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2 years ago

How can you tell on which two whole numbers an improper fraction would be located between on a number line?

2 answers:

3 0

The way you can tell if it is an improper fraction is when the top number is bigger then the bottom. hope that's what you are looking for :)

Margarita [4]2 years ago

3 0

The way you can tell the improper an proper fractions i think

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BIG IDEAS MATH

IgorC [24]

Answer:

106

Step-by-step explanation:

See attached

7 0

2 years ago

How many 1/3 liters are in 4 2/3 liters

jonny [76]

Answer:

Step-by-step explanation:

This is a division problem.

We want to find how many 1/3 liters are in 4⅔ liters.

We divide to obtain:

$\frac{4 \frac{2}{3} }{ \frac{1}{3} }$

$4 \frac{2}{3} \div \frac{1}{3}$

Convert to improper fraction.

$\frac{14}{3} \div \frac{1}{3}$

$\frac{14}{3} \times \frac{3}{1}$

This simplifies to

$14$

Therefore there are fourteen ⅓ liters in 4⅔ liters

5 0

3 years ago

Solve for each equation h divided by 4/9 for h = 5 1/3 A) 2 10/27 B) 10 C) 12

arsen [322]

Solve for each equation
h divided by 4/9 for h = 5 1/3

h 16/3
------- = -----------------
4/9 4/9
= 16/3 * 9/4
= 12
answer is C) 12

6 0

3 years ago

What is 11.52/128 show your work

grandymaker [24]

Answer:

The answer is

Step-by-step explanation:

11.5/128

First convert the decimal to an improper fraction

That's

$11.52 = \frac{288}{25}$

So we have

$\frac{288}{25} \div 128$

Change the division sign to multiplication sign and reverse the second number

That's

$\frac{288}{25} \times \frac{1}{128}$

Simplify

$\frac{288}{3200}$

Divide through by 32

The final answer is

$\frac{9}{100}$

Hope this helps you

6 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. We also know that this last pile was a multiple of 3 before a third was taken away by James.

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

2 years ago