivanzaharov [21]

Answer:

2/3

Step-by-step explanation:

The numbers greater then 2 are 3, 4, 5, 6

That's four numbers that are greater then two.

or 4/6, then you just reduce it by dividing by two on the top and bottom

4/2 =2

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4 0
3 years ago
Evaluate the surface integral:S
rjkz [21]
Assuming S does not include the plane z=0, we can parameterize the region in spherical coordinates using

\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v

Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v

So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
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