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2 years ago

What are 3 consecutive integers whose product is -120

1 answer:

5 0

Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero

x+6=0
x=-6

x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6

use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1

the numbers are -4,-5,-6

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2 years ago

the school newspaper has a circulation of 500 and sells for $.035 a copy. the students decide to raise the price to increase the

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Let x be the number of times they raise the price on the newspaper. Then the new cost of the newspaper is

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Then the income is given by the function

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no we equate the derivative to zero and solve for x.

$\begin{gathered} \frac{df}{dx}=0 \\ -2x+18=0 \\ -2x=-18 \\ x=\frac{-18}{-2} \\ x=9 \end{gathered}$

This means that we have an extreme value of the function when x=9. Now we need to find out if this value is a maximum or a minimum. To do this we need to take the second derivative of the function, then

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Since the second derivative is negative in the point x=9, we conclude that this value is a maximum of the function.

With this we conclude that the number of times that they should raise the price to maximize the income is 9. This means that they will raise the price of the newspaper (9)($0.05)=$0.45.

Therefore the price to maximize the income is $0.35+$0.45=$0.80.

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Marta_Voda [28]

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