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Usimov [2.4K]
3 years ago
8

William has been contracted to paint a school classroom. The classroom is 20 m long, 15 m wide and 5 m high. There are four wind

ows (2m by 3m) and a door (2m by 1m). Determine the cost of painting the ceiling at N$ 6.50/m²​
Mathematics
2 answers:
valina [46]3 years ago
3 0

Answer:

Step-by-step explanation:

l -> length

b -> width

h -> height

Find the area of four walls and ceiling. then subtract the area of four windows and a door form that area.

Area of four walls + <u>ceiling</u> = 2( lh + bh) +<u>lb</u>

 = 2*(20*5 + 15*5) + 20*15

= 2( 100 + 75) + 300

= 2* 175 + 300

= 350 +300

= 650 sq m

Area of window = 2 *3 = 6 sq.m

Area of four windows = 4*6 = 24 sq.m

Area of door = 2 * 1 =  2 sq.m

Area of four walls excluding 4 windows and door = 650 - 24 - 2 = 624 sq.m

Cost of painting = 624 * 6.50

                          = $ 4056

I am Lyosha [343]3 years ago
3 0

Answer:  1950 dollars to paint the ceiling only (ignoring the walls)

The cost to paint the walls only is 2106 dollars.

The cost to paint the walls and ceiling is 4056 dollars.

==================================================

Explanation:

It seems a bit strange how your teacher mentions the windows and doors, but then asks about the ceiling only. Perhaps this is a red herring, but I'm not sure.

Anyway, to directly answer the question, we'll need to find the area of the ceiling first. The ceiling is a rectangle of dimensions 20 m by 15 m, so its area is 20*15 = 300 square meters.

Since paint costs 6.50 dollars per square meter, the total cost for the ceiling alone is 6.50*300 = 1950 dollars

If your teacher only cares about the ceiling, then you can stop here (and ignore the next section below).

---------------------------

If you wanted to find the cost to paint the walls, then we need to find the area of the walls.

For now, ignore the windows and door. Two opposite walls have area of 20*5 = 100 m^2 each. That accounts for 2*100 = 200 m^2 of wall area so far.

The other pair of opposite walls have area 15*5 = 75 m^2 each. That's another 2*75 = 150 m^2 of wall area.

In all, the total wall area without considering the windows or door is 200+150 = 350 m^2.

Now we consider the windows. Each window is 2 m by 3 m, yielding an area of 2*3 = 6 m^2. Four such windows have a total area of 4*6 = 24 m^2.

The door is 2 m by 1 m, so its area is 2*1 = 2 m^2

We'll subtract the wall area and the combined window+door areas to get

wallArea - windowArea - doorArea = 350-24-2 = 324

So after accounting for the windows and door, the amount of wall to paint is 324 m^2, which leads to a cost of 6.50*324 = 2106 dollars.

Therefore, painting the walls and ceiling gets us a total cost of 1950+2106 = 4056 dollars

This section is entirely optional if your teacher only cares about the ceiling.

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Travka [436]

Answer:

# The solution x = -5

# The solution is x = 1

# The solution is x = 6.4

# The solution is x = 4

# The solution is 1.7427

# The solution is 0.190757

Step-by-step explanation:

* Lets revise some rules of the exponents and the logarithmic equation

# Exponent rules:

1- b^m  ×  b^n  =  b^(m + n) ⇒ in multiplication if they have same base

  we add  the power

2- b^m  ÷  b^n =  b^(m – n) ⇒  in division if they have same base we

   subtract  the power

3- (b^m)^n = b^(mn) ⇒ if we have power over power we multiply

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4- a^m × b^m = (ab)^m ⇒ if we multiply different bases with same  

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5- b^(-m) = 1/(b^m)  (for all nonzero real numbers b) ⇒ If we have

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6- If  a^m  =  a^n  ,  then  m  =  n ⇒ equal bases get equal powers

7- If  a^m  =  b^m  ,  then  a  =  b    or    m  =  0

# Logarithmic rules:

1- log_{a}b=n-----a^{n}=b

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3- log_{a}q+log_{a}p=log_{a}qp

4- log_{a}q-log_{a}p=log_{a}\frac{q}{p}

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∴ log_{2}x(x-3)=2

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OR

∵ x + 1 = 0 ⇒ subtract 1 from both sides

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