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Sophie [7]
3 years ago
7

Suppose you paid $600 to take a geology course, which meets 40 times for one

Mathematics
1 answer:
Sav [38]3 years ago
6 0

Answer:

$15

Step-by-step explanation:

$600 divided by 40 = $15

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Help me I'm in jeproady
olga nikolaevna [1]

Answer:

She subtracted 7x instead of dividing it. Personally, i wouldn't have done that. I would subtract the 14 first, then divide.

3 0
3 years ago
Nicole bought six pounds of apples at $1.50 per pound. The store had a discount of $2 off her total purchase. Nicole and her fri
Andru [333]

Answer:

Step-by-step explanation:

Total amount of apples that Nicole bought was 6 pounds.

Each pound of apple cost $1.50

This means that the total cost of the 6 pounds of apple that Nicole bought would be

6 × 1.5 = $9

The store had a discount of $2 off her total purchase. This means that the amount that was paid is

9 - 2 = $7

If Nicole and her friend then divided the cost of the purchase evenly, let x represent the amount that each of them would pay. Therefore, the expression that can be used to determine how much Nicole and her friend each paid for the apples would be

x = 7/2

6 0
3 years ago
I have no idea so just solve it for me
romanna [79]
731/17 is equal to 43


6 0
3 years ago
$1900 at 9 for 2 years
vekshin1

Answer:

$342

Step-by-step explanation:

Complete question

Find the simple interest on $1900 at 9% for 2 years.

Simple interest PRT/100

Given

Principal P = 1900

Rate R = 9%

Time T = 2years

Substitute

Simple interest = 1900*9*2/100

Simple interest  = 19*18

Simple interest  = $342

Hence the simple interest on $1900 at 9% for 2 years is $342

7 0
3 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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