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astra-53 [7]
2 years ago
6

16. J.D. opened a savings account with $425. After

Mathematics
1 answer:
snow_lady [41]2 years ago
8 0

Answer:

1.2%

Step-by-step explanation:

Solving our equation

r = 10.2 / ( 425 × 2 ) = 0.012

r = 0.012

converting r decimal to a percentage

R = 0.012 * 100 = 1.2%/year

The interest rate required to

accumulate simple interest of $ 10.20

from a principal of $ 425.00

over 2 years is 1.2% per year.

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Last week, Benjamin spent $82.80 on groceries and $132.48 on bills. The amount he spent on groceries
Genrish500 [490]

Answer:

amount spend on groceries = 82.80

percent spend in groceries = 15%

let total paycheck be x

15%of x= 82.80

x=552

percent spend on bills = 132.48/552 = 24%

Step-by-step explanation:

6 0
2 years ago
This is for a mastery
valkas [14]

Answer:b

Step-by-step explanation:

3 0
3 years ago
Use the net to compute the surface area of the three-dimensional figure. A) 210 units B) 225 units C) 240 units D) 280 units
11111nata11111 [884]
The answer would most commonly  be C only because a square is 360 degrees and the surface area would be a 6th of that which is 60 then i multiplied it by 4 because of the four corners signaling the height and or length of the 2D figure square which gives me C.
5 0
3 years ago
Answer this question for me please cause idk it
Greeley [361]

H=-4.9t^2+25t 0=t(-4.9t+25) T=25÷4.9 T= approx 5.1 which is closes to 5.0  So answer will be 5.0

8 0
2 years ago
A sphere of radius r is cut by a plane h units above the equator, where
Anika [276]
Consider the top half of a sphere centered at the origin with radius r, which can be described by the equation

z=\sqrt{r^2-x^2-y^2}

and consider a plane

z=h

with 0. Call the region between the two surfaces R. The volume of R is given by the triple integral

\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx

Converting to polar coordinates will help make this computation easier. Set

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

Now, the volume can be computed with the integral

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta

You should get

\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)
5 0
3 years ago
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