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AURORKA [14]
4 years ago
12

Write a word problem that involves combining three equal groups

Mathematics
2 answers:
ipn [44]4 years ago
7 0
Combining three equal groups means that we will mainly depend on multiplying the quantity by 3 to get the total

<u><em>Examples are shown below:</em></u>
1- Mrs Nadia teaches three classes. Each class has 25 student. How many students does Mrs Nadia teach in total?
<u>In this problem</u> we will be combining three equal groups of students where each group has 25 students, therefore:
Total number of students = 3 * 25 = 75 students

2- John has three bags of candies. Each bag contains 10 pieces of candies. How many candies does John has?
<u>In this problem</u> we will be combining three equal groups of candies where each group has 10 pieces, therefore:
Total number of candies = 3 * 10 = 30 candies

Hope this helps :)
Afina-wow [57]4 years ago
3 0
By "three equal groups" I'm going to assume you mean three sets of different things with the same amount in each.

Word problem:

Josh went to the grocery store to buy fruit for smoothies. He bought 6 oranges, 6 apples, and 6 pears. How many fruits did he buy all together? What fraction of all of the fruits are the oranges?
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Step-by-step explanation:

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Alex787 [66]

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Step-by-step explanation:

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3 years ago
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Solving Equations<br> ve<br> x + 14 = 7x + 32 - 3x
Mkey [24]

Answer:

-6

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

x+14=7x+32−3x

x+14=7x+32+−3x

x+14=(7x+−3x)+(32)(Combine Like Terms)

x+14=4x+32

x+14=4x+32

Step 2: Subtract 4x from both sides.

x+14−4x=4x+32−4x

−3x+14=32

Step 3: Subtract 14 from both sides.

−3x+14−14=32−14

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Step 4: Divide both sides by -3.

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5 0
3 years ago
Why is the answer to this integral's denominator have 1+pi^2
ss7ja [257]

It comes from integrating by parts twice. Let

I = \displaystyle \int e^n \sin(\pi n) \, dn

Recall the IBP formula,

\displaystyle \int u \, dv = uv - \int v \, du

Let

u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Then

\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn

Apply IBP once more, with

u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Notice that the ∫ v du term contains the original integral, so that

\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)

\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I

\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n

\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C

6 0
2 years ago
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