I don’t understand please help

Blake is playing a racing game on his computer. The game tracks the locations of objects using and y coordinates (see graph belo

Anvisha [2.4K]

<h3>Answer: 2.2 units</h3>

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Explanation:

I'll define these point labels

B = Blake's starting position
F = finish line
C = the third unmarked point of the triangle

The locations of the points are

B = (-8,1)
C = (-6,-3)
F = (4,-2)

Use the distance formula to find the distance from B to C

$B = (x_1,y_1) = (-8,1) \text{ and } C = (x_2,y_2) = (-6,-3)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-8-(-6))^2 + (1-(-3))^2}\\\\d = \sqrt{(-8+6)^2 + (1+3)^2}\\\\d = \sqrt{(-2)^2 + (4)^2}\\\\d = \sqrt{4 + 16}\\\\d = \sqrt{20} \ \text{ ... exact distance}\\\\d \approx 4.47214 \ \ \text{... approximate distance}\\\\$

Segment BC is roughly 4.47214 units long.

Following similar steps, you should find that segment CF is approximately 10.04988 units long.

If Blake doesn't take the shortcut, then he travels approximately BC+CF = 4.47214+10.04988 = 14.52202 units. This is the path from B to C to F in that order.

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Use the distance formula again to find the distance from B to F. This distance is about 12.36932 units. He travels this amount if he takes the shortcut.

Subtract this and the previous result we got

14.52202 - 12.36932 = 2.1527

That rounds to 2.2

This is the amount of distance he doesn't have to travel when he takes the shortcut.

In other words, the track is roughly 2.2 units shorter when taking the shortcut.

Side note: Replace "units" with whatever units you're working with (eg: feet or meters).

7 0

3 years ago

Ignoring leap days, the days of the year can be numbered 1 to 365. Assume that birthdays are equally likely to fall on any day o

faust18 [17]

Answer:

Follows are the solution to the given question:

Step-by-step explanation:

They can count the days of the year 1 to 365. The random project consists of drawing a sample of n objects from D where elements are n people's birth in a group but instead, D = {1,....365}. And then there's the issue.

$S=365^n$

This because the list of future birthdays of n people was its test point; therefore m points will be in the sequence so each point contains 365 distinct outcomes. The probability function P for \Omega is that any event is likely to happen in 365 days.

$P(x)=\frac{1}{365^{n}}$

if x is between 1 and 365 as well as the occurrence is just all similarly possible

In point i:

That somebody mentions their birthday throughout the party

Guess I was born on day b. Therefore the consequence of "x is in A" is "b is now in the series of x," which is to say, b = bk for some amount k approximately 1 and n.

In point ii:

Any 2 persons share the same birthday at this party". A result x is in B" means which "two of entries in x are same." This means that perhaps the outcome x is in B if or only if bj = bk is in B of two numbers j, and k of 1, of two. , no, n.

In point iii:

Many three students share the same birthday with both the party. The consequence is x at the level of C only when bj = bk = bl at three (different) indices, j, k, l, 1. , no, n.

6 0

3 years ago

Britney spent 115 minutes reading last week and 140 minutes reading this week.

Ann [662]

255? i not sure though

5 0

3 years ago

The length of a rectangle is 3 more inches than the width. Find the length and width of the rectangle is the perimeter of the re

n200080 [17]

Part A:

Let the width of the rectangle be = x inches

As given, the length of a rectangle is 3 more inches than the width so let the length of the rectangle be = x+3 inches