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Tems11 [23]
3 years ago
15

What are fractions used for?

Mathematics
2 answers:
mixas84 [53]3 years ago
8 0

Fractions are used for dividing things into equal portions.

user100 [1]3 years ago
5 0
Fractions are important because they tell you what portion of a whole you need, have, or want. fractions are used in baking to tell how much of an ingredient to use. fractions are used intelling time; each minute is a fraction of the hour.
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Answer:

y=-\frac{1}{9}x+\frac{5}{3}

Step-by-step explanation:

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3 years ago
What does negative 2 over 7 > −4 indicate about the position of negative 2 over 7 and −4 on the number line?
jeyben [28]
-4/7 > -2

If it is greater it goes to the right, and if it is less it goes to the left.

-4/6 is located to the right of -2
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Which of the following is a true statement about the rhombus shown?
Airida [17]

Answer:-

B

I hope it helps.

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Read 2 more answers
A group of people were asked if they had run a red light in the last year. 150 responded "yes", and 185 responded "no". Find the
Softa [21]

Answer:

150/335, or 30/67

Step-by-step explanation:

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Have a great day!

8 0
3 years ago
Please help with imaginary numbers worksheet!
mihalych1998 [28]

Problem 5

<h3>Answer:   6i</h3>

------------------

Explanation:

We have these four identities

  • i^0 = 1
  • i^1 = i
  • i^2 = -1
  • i^3 = -i

Notice how computing i^4 leads us back to 1. So i^4 = i^0. The pattern repeats every 4 terms. So we divide the exponent by 4 and look at the remainder. We ignore the quotient entirely. We can see that 28/4 = 7 remainder 0. Meaning that i^28 = i^0 = 1.

We can think of it like this if you wanted

i^28 = (i^4)^7 = 1^7 = 1

Then the sqrt(-36) becomes 6i

So overall, we end up with the final answer of 6i

=============================================

Problem 6

<h3>Answer:  -3i</h3>

------------------

Explanation:

We'll use the ideas mentioned in problem 5

46/4 = 11 remainder 2

i^46 = i^2 = -1

sqrt(-9) = 3i

The two outside negative signs cancel out, but there's still a negative from -1 we found earlier. So we end up with -3i

In other words, here is one way you could write out the steps

-i^{46}*-\sqrt{-9}\\\\-i^{2}*-3i\\\\i^{2}*3i\\\\-1*3i\\\\-3i\\\\

=============================================

Problem 7

<h3>Answer:   -1</h3>

------------------

Work Shown:

i^10 = i^2 because 10/4 = 2 remainder 2

i^19 = i^3 because 19/4 = 4 remainder 3

i^7 = i^3 because 7/4 = 1 remainder 3

Again, all we care about are the remainders.

i^{10}+i^{19} - i^{7}\\\\i^{2}+i^{3} - i^{3}\\\\i^{2}\\\\-1

=============================================

Problem 8

<h3>Answer:    -1 + i</h3>

------------------

Work Shown:

i^22*i^6 = i^(22+6) = i^28

Earlier in problem 5, we found that i^28 = i^0 = 1

So,

i^1 - \left(i^{22}*i^{6}\right)\\\\i^1 - i^{28}\\\\i^1 - i^{0}\\\\i - 1\\\\-1+i

8 0
2 years ago
Read 2 more answers
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