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Serjik [45]
3 years ago
14

A train travels 600 kilometers in 1 hour. What is the trains velocity in meters/seconds?

Mathematics
1 answer:
bixtya [17]3 years ago
8 0

Answer:

166.666667 m/ps

Step-by-step explanation:

600/60 will give us the amount of kilometers in a minute. 10 km/ph, now i'll divide more to find the seconds. 0.166666667 this is the amount of KILOMETERS in a second, we are trying to find in meters. multiply by 1000 and 166.666667. a fast fricking train if you ask me.

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Give B= {a,l,g,e,b,r} and C={m,y,t,h} find B U C
masha68 [24]

The union of a collection of sets is the set of all elements in the collection.

We have B = {a, l, g, e, b, r} and C = {m, y, t, h}

<h3>B ∪ C = {a, l, g, e, b, r, m, y, t, h}</h3>
6 0
3 years ago
which function rule is represented by the tablex 0 1 2 3y 0 -7 -14 -21a y= 7xb y= -7xc y= x-8d y= 7x-14
salantis [7]

The answer is y = -7x

Expalnation

Two points on the table are (0,0) and (1,-7)

Change in y = -7-0 = -7

Change in x = 1-0 = 1

Slope m of the function = -7/1= -7

Using y= mx + c

Picking point (0,0), x = 0, y = 0

y = mx + c becomes

0 = -7(0) + c

0 = 0 + c

c= 0

Hence, the equation is y = -7x + 0

which is y = -7x

Option b is the correct answer

8 0
1 year 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

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