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Vesna [10]
3 years ago
6

if 14 giant solar power plants generate 2.6 gigawatts of energy to power 1.8 million homes how many homes can 28 solar panels po

wer
Mathematics
1 answer:
user100 [1]3 years ago
8 0

28 solar panels would be able to power 3.6 million homes. I hope I helped!

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Andrew borrows $79,500 for 5 months on 6.30% interest rate in his saving account. calculate the simple interest
alukav5142 [94]
Equation i= p time r time T 79500 time .0630 time .42 Equal 21,035.7 simple interest Dos Doctor pepper on the interest, going back two time on the left which would become .0630 5 divide month of 12 which is .41666 Round to the nearest tenth which is .42
6 0
3 years ago
I will give brainliest to whoever helps me with this question. Also, if you could please explain to me how you solved it I would
krok68 [10]
Answer: I don’t have my calculator on me but I’ll explain

Step by step: the shape is a cone and a sphere so the volume of a sphere is 4/3 πr^3 but because it’s half a sphere, divide by 2 and the volume of a cone is 1/3h πr^2

Sphere= 4/3 π(10)^3 (plug that in ur calculator) then divide by 2

Cone= 1/3(24) π(10)^2
Then add them together
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3 years ago
Solve for y: 3y + 5 - 8y = 2y - 7 - 4y<br> A y = - 4<br> B y = 12/7<br> C y = 4<br> D y = 2/3
MrMuchimi
The answer for this question is C

8 0
2 years ago
Can someone plz help a bro out
Eddi Din [679]
F(-1) = -6
f(0) = -5
f(1) = -4
f(2) = -3
f(5) = 0
f(8) = 3
7 0
3 years ago
Find f. f ″(x) = x^−2, x &gt; 0, f(1) = 0, f(6) = 0
marin [14]

If you do in fact mean f(1)=f(6)=0 (as opposed to one of these being the derivative of f at some point), then integrating twice gives

f''(x) = -\dfrac1{x^2}

f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1

f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2

From the initial conditions, we find

f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0

f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)

Eliminating C_2, we get

(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))

-5C_1 = \ln(6)

C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)

Then

\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}

3 0
2 years ago
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