9. The Brightline train travels 204 miles in 3 hours, moving at a constant

1 answer:

8 0

The answer is D
3/204= 0.01470588
I know that is how much of an hour it will take to travel an hour so multiply that by the miles
544* 0.01470588= 8
8 Hours

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

stealth61 [152]

Answer:

BD = 8.8

Step-by-step explanation:

You have a right triangle.

The hypotenuse is BD

The side opposite the 27o angle is 4

Sin(27) = opposite / hypotenuse.

Sin(27) = 0.45399

opposite = 4

hypotenuse = ?

0.45399 = 4/hypotenuse

BD = 4/0.45399

BD = 8.811

The answer is 8.8

4 0

3 years ago

Read 2 more answers

2a. Determine the equation of the line connecting the points (0, -1) and (2, 3) *

Sauron [17]

First create the equation y=Mx+ B

B is the y intercept so when y = 0, B = -1.

Now we have y=Mx - 1

The slope is M. We can calculate this by using the formula M = (y2 -y1) / (x2 - x1)

Use the points (0, -1) and (2,3) for these values. So y2 = 3, y1 = -1, x2 = 2, x1 = 0

Plug them into the equation and solve

M = (3 + 1) / (2 - 0)
M = 4/2 = 2

Now we have the equation y = 2x - 1

Next to figure out if the points given are on the line you take the values and plug them into your equation like so: 4 = 2(-2) - 1

2(-2) - 1 does not equal 4 so this point does not fall on this line. Follow this same procedure for the next point given.

8 0

2 years ago

Two Social Security numbers (see Exercise 8.12) match zeros if a digit of one number is zero iff the corresponding digit of the

Roman55 [17]

Answer:

Proved

Step-by-step explanation:

From the given parameters, we have:

$n = 9$ i.e. the length of each security numbers

$r = 2$ i.e. 2 security numbers

Required

In 513 security numbers, 2 must have matching zeros

To do this, we make use of Pigeonhole principle.

First, we calculate the number of all security numbers not having matching zeros.

Each of the 9 digits can be selected in 2 ways.

2 ways implies that each digit is either 0 or not

So, total selection is:

$Total = 2^9$

$Total = 512$

Apply Pigeonhole principle

The principle states that: suppose there are n items in m containers, where $n>m$ , then there is at least one container that contains more than 1 item.