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3 years ago

Evaluate the variable expression xyz for the given values of x, y, and z. x = 4/5 , y = −15, z = 7/8

Mathematics

2 answers:

Georgia [21]3 years ago

7 0

Answer:

xyz = - 21 / 2

Step-by-step explanation:

xyz = (4/5)(-15)(7/8) = (4×-15×7)/(5×1×8)

= (4×-3×5×7)/(5×1×4×2)...... because: 15 = 5×3 and 8 = 4×2

xyz = - 21 / 2

Sav [38]3 years ago

6 0

Answer:

The answer is -10.5

Step-by-step explanation:

You can just simply multiply the numbers above.

xyz=(4/5)*(-15)*(7/8)=-10.5

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3 years ago

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3 years ago

May runs 10/4 miles everyday after school. Beth says she runs further because she runs 5/2 miles everyday after school. Is her s

Andrew [12]

Answer:

Her statement is not correct.

Step-by-step explanation:

It is not correct because if you simplify 10/4 it comes out to 5/2 therefore 5/2=5/2.

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3 years ago

In a rational function, is the horizontal shift represented by the vertical asymptote?

ziro4ka [17]

<h3>Short Answer: Yes, the horizontal shift is represented by the vertical asymptote</h3>

A bit of further explanation:

The parent function is y = 1/x which is a hyperbola that has a vertical asymptote overlapping the y axis perfectly. Its vertical asymptote is x = 0 as we cannot divide by zero. If x = 0 then 1/0 is undefined.

Shifting the function h units to the right (h is some positive number), then we end up with 1/(x-h) and we see that x = h leads to the denominator being zero. So the vertical asymptote is x = h

For example, if we shifted the parent function 2 units to the right then we have 1/x turn into 1/(x-2). The vertical asymptote goes from x = 0 to x = 2. This shows how the vertical asymptote is very closely related to the horizontal shifting.

7 0

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