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2 years ago

Lim f(x) = 19, find the value of A and B, if the lim f(x) = Ax – 3B and lim f(x) = **+ B

Mathematics

1 answer:

podryga [215]2 years ago

6 0

Answer:

A = 4, B = 3

Step-by-step explanation:

In order for the limit to exist and have a value of 19, the left limit and the right limit must both have that value at x=7.

left limit = 19

right limit = 19

4A(7)/7 +B = 19

A(7) -3B = 19

Adding 3 times the first equation to the second gives ...

3(4A +B) +(7A -3B) = 3(19) +(19)

19A = 4(19) . . . . . simplify

A = 4 . . . . . . . . divide by 19

Substituting into the first equation gives ...

4(4) +B = 19

B = 3 . . . . . . . subtract 16

A is 4, B is 3.

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3 years ago

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Apologiabiology Ambitious

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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

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2 years ago