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LenKa [72]
4 years ago
10

The power absorbed by a solar panel varies directly with its area. Suppose an 8 square meter panel absorbs 8,160 watts of power.

How much power does a 12 square meter solar panel absorb?
Mathematics
2 answers:
Verdich [7]4 years ago
4 0
Write them equations: 8/8160 = 12/x. What's x? 

Cross multiplying yields: 8x = 12*8160 --> 1020*12 --> x = 12240.

So the 12 square meter solar panel absorbs 12240 watts of power.
olga_2 [115]4 years ago
3 0
Answer is 12240 watts of power that a 12 square meter solar panel absorb. 
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Exact = 50 Approximate = 45
Nikitich [7]

Answer:

Approximate of error = 11.11 % (Approx.)

Step-by-step explanation:

Given:

Exact value = 50

Approximate value = 45

Find:

Approximate of error

Computation:

Approximate of error = [(Exact value - Approximate value)/Approximate value]100

Approximate of error = [(50 - 45)/45]100

Approximate of error = [(5)/45]100

Approximate of error = [0.11111]100

Approximate of error = 11.11 % (Approx.)

3 0
3 years ago
Help 10 points and explain how you get the answer.
Alenkinab [10]

you solve the square root of 54a*7b*3

6 0
4 years ago
HELP HELP HELP
Alik [6]
The answer would be 7
7 0
3 years ago
a cube has sides with the following number: 3,3,6,9,,7,8. what is the probability of getting a factor of 9 when rolled?
muminat

Answer:

The probability of rolling a factor of 9 is 1/2, or 50%.

Step-by-step explanation:

The number 9's factors are 1, 3, and 9. Three of the six sides of the cube have a factor of 9. Therefore 3/6 of the sides are factors of nine.

Simplify 3/6 to 1/2 to get a probability of 50%.

5 0
2 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
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