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Tcecarenko [31]
3 years ago
8

aron lei is making identical balloon arrangements for a party. He has 32 maroon balloons and 24 white balloons, he wants each ar

rangements to have the same number of each color. what is tha greatest number of arrangements that he can make if every balloon is used?​
Mathematics
1 answer:
alexandr402 [8]3 years ago
5 0

Answer:

The greatest number of arrangements that he can make if every balloon is used is 8.

Step-by-step explanation:

The greatest number of arrangements will be the greatest common factor between 24 and 32.

GCF of 24 and 32:

We keep factoring both numbers by prime factors, while they can both be divided by the same number. So

24 - 32|2

12 - 16|2

6 - 8|2

3 - 4

There is no factor for which we can divide both 3 and 4. So the GCF is 2*2*2 = 8.

This means that the greatest number of arrangements that he can make if every balloon is used is 8.

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Write 9/10 - 3/10 as a mixed number
jeka57 [31]
You do 9/10. Which here it only fits once. So this is 1 1/10. The other one fits 3 times. So its 3 1/10.

If you want the subtraction also done its 1 4/10.
3 0
3 years ago
There are 158 students registered for American History classes. There are twice as many students registered in second period as
Margarita [4]

There are 28 students in first period and 56 students in sceond period and 74 students in third period

<em><u>Solution:</u></em>

Let the number of students in first period be "x"

Let the number of students in second period be "y"

Let the number of students in third period be "z"

<em><u>There are 158 students registered for American History classes.</u></em>

Therefore,

x + y + z = 158 ---------- eqn 1

<em><u>There are twice as many students registered in second period as first period</u></em>

number of students in second period = twice of number of students in first period

y = 2x ------- eqn 2

<em><u>There are 10 less than three times as many students in third period as in first period</u></em>

number of students in third period = 3 times number of students in first period - 10

z = 3x - 10 ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

x + 2x + 3x - 10 = 158

6x = 168

<h3>x = 28</h3>

<em><u>Substitute x = 28 in eqn 2</u></em>

y = 2(28)

<h3>y = 56</h3>

<em><u>Substitute x = 28 in eqn 3</u></em>

z = 3(28) - 10

z = 84 - 10

<h3>z = 74</h3>

Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period

5 0
3 years ago
Please help this is due in 10 minutes
mafiozo [28]

Answer:

the answer is 4.

4 0
3 years ago
How do I find the variable?<br><br> N x 1/5 = 2/15<br><br> N / 1/5 = 1/3
algol [13]
First you need to get all variables on one side. You do that by multiplying the reciprocal of the fraction to everything.
N(5/1) x 1/5(5/1)=2/15(5/1)
5n=2/15 x 5/1
Then solve the multiplication problem
5n= 10/15
Then you should reduce the fraction
5n=2/3
Then divide both sides by 5
5n/5=N
2/3 divided by 5 =2/15

N=2/15




4 0
3 years ago
Question
maria [59]

Answer:

y = m x + b equation of a straight line

m m' = -1    condition for perpendicular lines

If y = 4 x - 7     then m = 4 so m' = -.25

Y = -.25 X + A      we need to find A

A = Y + .25 * 8 = 2 + 2 = 4

Y = -.25 X + 4

Check:

2 = -.25 * 8 + 4 = -2 + 4 = 2

3 0
3 years ago
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