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irina1246 [14]
3 years ago
8

The dimensions of a hall are 20m×15m×8m. Find the cost of white washing it's walls and ceiling at the rate of rupees 20per 10m^2

. Also find the volume of air present in the hall
The answer is 1360 and 2400m^3
But I need to know the explanation
I will mark u as brainlest if u answer these question
Mathematics
1 answer:
g100num [7]3 years ago
6 0

Answer:

Step-by-step explanation:

Area of wall = lh + lh + bh + bh = 2lh + 2bh = 2h *(l + b)

Area of ceiling = lb

Area of wall and ceiling = 2h(l +b) + lb

=  2*8* (20 +15) + 20*15 = 16*35 + 20*15

= 560 + 300 = 860

Cost of painting per m^2 =20/10 = 2

Cost of painting 860 m^2 = 860 * 2 =  1720

Volume = l * b *h = 20 * 15 * 8 = 2400 m^3

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I think it would be 100°

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3 years ago
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What is 2,034,627 rounded to the nearest ten thousand
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Answer:

rounded to the nearest ten thousand

2,034,627

2,030,000

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10x114x A8- О<br> Simplify
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The simplified answer would be 9120ax
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2 years ago
This hyperbola is centered at theorigin. Find its equation.Foci: (-2,0) and (2,0)Vertices: (-1,0) and (1,0)
beks73 [17]

SOLUTION

From the question, the center of the hyperbola is

\begin{gathered} (h,k),\text{ which is } \\ (0,0) \end{gathered}

a is the distance between the center to vertex, which is -1 or 1, and

c is the distance between the center to foci, which is -2 or 2.

b is given as

\begin{gathered} b^2=c^2-a^2 \\ b^2=2^2-1^2 \\ b=\sqrt[]{3} \end{gathered}

But equation of a hyperbola is given as

\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1

Substituting the values of a, b, h and k, we have

\begin{gathered} \frac{(x-0)^2}{1^2}-\frac{(y-0)^2}{\sqrt[]{3}^2}=1 \\ \frac{x^2}{1}-\frac{y^2}{3}=1 \end{gathered}

Hence the answer is

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5 0
11 months ago
The diameter of a circle is 8 feet. What is the angle measure of an arc a feet long?
fiasKO [112]

Answer:

Given: circle

diameter = 10 cm => radius (R) = 5 cm

Find: measure of angle bounding sector = 11 π sq. cm.

Plan: determine what part of the circle’s total area equals the sector’s area.

Total Area of Circle A = π R^2 = π 5^2 = 25 π sq. cm.

Therefore: Sector Area = 11 π cm^2/25 π cm^2 = 11/25

Since the sector is 11/25 th of the circles area, the sector angle will measure 11/25 th of the circle’s circumference. They are proportional.

C = 2 π R = 2 π (5) = 10 π cm

Sector Arc = measure of sector angle = 11/25 (10 π) =

22π/5 radians

Answer: Sector Arc = 22π/5 Radians

4 0
2 years ago
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