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3 years ago

How many hours would someone who earns 7.50 per hour have to work to earn $116.25

Mathematics

2 answers:

belka [17]3 years ago

8 0

Answer:

15.5 hours, 15 hours and 30 minutes.

Step-by-step explanation:

116.15 divided by 7.50 = 15.5

givi [52]3 years ago

6 0

They would have to work 15 hours and 50 minutes

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At a high school movie night, the refreshments stand sells popcorn and soft drinks. Of the 100 students who came to the movie, 6

RUDIKE [14]

Independent events. Someone does not have to buy popcorn to buy a drink and someone does not have to buy a drink to buy popcorn.

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3 years ago

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How to solve x + y<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{d

nadya68 [22]

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The answer is {D}

Step-by-step explanation:

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2 years ago

Can some help please?!?!

Alexandra [31]

Answer:

<em>Choice D.</em>

Step-by-step explanation:

Vertical line test for functions:

If a vertical line drawn anywhere on the graph intersects the graph in more than one point, it is not a function.

If you draw a vertical line passing through any point between x = -4 and x = 4, the line will intersect two points on the circle. The relation fails the vertical line test. That shows that this relation is not a function.

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3 years ago

— 3х – 19 <-18<br> What is x

kondaur [170]

Answer:

x>-1/3

Step-by-step explanation:

−3x<−18+19

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2 years ago

A 500500-gallon tank initially contains 200200 gallons of brine containing 100100 pounds of dissolved salt. brine containing 11

mixer [17]

Let $A(t)$ denote the amount of salt in the tank at time $t$ . We're given that the tank initially holds $A(0)=100$ lbs of salt.

The rate at which salt flows in and out of the tank is given by the relation

$\dfrac{\mathrm dA}{\mathrm dt}=\underbrace{\dfrac{11\text{ lb}}{1\text{ gal}}\times\dfrac{44\text{ gal}}{1\text{ min}}}_{\text{rate in}}-\underbrace{\dfrac{A(t)}{200+(44-11)t}\times\dfrac{11\text{ gal}}{1\text{ min}}}_{\text{rate out}}$
$\implies A'(t)+\dfrac{11}{200+33t}A(t)=484$

Find the integrating factor:

$\mu(t)=\exp\left(\displaystyle\int\frac{11}{200+33t}\,\mathrm dt\right)=(200+33t)^{1/3}$

Distribute $\mu(t)$ along both sides of the ODE:

$(200+33t)^{1/3}A'(t)+11(200+33t)^{-2/3}A(t)=484(200+33t)^{-1/3}$
$\bigg((200+33t)^{1/3}A(t)\bigg)'=484(200+33t)^{-1/3}$
$A(t)=484\displaystyle\int(200+33t)^{-1/3}\,\mathrm dt$
$A(t)=22(200+33t)^{2/3}+C$

Since $A(0)=100$ , we get

$100=22(200)^{2/3}+C\implies C\approx-652.39$

so that the particular solution for $A(t)$ is

$A(t)=22(200+33t)^{2/3}-652.39$

The tank becomes full when the volume of solution in the tank at time $t$ is the same as the total volume of the tank:

$200+(44-11)t=500\implies 33t=300\implies t\approx9.09$

at which point the amount of salt in the solution would be

$A(9.09)\approx733.47\text{ lb}$

4 0

3 years ago