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olga2289 [7]
3 years ago
9

Eric created a rectangular patio using 1 foot square paving stones,which are sold in batches by the dozen. The patio measures 7

feet by 8 feet. How many batches of paving stones did Eric need?
Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0
I believe it is 56 because 7•8 is 56 and 1•56 is 56
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Correlation between x & y is 0.6125.

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1 year ago
Alex learning algebra 2 quiz
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5 0
3 years ago
If 5 boxes of cereal cost $27.50, how much does one box of cereal cost? a) $5.50 b) $5.05 c) $550 d) .55​
olga_2 [115]

}

Answer:B

Step-by-step explanation: I said that because you have 27 and 50 Cent so basically you won't have that $5 in that picture because you still going to have that $0.04 and then you still have to have that much so I I think I am right but if I'm not I'm so sorry but I'm taking not even to get I'm trying to get back on track but I think is b for a reason like I think you could be because you have 27 and 50 Cent so if you got twenty-seven fifty cents you just rounding it down

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2 years ago
Read 2 more answers
HELP PLEASEEEE !!!!!!!!!!! Will mark BRIANLIEST !!!!!
Ray Of Light [21]

Answer:

<QRS= (10.x - 1) = (10.6 - 1) = 59°

Step-by-step explanation:

the triangle is not accurate but drawing one is truly useful in calculating. Besides, I don't know how it is called in English but it is something about the exterior angles (maybe)

7 0
3 years ago
A large tank is filled to capacity with 600 gallons of pure water. brine containing 4 pounds of salt per gallon is pumped into t
nignag [31]
If A(t) is the amount of salt in the tank at time t, then the rate at which the amount of salt in the tank changes is given by

\dfrac{\mathrm dA(t)}{\mathrm dt}=\dfrac{4\text{ lbs}}{1\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}-\dfrac{A(t)\text{ lbs}}{600\text{ gal}}\dfrac{6\text{ gal}}{1\text{ min}}
\dfrac{\mathrm dA}{\mathrm dt}=24\dfrac{\text{lb}}{\text{min}}-\dfrac{A(t)}{100}\dfrac{\text{lb}}{\text{min}}

Let's drop the units for now. We have

\dfrac{\mathrm dA(t)}{\mathrm dt}+\dfrac{A(t)}{100}=24
e^{t/100}\dfrac{\mathrm dA(t)}{\mathrm dt}+e^{t/100}\dfrac{A(t)}{100}=24e^{t/100}
\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/100}A(t)\right]=24e^{t/100}
e^{t/100}A(t)=\displaystyle24\int e^{t/100}\,\mathrm dt
e^{t/100}A(t)=2400e^{t/100}+C
A(t)=2400+Ce^{-t/100}

We're given that the water is pure at the start, so A(0)=0, giving

A(0)=0=2400+Ce^{-0/100}\implies C=-2400

So the amount of salt in the tank (in lbs) at time t is

A(t)=2400\left(1-e^{-t/100}\right)
4 0
3 years ago
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