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2 years ago

15

A virus has been spreading in a computer network. This function can be used to determine v(x). the number of computers affected

by the virus after x days.
v(x) =2p^x

what does the variable p represent in the equation?

A. the number of computers in the network

B. the number of computers affected each day

C. the rate of decay in the number of computers affected.

D. the rate of growth in the number of computers affected

please only respond if you know the answer :)

1 answer:

sashaice [31]2 years ago

5 0

Step-by-step explanation:

it will be D because it sounds correct and it looks write

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You offer senior citizens a 20% discount on their tune-ups at your gas station. Assuming that an average of 50 senior citizens g

GREYUIT [131]

X over 49.5 and 20 over 100 then you cross multiply and you get 100x is equal to 990 you divide 100x and 990 both by 100 and get x and 9.9 then you have 50 seniors a month that you give the discount to so you multiply 9.9 by 50 and get 495 so the total discount you give on an average monthly basis is $495.00 Hope this helped

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3 years ago

I forgot how to find x

GuDViN [60]

You would do 2x-3+x+3x-15=360
combine like terms which then gives you
5x-18=360 then you would add 18 to both sides giving you
5x=378
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3 years ago

5x - 9 = -24 + 10x

lys-0071 [83]

Answer:

One Solution: 3

Step-by-step explanation:

5x - 9 = -24 + 10x

Subtract 5x from both sides. -9 = -24 + 5x

Add 24 to both sides. 15 = 5x

Dived 15/5. 3 = x

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3 years ago

John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m

Dmitry [639]

Answer:

The minimum possible initial amount of fish: $52$

Step-by-step explanation:

Let's start by saying that

$x$ = is the initial number of fishes

John:

When John arrives:

he throws away one fish from the bunch

$x-1$

divides the remaining fish into three.

$\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}$

takes a third for himself.

$\dfrac{x-1}{3} + \dfrac{x-1}{3}$

the remaining fish are expressed by the above expression. Let's call it John

$\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}$

and simplify it!

$\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}$

When Joe arrives:

he throws away one fish from the remaining bunch

$\text{John} -1$

divides the remaining fish into three

$\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}$

takes a third for himself.

$\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

the remaining fish are expressed by the above expression. Let's call it Joe

$\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}$

and simiplify it

$\text{Joe}=\dfrac{2}{3}(\text{John}-1)$

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

$\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)$

$\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}$

When James arrives:

We're gonna do this one quickly, since its the same process all over again

$\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}$

$\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)$

$\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}$

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say $n$ ), a third is taken away by James. the remaining bunch would be $\frac{n}{3}+\frac{n}{3}$

hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

$\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

$\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}$

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

$14=\dfrac{8x}{27} - \dfrac{38}{27}$

$x=52$

This is minimum possible amount of fish before John threw out the first fish

8 0

3 years ago

If a line falls on the points (11, 13) and (4, 12), what is its slope? Enter your answer as a fraction in lowest terms. Use a sl

loris [4]

2/14= 1/7

I just graphed them then did the slope which was rise over run

8 0

3 years ago