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Math help! It's a tall order....?

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So I totally understand if you can not answer all of these...if you can then you are a god = D but any help i can get i will take thankfully!!!!!

1) which of the following expressions are identical?

a. cos^2 x b. (cosx)^2 c. cosx^2

2) Solve for x on the interval [0,2pi]

a. tan^2x-1 = 0 b. cos^2x + 3 cosx +2 = 0 c. 2sin^2x - sin x = 1

3)Factor

a. e^-x -xe^x + (2x^2) (e^2x)

b. sin x + tan x

4)The height of a cylinder is twice its diameter. Express the total surface area S as a function of the height h.

5) a light 3 meters above the ground causes a boy 180 cm tall to cast a shadow s centimeters long, measured along the ground. Express s as a function of his distance d from the light.

6) the surface area of a box with a square base and top is 3 square meters. Express the volume V of the box as a function of width, w, of the base.

7) A cylindrical can has a volume of 400pi cubic centimeters. The material for the top and bottom costs 3 cents per square centimeter. The material used for the vertical surface costs 2 cents per square centimeter. Express the cost of the materials used to make the can as a function of the radius r.

LAST ONE!!! sorry guys but thanks again!

8) a power station and a factory are on opposite sides of a river 2 km wide. A power line must be run from the power station. It costs $25 per meter to run the cable in the river and $20 per meter on land. Express the total cost as a function of x.

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  1. 1) which of the following expressions are identical?

    a. cos^2 x b. (cosx)^2 c. cosx^2

    A and B are identical.  

    2) Solve for x on the interval [0,2pi]

    a. tan^2x-1 = 0 b. cos^2x + 3 cosx +2 = 0 c. 2sin^2x - sin x = 1

    a. tan^2x=1, tanx=+-1, take inverse tan...

    3)Factor

    a. e^-x -xe^x + (2x^2) (e^2x)

    b. sin x + tan x

    b. sinx (1 + secx)

    4)The height of a cylinder is twice its diameter. Express the total surface area S as a function of the height h.

    The formula for surface area of cylinder is:

    SA = 2(pi)r^2+2(pi)rh

    If h=2d and d=2r (where r is the radius) then h=4r and plug in:

    SA = 2(pi)r(r + h) where SA = 2(pi)(h/4)((h/4)+h) and reduce

    SA = 2(pi)(5h^2/16) or (pi)(5h^2/8)

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