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melisa1 [442]
3 years ago
14

PLEASE HELP PLEASE AND EXPLAIN THANK YOU​

Mathematics
1 answer:
Ainat [17]3 years ago
3 0

Answer: 56

Step-by-step explanation:

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A building is 200 ft long and 140 ft wide a 1/500 model is built of the building how long and how wide is the model
Luba_88 [7]
It will be 2/5 ft long and 7/25 ft wide. 200•1/500=2/5
140•1/500=7/25
8 0
4 years ago
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A tool bench is 35 inches long and 19. Inches wide how many square inches of the basement floor does it cover
konstantin123 [22]

Answer:

665

Step-by-step explanation:

The tool bench which has the shape of a rectangle, a 4-sided shape with lengths and width.

The area of a rectangle A is given by the formula;

A = L × B

Where L is the length, B is the width

Given the bench is 35 inches long and 19 Inches wide

L = 35 inches, B = 19 inches

A = 35 × 19 = 665 square inches

The tool bench will cover 665 square inches of the basement floor .

7 0
3 years ago
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Zaria wants you to solve this puzzle: “I am thinking of a number. If you divide my number by 2 and subtract 4, you will get 2. W
ycow [4]
Your answer should be 12 because you would take 2 ( because that is the ending number) and add 4. Then multiply by 2 and it is 12.
    step 1)  2+4=6
    step 2) 6 x 2=12
    step 3) 12=12 you're answer
6 0
4 years ago
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The equation of the graphed line is 2x - 3y = 12.
Kazeer [188]

Answer:

x- intercept = 6

Step-by-step explanation:

to find the x- intercept, let y = 0 in the equation and solve for x

2x - 3y = 12

2x - 3(0) = 12

2x - 0 = 12

2x = 12 ( divide both sides by 2 )

x = 6 ← is the x- intercept

7 0
1 year ago
Find the volume of the following solid. Using cylindrical coordinates
Reika [66]

In cylindrical coordinates, the equations of the surfaces become

z = 4 - 4r^2 \\\\ z = (r^2)^2 - 1 = r^4 - 1

These surfaces intersect on the cylinder of radius 1 with cross sections parallel to the x,y-plane:

4 - 4r^2 = r^4 - 1 \\\\ \implies r^4 + 4r^2 - 5 = (r-1)(r+1)(r^2+5) = 0 \\\\ \implies r=1

Then in cylindrical coordinates, the volume of the space bounded by these surfaces is

\displaystyle \int_0^{2\pi}\int_0^1\int_{r^4-1}^{4-4r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^1 \int_{r^4 - 1}^{4-4r^2}r\,\mathrm dz\,\mathrm dr \\\\ = \pi \int_0^1 (4-4r^2)^2 - (r^4-1)^2 \,\mathrm dr \\\\ = \pi \int_0^1 (15-32r^2+18r^4-r^8)\,\mathrm dr = \boxed{\frac{352\pi}{45}}

7 0
3 years ago
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