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Nutka1998 [239]

3 years ago

15

Andrew has a summer job doing yard work. He is paid $5 per hour and a $20 bonus when he completes the yard. He was paid $85 for

completing one yard. Write an equation to represent the amount of money he earned

1 answer:

zheka24 [161]3 years ago

5 0

Answer: 65$

He worked for 65 dollars worth of work.

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R(r1+r2)=r1r2 solve for R

romanna [79]

We need to solve for R, This is really simple.

The original expression is:
R (r1 + r2) = r1r2

To solve for a certain variable, we need to get this variable alone on one side of the equation and equate it with the other side.

In the given expression, to get R alone on one side we have to eliminate (r1 + r2).
In order to do this, we will divide both sides by (r1 + r2).
Doing this, we get the solution as follows:
R = (r1r2) / (r1 + r2)

4 0

3 years ago

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3x=6y. i need help to turn that into y-intercept form and find tge y intercept. Help please

evablogger [386]

3x = 6y

to get this into slope-intercept form, you simply need to solve for y. remember that slope-intercept form is y = mx + b, where m is your slope and b is your y-intercept.

3x = 6y ... divide both sides by 6
1/2x = y
y = 1/2x is your equation in slope-intercept form. because no "b" value is present, your intercept will simply be 0.

8 0

3 years ago

Read 2 more answers

In April 2011, an online social networking site had approximately 200 million registered accounts. In August, the number of regi

fenix001 [56]

All you have to do is move your decimal point over eight places to the right and then subtract that from the first number and you have your answer.

6 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

Find the missing dimension of the solid. Round your answer to the nearest 10th

Svetach [21]

Answer:

the answer is 9.9997 ft

Step-by-step explanation:

the equation to find the radius is

r = √(3v / πh)

5 0

3 years ago