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maksim [4K]
3 years ago
9

Paul has $2,400 in a savings account earning 3.4% simple interest. How much more interest would he earn in 5 years if the intere

st were compounded annually?
Mathematics
1 answer:
bija089 [108]3 years ago
7 0
Amount using compound interest = 2400(1 + 0.034)^5 = 2400(1.034)^5 = 2400(1.1820) = $2,836.70

Amount using simple interest = 2400(1 + 5 x 0.034) = 2400(1 + 0.17) = 2400(1.17) = $2,808

Difference = $2,836.70 - $2,808 = $28.70

Therefore, <span>he will earn $28.70 more interest in 5 years if the interest were compounded annually</span> .
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Cómo se resuelve la ecuación raíz cuadrada (×) -4=8
denis23 [38]
X=12 that's is the answers
4 0
3 years ago
Robbie was given $80 to spend at the Fair. His admission to the park costs $15.50 and each ride cost $5. He anticipates the cost
Ad libitum [116K]

Answer:

7 or 8

Step-by-step explanation:

Robbie was given $80.

His admission costs $15.50

80 - 15.50 = $64.50

He anticipates the cost of food at the fair will be $25.

64.50 - 25 = $39.50

What is the maximum number of rides he can take at the Fair?

(Total $ after admission and cost of food/ ride cost)

$39.50 / $5 = 7.9

4 0
2 years ago
Read 2 more answers
PLs solve i will award brainlist
ollegr [7]

Answer:

x = 120 degrees

Step-by-step explanation:

angle below x is 40 degrees because it is a corresponding angle.

x = 180 - 40 - 20 = 120 degrees

7 0
3 years ago
Read 2 more answers
Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
Sophie [7]
Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
=-2\pi

So, the flux over the paraboloid alone is

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
6 0
3 years ago
How do i simplify (3/5 - 1) / 1 2/3?
Tpy6a [65]
I hope this helps you

4 0
3 years ago
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