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Studentka2010 [4]

3 years ago

8

A gym charges $45 per month and a $75 one-time starting fee. Which of these will show how much it costs to join the gym for one

year?

2 answers:

Rufina [12.5K]3 years ago

8 0

Answer:

615

Step-by-step explanation:

u times 45 for 12 months and add the 75 fee

Sauron [17]3 years ago

4 0

Answer: 615

Step-by-step explanation: you take the 45 times 12 months plus the 75 for the one time fee.

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Words problems involving length and money

stiv31 [10]

Answer:

12

Step-by-step explanation:

Since 8 poster takes 48 inches of tape

48/8 = 6

each poster will take 6

if you have two posters

6x2=12

so 12

5 0

2 years ago

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carly's baby sister like to stack her blocks to make patterns. how do she demostrate the relationship between the stack number a

Svetlanka [38]

She counts both piles.

6 0

3 years ago

A game increased in price by 1/2 . After the increase it was priced at £27. What was the original price of the game? (Hegarty Ma

lukranit [14]

Answer:

£18

Step-by-step explanation:

Let

x = original price of the game

Increase in price = 1/2

New price = £27

x + 1/2x = £27

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3/2x = 27

x = 27 ÷ 3/2

= 27 × 2/3

= 54 / 3

x = £18

Therefore, the original price of the game is £18

4 0

2 years ago

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Mr. Bridges paid $582.40 for a set of four tires for a truck about how much was the cost per tire

skelet666 [1.2K]

$582.40 / 4 tires = $145.60

Answer: $145.60/tile

5 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