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pishuonlain [190]

3 years ago

7

The area of a rectangular classroom is given by the trinomial a - 4a - 21. The length of the rectangle is a+3. What is the expre

ssion for the width of the classroom?

1 answer:

djverab [1.8K]3 years ago

3 0

I am assuming you wrote the problem wrong. Is it a^2?
A^2 -4A-21=0
(A-7)(A+3)=0
A=7 A=-3, ignore the negative number because it doesn't make sense if you plug in for a.
length=a+3=7+3=10
Width=7=a

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$\{(1,55)\ (2, 45)\ (3, 35)\ (4, 25)\}$

Solving (8): Arithmetic or Geometric

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<em>Because the common difference is equal, then it is an arithmetic progression</em>

<em></em>

Solving (8):

$f(n) = f(1) + f(n-1)$

To find f(9), we substitute 9 for n

$f(9) = f(1) + f(9-1)$

$f(9) = f(1) + f(8)$

We need to solve for f(8); substitute 8 for n

$f(8) = f(1) + f(8 - 1)$

$f(8) = f(1) + f(7)$

We need to solve for f(7); substitute 7 for n

$f(7) = f(1) + f(7 - 1)$

$f(7) = f(1) + f(6)$

We need to solve for f(6); substitute 6 for n

$f(6) = f(1) + f(6 - 1)$

$f(6) = f(1) + f(5)$

We need to solve for f(5); substitute 6 for n

$f(5) = f(1) + f(5 - 1)$

$f(5) = f(1) + f(4)$

From the function, f(4) = 25 and f(1) = 55.

So:

$f(5)=55 + 25$

$f(5)=80$

$f(6) = f(1) + f(5)$

$f(6) = 55 + 80$

$f(6) = 135$

$f(7) = f(1) + f(6)$

$f(7) = 55 + 135$

$f(7) = 190$

$f(8) = f(1) + f(7)$

$f(8) = 55 + 190$

$f(8) = 245$

$f(9) = f(1) + f(8)$

$f(9) = 55 + 245$

$f(9) = 300$

8 0

3 years ago