A wall is 4m high and 40 cm wide it contain two Windows of dimension 2.5 X 1 metre and gateway of dimension 2 metre multiply 3 if 24560 cubical bricks of length 20 cm are required to construct the wall, find the length of the wall. solve it

125.55 m.

Step-by-step explanation:

We need to find the volume of the bricked wall.

= total volume - volume of the windows - volume of the gateway.

= 4 * 0.4 * L - 2 * 2.5 * 1 * 0.4 - 2 * 3 * 0.4  m^3.   (where L is the length of the wall).

Also this volume = number of bricks * volume of 1 brick

= 24560* 0.2^3.

So

4 * 0.4 * L - 2 * 2.5 * 1 * 0.4 - 2 * 3 * 0.4 = 24560* 0.2^3  

1.6L - 2 - 2.4 = 196.48

1.6L = 200.88

L = 125.55 m.

3. (5x + 1) - (-10x + 6)

simplify the answer choice. combine like terms. remember to distribute the negative to the terms inside the second parenthesis. also remember that two negatives = one positive, and one of each sign = negative.

- (-10x + 6) = + 10x - 6

simplify. combine like terms:

5x + 1 + 10x - 6

5x + 10x + 1 - 6

15x - 5

∴ it is your answer

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1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.