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Paul [167]
3 years ago
14

What is the unit rate for meters per second if a car travels 374 meters in 17 seconds?

Mathematics
2 answers:
Vikentia [17]3 years ago
8 0

Answer:

22 meters per second.

Step-by-step explanation:

374 divided by 17 = 22.

erastova [34]3 years ago
6 0

Answer:

22

374meters. 17

(/). divided by (/)

17seconds 17

Equals 22 meters per second

Step-by-step explanation:

hope this helps and if ur feeling genous mark brainliest

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2 years ago
Write an algebraic expression for the phrase. <br><br> the product of p and 9
Cerrena [4.2K]

Answer:

9p

Step-by-step explanation:

Product means multiplication

So 9*p equals 9p

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4 0
3 years ago
Given z1 = 2 +StartRoot 3 EndRoot i and z2 = 1 – StartRoot 3 EndRoot i, what is the sum of z1 and z2?
Fofino [41]

Answer:

z1 + z2 = 3

Step-by-step explanation:

Since we are given z1 = 2 + √(3)i and z2 = 1 – √(3)i. The sum of z1 + z2 would be:

(2 + √(3)i) + (1 – √(3)i) = 2 + √(3)i + 1 – √(3)i = 2 + 1 + √(3)i – √(3)i = 3

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8 0
3 years ago
Read 2 more answers
A rectangular parking area measuring 5000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5
kaheart [24]

Answer:

Dimensions: 75.3778 ft and 66.3325 ft

Minimum price: $1658.31

Step-by-step explanation:

Let's call the length of the parking area 'x', and the width 'y'.

Then, we can write the following equations:

-> Area of the park:

x * y = 5000

-> Price of the fences:

P = 2*x*5.5 + y*5.5 + y*7

P = 11*x + 12.5*y

From the first equation, we have that y = 5000/x

Using this value in the equation for P, we have:

P = 11*x + 12.5*5000/x = 11*x + 62500/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

dP/dx = 11 - 62500/x^2 = 0

x^2 = 65000/11

x = 250/sqrt(11) = 75.3778 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x*y = 5000

y = 5000/75.3778 = 66.3325

P = 11*x + 62500/x = $1658.31

5 0
3 years ago
What is 7^5 equal to
Alona [7]
The answer to 7^5 = 16807
5 0
2 years ago
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