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Fed [463]
3 years ago
7

Natalie walked 3/5 mile in 1/2 hour. How many miles did she walk per hour

Mathematics
1 answer:
Marizza181 [45]3 years ago
5 0

Answer:

She walked 1  1/5 miles per hour.

Step-by-step explanation:

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8xy^3*xy ^8 plsssssssssssssssssssss help
Sindrei [870]

Answer:

x • (x^3 - 3x^2y + 3xy^2 + 7y^3)

Step-by-step explanation:

(8x • (y^3)) +  x • (x - y)^3

2^3xy^3 +  x • (x - y)^3

Evaluate :  (x-y)^3   =   x^3-3x^2y+3xy^2-y^3

Pull out like factors :

x^4 - 3x^3y + 3x^2y^2 + 7xy^3  =

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x^3 - 3x^2y + 3xy^2 + 7y^3 is not a perfect cube

4 0
2 years ago
A printer cost $60 at the big box store, but I have a 25% off coupon. The tax rate is 6.75%.
Andrews [41]

Answer:

60 \times \frac{25}{100}

= 15

B) 15

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4 0
2 years ago
How does the graph show that the rate of change is constant?
muminat

A graph has a constant rate of change when the graph is linear and portrays a straight line. The slope will remain constant which forms the straight line in the graph.

4 0
3 years ago
Read 2 more answers
Volunteers at an animal shelter are building a rectangular dog run so that one shorter side of the rectangle is formed by the sh
Shkiper50 [21]

Write and solve a compound inequality to model the possible length of the dog run.

The inequality to model the possible length of the dog run is;. 100 ≤ 2.50x ≥ 200

And the possible length of the dog run is 80ft.

Minimum spending = $100

Maximum spending = $200

Cost per square feet = $2.50

let

x = possible number of square feet

The inequality:

100 ≤ 2.50x ≥ 200

This means possible number of square feet constructed is greater than or equal to $100 or less than or equal to $200

solve:

100 ≤ 2.50x ≥ 200

divide the inequality into 2

100 ≤ 2.50x

x ≤ 100/2.5

x ≤ 40

the other part:

2.50x ≥ 200

x ≥ 200/2.50

x ≥ 80

Therefore,

the possible length of the dog run is 80 feet

Read more:

brainly.com/question/9135876

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2 years ago
The radius of a circle is 5 meters. What is the angle measure of an arc 2π meters long
forsale [732]

Answer:

try using an app called photomath, it is very accurite and helpful

Step-by-step explanation:

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