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balandron [24]
3 years ago
8

Should a quadrilateral have at least one pair of parallel sides?

Mathematics
1 answer:
Taya2010 [7]3 years ago
6 0

Answer:

it have two parallel sides

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Kay [80]
Answer: 7
Explanation: both triangles are congruent, so the mid segment is half of 14
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How do I solve this equation?
Norma-Jean [14]

Answer:

1. move the constant to the right hand side to change its sign.

2.add the numbers.

3. using the absolute value definition rewrite the absolute value equation as two separate equations.

4.slove the equation for X

Step-by-step explanation:

it has two solutions

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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

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Answer:

It was reflected down and to the right I think.

Step-by-step explanation:

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