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madreJ [45]
1 year ago
10

A community hall is in the shape of a cuboid the hall is 40m long 15m high and 3m wide. 10 litre paint covers 25m squared costs

?10. 1m squared floor tiles costs ?3. Work out the total costs of tiles and paints
Mathematics
1 answer:
Darina [25.2K]1 year ago
6 0

Answer:

Total cost for tiles and paints is $924.  

Step-by-step explanation:

We have been given that a community hall is in the shape of a cuboid. The hall is 40m long 15m high and 3m wide.

The paint will be required for 4 walls and ceiling.

Let us find area of walls and ceiling.

\text{Area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)

\text{Area of walls and ceiling}=1200+90+120

\text{Area of walls and ceiling}=1410

Therefore, the area of walls and ceiling is 1410 square meters.

Given: Cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. Therefore,  

\text{ The total painting cost}=10*(\frac{1410}{25})

\text{ The total painting cost}=10*56.4=564

Therefore, the total painting cost is $564.  

Tiles will be required for floor. Let us find the area of floor.

\text{Area of floor} = 40*3\text{ square meters}

\text{Area of floor} =120\text{ square meters}

Given: 1m squared floor tiles costs $3. So,  

\text{Total cost for tiles} = 3*120 = 360

Therefore, the total tiles cost is $360.

Now let us find combined total cost of tiles and paint.

\text{Combined total cost}= 564+360 = 924

Therefore, the combined total cost of tiles and paint is $924.

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Viefleur [7K]

Answer:

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

Step-by-step explanation:

See procedure below;

First let us make these two functions ( y = -x + 1 , y = x^2 + x - 2 ) equivalent;

- x + 1  = x^2 + x - 2,

-x - x + 1 + 2 - x^2 = 0,

- 2x + 3 - x^2 = 0,

x^2 + 2x - 3 = 0,

Now factor the simplified equation;

x^2 + 2x - 3 = 0,

( x^2 - x ) + ( 3x - 3 ) = 0,

x ( x - 1 ) + 3 ( x - 1 ) = 0,

( x - 1 )( x + 3 ) = 0,

And solve for x;

x - 1 = 0, and x + 3 = 0,

x = 1, and x = - 3

Now substitute this value of x into the two functions as to receive the y  values for each x - value;

y = - ( 1 ) + 1, <em>y = 0 for x = 1</em>,

y = ( - 3 )^2 - 3 - 2, y = 9 - 3 - 2, <em>y = 4 for x = - 3</em>,

<em>Solution; ( 1, 0 ), and ( - 3, 4 )</em>

8 0
1 year ago
I am a factor of 18.the other factor is 9.what number am i?
Liula [17]
The answer to this question is 2

3 0
1 year ago
What are the real zeros of the function g(x) = x^3 + 2x^2 − x − 2?
PSYCHO15rus [73]
If there are real roots to be found for this polynomial, the Rational Root Theorem and synthetic division are the best way to find them. I teach from a book that uses c and d for the possible roots of the polynomial.  C is our constant, 2, and d is the leading coefficient, 1.  The factors of 2 are +/- 1 and +/-2.  The factors for 1 are +/-1 only.  Meaning, in all, there are 4 possibilities as roots for this polynomial.  But there are only 3 total (because our polynomial is a third degree), so we have to find the first one, at least, from our possibilities above.  Let's try x = -1, factor form (x + 1).  If there is no remainder when we do the synthetic division, then -1 is a root.  Put -1 outside the "box" and the coefficients from the polynomial inside: -1  (1  2  -1  -2).  Bring down the first coefficient of 1 and multiply it by the -1 outside to get -1.  Put that -1 up under the 2 and add to get 1.  Multiply 1 times the -1 to get -1 and put that -1 up under the -1 and add to get -2.  -1 times -2 is 2, and -2 + 2 = 0.  So we have our first root of (x+1).  The numbers we get when we do the addition along the way are the coefficients of our new polynomial, the depressed polynomial (NOT a sad one cuz it hates math, but a new polynomial that is one degree less than that of which we started!).  The new polynomial is x^{2} +x-2=0.  That can also be factored to find the remaining 2 roots.  Use standard factoring to find that the other 2 solutions are (x+2) and (x-1).  Our solutions then are x = -2, -1, 1, choice B from above.
6 0
2 years ago
Read 2 more answers
On a coordinate plane, a parallelogram has points A (negative 3, 4), B (3, 4), C (1, negative 2), and D (negative 5, negative 2)
vladimir2022 [97]

Answer:

(1,-12)

Step-by-step explanation:

d starts at (-5,-2) if you add 6 to -5 and minus -2 by 10 you will end up with (1,-12)

3 0
1 year ago
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In △ABC,c=71, m∠B=123°, and a=65. Find b.<br><br> A. 101.5<br> B. 117.8<br> C. 123.0<br> D. 119.6
tia_tia [17]

Answer:

Option D

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the two sides that are given are adjacent to each other the given angle is the included angle. This means that the angle B is formed by the intersection of the lines a and c. Therefore, cosine rule will be used to calculate the length of b. The cosine rule is:

b^2 = a^2 + c^2 - 2*a*c*cos(B).

The question specifies that c=71, B=123°, and a=65. Plugging in the values:

b^2 = 65^2 + 71^2 - 2(65)(71)*cos(123°).

Simplifying gives:

b^2 = 14293.0182932.

Taking square root on the both sides gives b = 119.6 (rounded to the one decimal place).

This means that the Option D is the correct choice!!!

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2 years ago
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