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Vesna [10]
3 years ago
9

Need math help plsssssss

Mathematics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

The answer is a

Step-by-step explanation:

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Can you think of any other real-life scenarios where parallel lines and transversals<br> exist ?
RoseWind [281]

Answer:

on the road the 2 yellow line on the road

Step-by-step explanation:

8 0
4 years ago
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The length of a rectangle is three less than four times the width. The perimeter of the rectangle is 54 feet. Find the width of
kirill115 [55]

Answer:

6 ft

Step-by-step explanation:

Let the width = w

Then the length = 4w - 3

perimeter = 2(length + width)

perimeter = 2(4w - 3 + w)

perimeter = 2(5w - 3)

perimeter = 10w - 6 = 54

10w - 6 = 54

10w = 60

w = 6

Answer: The width is 6 ft

4 0
3 years ago
Mrs. Magdalino kept records on how much she spent on gasoline and the maintenance of her car. She found that it cost $485 to dri
algol [13]

Answer:

  • C = 0.97m
  • $1164 for 1200 miles
  • 845 miles for $820

Step-by-step explanation:

Given a car's cost of operation is $485 for 500 miles, you want an equation relating cost for m miles, and solutions to that equation for 1200 miles, and for a cost of $820.

<h3>Cost per mile</h3>

The cost per mile is found by dividing the cost by the associated number of miles:

  $485/(500 mi) = $0.97 /mi

<h3>Equation</h3>

The equation for the cost will show the cost as the cost per mile multiplied by the number of miles:

  C = 0.97m . . . . . where C is cost in dollars for m miles driven

<h3>1200 miles</h3>

The cost for driving $1200 miles will be ...

  C = 0.97(1200) = $1164

The cost of driving 1200 miles is $1164.

<h3>$820</h3>

The number of miles that can be driven for a cost of $820 is ...

  820 = 0.97m

  m = 820/0.97 = 845.36

About 845 miles are driven for a cost of $820.

3 0
1 year ago
A car travels 1 092 km in 21 days what is the average distance show how you got your answer
spin [16.1K]
Maybe try dividing 1092 by 21 I got 52
7 0
4 years ago
Please answer the question ​
nataly862011 [7]

Answers:

A) 2,600 cubes (13 x 40 x 5 = 2,600)

B) 6,600 cubes (11 x 20 x 30 = 6,600)

C) Boxes A and C or boxes B and C

D) 11,200 cubes (13 x 40 x 5) + (11 x 20 x 30) + (8 x 10 x 25) = 11,200

5 0
2 years ago
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