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Ugo [173]
3 years ago
5

Chase is campaigning for class president and plans to distribute some campaign materials: 8 flyers and 12 buttons. He wants each

classroom to receive an identical set of campaign materials, without having any materials left over. What is the greatest number of classrooms Chase can distribute materials to?
Mathematics
1 answer:
rjkz [21]3 years ago
6 0
Answer: 4 classrooms

Explanation:
The greatest common factor of 8 and 12 is 4, so the answer would be 4 because Chase doesn’t want any leftover supplies.
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Very hard problem someone help
miv72 [106K]

Answer:

Everything in the box currently is correct but the bottom two numbers are just -3 and -4

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1 year ago
What is an equation of the line with slope 5 and y-intercept 8?
jeka94

Answer:

y=5x+8

Step-by-step explanation:

You need have the slope-intercept formula:

y=mx+b

m=slope

b=y-intercept

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8 0
3 years ago
The first two terms in an arithmetic progression are -2 and 5. The last term in the progression is the only number in the progre
Rudik [331]

Given:

The first two terms in an arithmetic progression are -2 and 5.

The last term in the progression is the only number in the progression that is greater than 200.

To find:

The sum of all the terms in the progression.

Solution:

We have,

First term : a=-2

Common difference : d = 5 - (-2)

                                      = 5 + 2

                                      = 7

nth term of an A.P. is

a_n=a+(n-1)d

where, a is first term and d is common difference.

a_n=-2+(n-1)(7)

According to the equation, a_n>200.

-2+(n-1)(7)>200

(n-1)(7)>200+2

(n-1)(7)>202

Divide both sides by 7.

(n-1)>28.857

Add 1 on both sides.

n>29.857

So, least possible integer value is 30. It means, A.P. has 30 term.

Sum of n terms of an A.P. is

S_n=\dfrac{n}{2}[2a+(n-1)d]

Substituting n=30, a=-2 and d=7, we get

S_{30}=\dfrac{30}{2}[2(-2)+(30-1)7]

S_{30}=15[-4+(29)7]

S_{30}=15[-4+203]

S_{30}=15(199)

S_{30}=2985

Therefore, the sum of all the terms in the progression is 2985.

6 0
3 years ago
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