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2 years ago

Mark had 45 paper clips. He evenly placed them into 5 different containers. How many paper clips were in each container?

Mathematics

1 answer:

Marat540 [252]2 years ago

7 0

Answer:

9 containers

Step-by-step explanation:

You have to evenly divide 45 paper clips into 5 different containers.

45÷5=9

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nlexa [21]

Answer:

5x7=35

5x7=5x(5+2)

5x7=(5x3)+(5x4)

5x7=20+15

5x7=35

Step-by-step explanation:

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3 years ago

Draw or add images here

Elodia [21]

Answer:

jonathan

Step-by-step explanation:

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3 years ago

Point M is the midpoint of segment QR. If QM = 16 + x and MR = 2(x + 2), find the length of QM.

Mrrafil [7]

Since M is the midpoint then QM and MR are equal to each other. So by setting them equal to each other we can find the value of x first.

16 + x = 2(x + 2)
16 + x = 2x + 4
16 = x + 4
12 = x

Now we can substitute in the value of x to find the length if QM.

16 + x
16 + 12
28

Hope this helps :)

7 0

3 years ago

Read 2 more answers

Jose bought 750 bags of peanuts for $375.00.he intends to sell each bag for $0.15 more than he paid. How much should he charge f

tatyana61 [14]

Jose should charge $0.65 for each bag if he intends to sell the bag for $0.15 more than he paid.

<h3>How do we solve word problems?</h3>

Mathematical problems dealing with word problems can be solved by arranging data given into variables and using the appropriate arithmetic operation to solve them.

From the given information:

If Jose bought a 750 bag of peanuts for $375.00. It means that he bought each bag(1 bag) for:

$\mathbf{=\$\dfrac{375}{750}}$

= $0.5

However, he intends to sell each bag for $0.15 more than he paid; Then the amount he should charge for each bag is:

=$0.5 + $0.15

=$0.65

Learn more about solving word problems here:

brainly.com/question/27893239

#SPJ1

3 0

2 years ago

If we assume the population of Grand Rapids is growing at a rate of approximately 4% per decade, we can model the population fun

saw5 [17]

Answer:

The average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

Step-by-step explanation:

The given function is

$P(t)=181843(1.04)^{(\frac{t}{10})}$

where, P(t) is population after t years.

At t=5,

$P(5)=181843(1.04)^{(\frac{5}{10})}=185444.20$

At t=6,

$P(6)=181843(1.04)^{(\frac{6}{10})}=186172.95$

At t=7,

$P(7)=181843(1.04)^{(\frac{7}{10})}=186904.57$

At t=8,

$P(8)=181843(1.04)^{(\frac{8}{10})}=187639.06$

At t=9,

$P(9)=181843(1.04)^{(\frac{9}{10})}=188376.44$

At t=10,

$P(10)=181843(1.04)^{(\frac{10}{10})}=189116.72$

The rate of change of P(t) on the interval $[x_1,x_2]$ is

$m=\frac{P(x_2)-P(x_1)}{x_2-x_1}$

Using the above formula, the average rate of change of the population on the intervals [ 5 , 10 ] is

$m=\frac{P(10)-P(5)}{10-5}=\frac{189116.72-185444.20}{5}=734.504$

The average rate of change of the population on the intervals [ 5 , 9 ] is

$m=\frac{P(9)-P(5)}{9-5}=\frac{188376.44-185444.20}{4}=733.06$

The average rate of change of the population on the intervals [ 5 , 8 ] is

$m=\frac{P(8)-P(5)}{8-5}=\frac{187639.06-185444.20}{3}=731.62$

The average rate of change of the population on the intervals [ 5 , 7 ] is

$m=\frac{P(7)-P(5)}{7-5}=\frac{186904.57-185444.20}{2}=730.185$

The average rate of change of the population on the intervals [ 5 , 6 ] is

$m=\frac{P(6)-P(5)}{6-5}=\frac{186172.95-185444.20}{1}=728.75$

Therefore the average rate of change of the population on the intervals [ 5 , 10 ] , [ 5 , 9 ] , [ 5 , 8 ] , [ 5 , 7 ] , and [ 5 , 6 ] are 734.504, 733.06, 731.62, 730.185 and 728.75 respectively.

7 0

4 years ago