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Genrish500 [490]
3 years ago
12

There are approximately 39.37 iches in 1 meter. how many meters are in 170 inches

Mathematics
1 answer:
mel-nik [20]3 years ago
6 0
170 divided by 39.37= 4.32 meter
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XZ is a common external tangent to circles W and Y. What is the distance between the two centers of the circles? Round to the ne
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You can Draw a segment WY connecting the centers<span> of the </span>two circles<span>, and then draw a segment, WS, so that YS + SZ = YZ and WS ⊥ YZ).</span>
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3x-5+6y-9+10x
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Variables- x,y
terms- 4
coefficient-3x, 6y, 10x
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S=L (1 - r), for r
maria [59]
OK. Then let's solve for ' r '.  That means you have to come up with an equation that says     r =    everything else.

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and there you have it.
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3 0
3 years ago
A list of numbers is given <br><br><br> 32,16,8,4,2,1<br><br> Write a rule for the pattern
Sphinxa [80]
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5 0
3 years ago
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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

8 0
2 years ago
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