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Vilka [71]
3 years ago
8

A bowling ball is rolled down the alley with a constant velocity of 1.4 m/sec at an angle of 86 degrees to the starting line. Th

e position of the person throwing the bowling ball can be represented by the point (0,0). Where is the ball after 8 seconds?
Mathematics
2 answers:
CaHeK987 [17]3 years ago
7 0

Answer:(0.6,10.6) after 8 seconds.

Step-by-step explanation:

Aleonysh [2.5K]3 years ago
6 0

Answer:

The position of ball is (0.78, 11.17) after 8 seconds.

Step-by-step explanation:

A bowling ball is rolled down the alley with a constant velocity of 1.4 m/sec at an angle of 86 degrees to the starting line.

Horizontal Position:

x=v\cos\theta \times t

Vertical Position:

y=v\sin\theta \times t

v=1.4\ m/s

\theta=86^\circ

t=8\ sec

Substitute the value into formula

x=1.4\cos86^\circ \times 8

x=0.78\ m

y=1.4\sin86^\circ \times 8

y=11.17\ m

Hence, The position of ball is (0.78,11.17)

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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
3 years ago
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X=-1 is pretty sure
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Completely Factoring<br> 3x2 – 10x – 8
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Answer:

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Step-by-step explanation:

Dimensions are missing.

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