SOLUTION: w..8 I need help understanding step by step i got them wrong for my homework so what is the right way to solve them. Problem #6 Geometry. The length of a rectangle i

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Question 75800: w..8
I need help understanding step by step i got them wrong for my homework so what is the right way to solve them.
Problem #6
Geometry. The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?Give all answers to the nearest thousandth.

Problem #22
Science and medicine. The equation gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to
reach a height of 180 ft.Give all answers to the nearest thousandth.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 4 cm, what are the dimensions (the length and the width) of the rectangle?
:
Think of the right triangle formed by the diagonal and the two sides:
Let the width = x; then the length = (x+1), given the hypotenuse = 4
:
Remember pythagorus: a^2 + b^2 = c^2
In our problem that would be:
x^2 + (x+1)^2 = 4^2
:
x^2 + (x^2+2x+1) = 16; FOILed (x+1)(x+1)
:
x^2 + x^2 + 2x + 1 - 16 = 0
:
2x^2 + 2x - 15 = 0; a quadratic equation
:
Find x using the quadratic formula; a = 2; b = 2; c = -15
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
:
x+=+%28-2+%2B-+sqrt%28+2%5E2+-+4+%2A+2+%2A+-15+%29%29%2F%282%2A2%29+
:
x+=+%28-2+%2B-+sqrt%28+4+-+%28-120%29+%29%29%2F%284%29+
:
x+=+%28-2+%2B-+sqrt%28+124+%29%29%2F%284%29+
:
x+=+%28-2+%2B11.1355%29%2F%284%29+; we only need the positive solution
:
x+=+%2B9.1355%2F4
:
x = 2.284 wide and 3.284 long
:
Check our solution on a calc: Sqrt(2.284^2 + 3.284^2) = 4.0000
:
:
:
Find x using the quadratic equation: a = 2; b = 2; c = -15
22.
The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 ft.
:
Substitute 180 for h in the given equation: -16t^2 + 112t = h
:
-16t^2 + 112t = 180
:
-16t^2 + 112t - 180 = 0; subtract 180 from both sides, gives us a quadratic eq:
:
Simplify divide equation by -4, that changes the signs and gives you:
4t^2 - 28t + 45 = 0
:
Factor this to:
(2t - 5)(2t - 9) = 0
:
2t = +5
t = 2.5 sec (on the way up)
and
2t = +9
t = 4.5 sec (on the way down)
:
:
Check solution using t = 2.5, in the original equation:
-16(2.5^2) + 112(2.5) =
-16(6.25) + 280 =
-100 + 280 = 180
:
You can check it using the t = 4.5 solution
:
Any questions here?