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3 years ago

Divide 33 photos into groups so the ratio is 4 to 7

Mathematics

1 answer:

Vikki [24]3 years ago

7 0

4:7....added = 11

4/11(33) = 132/11 = 12 <== one group
7/11(33) = 231/11 = 21 <== the other group

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Triangle A B C is shown. Angle B C A is a right angle. The length of hypotenuse B A is 9.8 inches, the length of C B is 6.3 inch

german

Answer:

$\angle B=\cos^{-1}\left(\dfrac{6.3}{9.8}\right)\\ \\\angle B=\sin^{-1}\left(\dfrac{7.5}{9.8}\right)$

Step-by-step explanation:

Consider right triangle ABC with right angle C.

In this triangle,

BA = 9.8 in
CB = 6.3 in
CA = 7.5 in

Write trigonometric functions:

$\cos \angle B=\dfrac{BC}{AB}=\dfrac{6.3}{9.8}\\ \\\sin \angle B=\dfrac{AC}{AB}=\dfrac{7.5}{9.8}$

Hence,

$\angle B=\cos^{-1}\left(\dfrac{6.3}{9.8}\right)\\ \\\angle B=\sin^{-1}\left(\dfrac{7.5}{9.8}\right)$

7 0

3 years ago

Read 2 more answers

What is 28:36 in simplest form

xeze [42]

It would be 7:9 because if you divide each side by 4 then you can get it in simplest form

6 0

3 years ago

Read 2 more answers

Rosa and Albert receive the same amount of allowance each week . The table shows what part of their allowance they each spent on

Anettt [7]

1/2 x > 0.4x

where Robert's spending is greater than Rosa's spending.

Allowance of Rosa = Allowance of Robert

let x = allowance

Rosa : video games = 0.4(x) ; pizza = 2/5 (x)
Robert : video games = 1/2 (x) ; pizza = 0.25 (x)

let us assume that their allowance is 100 each week. so, x = 100

Rosa : video games = 0.40(100) = 40
pizza 2/5 (100) = 200/5 = 40
total spending: 40 + 40 = 80

Robert : video games = 1/2 (100) = 50
pizza 0.25 (100) = 25
total spending: 50 + 25 = 75

Spending on video games
Rosa = 40
Robert = 50

Robert spent more of his allowance on video games than Rosa.

1/2 x > 0.4x

7 0

3 years ago

In a circle, which ratio is equivalent to π?

garri49 [273]

A - Radius to Area

π r² is the formula to find the area in a circle

5 0

3 years ago

Read 2 more answers

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