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worty [1.4K]
3 years ago
7

The distance between Bay Town and Oak Glen is 175 miles. If the equation y = -x +175 represents the distance left to travel to O

ak Glen, what does the domain represent? What is the domain?
A) time left to reach Oak Glen; x ≥ 0
B) distance left to reach Oak Glen; x ≥ 0
C) time it takes to travel to Oak Glen; x ≥ 0
Eliminate
D) distance traveled since leaving Bay Town; x ≥ 0

Mathematics
1 answer:
denis-greek [22]3 years ago
5 0
My answer is isn't on the choices and I can't show you how I got it. My answer is in the attached files.

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Find the distance between each pair of points E(-1, 0),F(12, 0)
Olegator [25]

Answer:

The answer is

<h2>13 \:  \: units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3>d =  \sqrt{ ({x1 - x2})^{2}  +  ({y1 - y2})^{2} }  \\ </h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

E(-1, 0) and F(12, 0)

The distance between them is

<h3>|EF|  =  \sqrt{ ({ - 1 - 12})^{2}  +  ({0 - 0})^{2} }  \\  =  \sqrt{ ({ - 13})^{2} }  \\  =  \sqrt{169}</h3>

We have the final answer as

<h3>|EF| = 13 \:  \: units</h3>

Hope this helps you

5 0
3 years ago
HELP ME I NEED HELP!!!!
eduard

Answer:

D

Step-by-step explanation:

Since they are the same variables

7 0
3 years ago
The speed of sounds in water is 1.46 x 10^3 meters per second. The speed of sound in air is 3.31 x 10^2 meters per second. How m
dlinn [17]

Answer:

Hey there!

1.46x10^3-3.31x10^2

14.6x10^2-3.31x10^2

11.29x10^2

1.129x10^3

J is the correct answer.

Hope this helps :)

6 0
3 years ago
The x-coordinate for an ordered pair in the solution set of x+2y=11 is 3.Find the y-coordinate
cestrela7 [59]

Answer:4

Step-by-step explanation:

8 0
3 years ago
A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui
Sedbober [7]

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= \frac{dx}{dt}=1. 8 m/s

\triangle ABD\sim\triangle ECD

Therefore, \frac{AB}{EC}=\frac{BD}{CD}

\frac{y}{2}=\frac{12}{x}

xy=24

Differentiate w.r.t t

x\frac{dy}{dt}+y\frac{dx}{dt}=0

x\frac{dy}{dt}=-y\frac{dx}{dt}

\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}

When the man is 4 m from  the building

Then, we have x=12-4=8 m

\frac{dx}{dt}=1.8 m/s

Substitute the values in above equation then, we get

8y=24

y=\frac{24}{8}=3

Substitute the values then we get

\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.

8 0
3 years ago
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