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Kipish [7]
3 years ago
15

A car travels 2 5/8 miles in 3 1/2 minutes at a constant speed. Which equation represents the distance, d, that the car travels

in m minutes?
Mathematics
1 answer:
Verizon [17]3 years ago
3 0

Answer:

d=0.75m

Step-by-step explanation:

Let

d------> the distance in miles

m----> the time in minutes

we know that

The speed is equal to divide the distance by the time

so

speed=d/m

we have

d=2\frac{5}{8}\ miles=\frac{2*8+5}{8}=\frac{21}{8}\ miles

m=3\frac{1}{2}\ minutes=\frac{3*2+1}{2}=\frac{7}{2}\ minutes

we know that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form y/x=k or y=kx

so

In this problem the speed is the constant of proportionality

d=km

Find the value of k

k=\frac{(21/8)}{(7/2)} =0.75\frac{miles}{minute}

d=0.75m ----> linear equation that represent the distance, d, that the car travels in m minutes

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A. f(x) = l-2xl - 3<br> Domain: {1, 2, 3}<br> Find the range
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Hello there!

We are given the function:

\displaystyle \large{ f(x) =  | - 2x|  - 3} \\

To find the range, we know that domain is the set of all x-values and also called 'input' while range is the set of all y-values and also called 'output'.

Basic Function - you add the input, you get the output. You add x-value in a function, you get y-value. You add domain, you get range.

So, we substitute x = 1,2 and 3 in the function.

<u>x</u><u> </u><u>=</u><u> </u><u>1</u>

\displaystyle \large{ f(1) =  | - 2(1)|  - 3} \\   \displaystyle \large{ f(1) =  | - 2|  - 3} \\

Recall that any numbers in absolute value are always positive.

\displaystyle \large{ f(1) =  2 - 3} \\   \displaystyle \large{ f(1) =   - 1} \\

<u>x</u><u> </u><u>=</u><u> </u><u>2</u>

\displaystyle \large{ f(2) =  | - 2(2)|  - 3} \\   \displaystyle \large{ f(2) =  | - 4 |   - 3} \\   \displaystyle \large{ f(2) =  4 - 3} \\   \displaystyle \large{ f(2) =  1}

<u>x</u><u> </u><u>=</u><u> </u><u>3</u>

\displaystyle \large{ f(3) =  | - 2(3)|  - 3} \\   \displaystyle \large{ f(3) =  | - 6|   - 3} \\   \displaystyle \large{ f(3) =  6 - 3} \\   \displaystyle \large{ f(3) =  3}

Therefore, Range: {-1,1,3}

Let me know if you have any questions!

Topic: Absolute Value Function / Modulus Function

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