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ddd [48]
2 years ago
8

Please do part A and part B on number 9!

Mathematics
1 answer:
Mashcka [7]2 years ago
7 0

Answer:

A=5

B=4

Step-by-step explanatation

so the 3ft at the front and the 8ft at the back, you do 8-3 which is 5 which will be A and on the side, it is 8ft and since the corner angle is 4ft then B will be 8-4 which is 4

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X+10=1/3(5x+10)<br><br> How do you solve this equation?
Nikitich [7]

Answer:

x = 10

Step-by-step explanation:

First, subtract 10 from both sides

Next, Simplify

Then, subtract 5/3x from both sides

After, simplify


Lastly, multiply both sides by 3


Then finally simplify for your final answer of x = 10


Hope this helps! :)

3 0
2 years ago
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How to graph -5\3 on a number line
Alex_Xolod [135]
<span>Number line is a way of expressing an integer into a line. 
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8 0
3 years ago
Show 1,700 using large cubes and flats
IgorC [24]

Answer:

How will the size of the model for 100,000 compare to the size of the model for 10,000? Possible answer: the pattern shows cube, long, flat, cube. So, the shape of the model for 10,000 will be long.

Step-by-step explanation:

3 0
3 years ago
If anyone could help that would be great / right answer gets brainilest
Keith_Richards [23]
2.69 cm would be the answer
7 0
3 years ago
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A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t
Licemer1 [7]
If A(t) is the amount of salt in the tank at time t, then the rate at which this amount changes over time is given by the ODE

A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}

A'(t)+\dfrac1{100}A(t)=15

We're told that the tank initially starts with no salt in the water, so A(0)=0.

Multiply both sides by an integrating factor, e^{t/100}:

e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}
\left(e^{t/100}A(t)\right)'=15e^{t/100}
e^{t/100}A(t)=1500e^{t/100}+C
A(t)=1500+Ce^{-t/100}

Since A(0)=0, we have

0=1500+C\implies C=-1500

so that the amount of salt in the tank over time is given by

A(t)=1500(1-e^{-t/100})

After 10 minutes, the amount of salt in the tank is

A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}
8 0
3 years ago
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