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myrzilka [38]
3 years ago
6

What shape best describes the cross-section cut at an angle to the base of a right rectangular prism? Trapezoid Parallelogram Sq

uare Rectangle

Mathematics
2 answers:
Dovator [93]3 years ago
4 0

Answer:

Parallelogram.

step by step explanation

A rectangular prism has a rectangle as its base .

-Dominant- [34]3 years ago
3 0

Answer:

Parallelogram

Step-by-step explanation:

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The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar​
Alexus [3.1K]

Question:

The cost of 12 kg sugar is $240 what will be the cost of 3 kg sugar.

Given:

  • Cost of 12 kg sugar = $240

To find?

  • cost of 3 kg sugar

Answer :

<u>To find </u><u>c</u><u>o</u><u>s</u><u>t</u><u> </u><u>of</u><u> </u><u>3kg sugar first we have to find cost of 1 kg sugar</u><u>.</u>

  • Cost of 1 kg sugar = Cost of total no. of sugar ÷Total sugar
  • Cost of 1 kg sugar = $240/12
  • Cost of 1 kg sugar = $20

<u>Now Let's find cost of 3 kg sugar</u>

Cost of 3kg sugar = total no of sugar ×Cost of 1 kg sugar

Cost of 3kg sugar = 3×$20

Cost of 3kg sugar = $60

3 0
2 years ago
The braking distance, in feet of a car a Travling at v miles per hour is given.
irakobra [83]

The braking distance is the distance the car travels before coming to a stop after the brakes are applied

a. The braking distances are as follows;

  • The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
  • The braking distance at 55 mph,  is approximately <u>298.35 ft.</u>
  • The braking distance at 85 mph,  is approximately <u>708.92 ft.</u>

b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s

Reason:

The given function for the braking distance is D = 2.6 + v²/22

a. The braking distance if the car is going 25 mph is therefore;

25 mph = 36.66339 ft./s

D = 2.6 + \dfrac{36.66339^2}{22} = 63.7 \ ft.

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>

At 55 mph, the braking distance is given as follows;

55 mph = 80.65945  ft.s

D = 2.6 + \dfrac{80.65945^2}{22} \approx 298.35 \ ft.

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>

At 85 mph, the braking distance is given as follows;

85 mph = 124.6555 ft.s

D = 2.6 + \dfrac{124.6555^2}{22} \approx 708.92 \ ft.

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>

b. The speed of the car when the braking distance is 450 feet is given as follows;

450 = 2.6 + \dfrac{v^2}{22}

v² = (450 - 2.6) × 22 = 9842.8

v = √(9842.2) ≈ 98.211 ft./s

The car was moving at v ≈ <u>98.211 ft./s</u>

Learn more here:

brainly.com/question/18591940

8 0
2 years ago
Subtract. Write your answer in simplest form. 8 7/10- 3 4/10
Misha Larkins [42]

Answer:

Exact form: 53/10 , Decimal form: 5.3 , Mixed number form: 5 3/10

Step-by-step explanation:

7 0
3 years ago
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How many miles is it from Turtle Bay to Waialua Bay​
lidiya [134]
12.9 miles or 27 minutes

4 0
3 years ago
Write the following numbers in order from least to greatest,
PIT_PIT [208]

Answer: D

Step-by-step explanation:

Look at the exponents first and order them from least to greatest

5 0
3 years ago
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