Algebra Final help please?

Question

I am having trouble in my Algebra class, and have a final tommorow, can someone please help explain these problems to me?

1. The length of a rectangle is 2ft longer than the width. If the area is 16ft^2, then what are the length and width?

2. A new computer can process a company's monthly payroll in 1 hour less time than the old computer. To really save time, the manager used both computers and finished the payroll in 3 hours. How long would it take the new computer to do the payroll by itself?

3. The height in feet for a ball thrown upward at 48 feet per second is given s(t) = -16t^2 + 48t, where t is the time in seconds after the ball is tossed. What is the maximum height that the ball will reach?

Thank you so much, to anyone who helps...it really means a lot...

Answer 1

1. l=w+2 lw=16 or w(w+2)=16 so w^2+2w=16
so w^2+2w-16=0 so [4+-sqrt(4+64)]/2
so [2+-sqrt(17)]
w=2+sqrt(17) and l=4+sqrt(17)

Answer 2

Let's tackle these one at a time:
1. Let x be the width and x+2 the length.
Then x(x+2) = 16.
x²+2x-16 = 0.
Now b²-4ac = 4+64 = 68,
so the lhs is not factorable. So we use the
quadratic formula:
x = ½(-2 + √68) = ½(-2 + 2√17) = -1 +√17--the width,
since x must be positive.
x+ 2 = 1 + √17--the length.

2. Let x be the time for the old computer and x-1 the
time for the new one.
The key is to figure out what portion each machine
can do in 1 hour.
Thus the old computer can do 1/x of the job
in 1 hour and the new one 1/(x-1) of it in 1 hour.
So our equation for this problem is
1/x + 1/(x-1) = 1/3
(2x-1)/(x²-1) = 1/3, combining the fractions.
This gives
x²-1 = 6x-3
x²-6x+2 = 0.
Again, b² -4ac= 28, so we can't factor the lhs.
So we use the quadratic formula.
Since x must be positive and greater than 3
(each machine takes longer to do the job
than both working together)
x = ½(6 + √28) = ½(6 + 2√7) = 3 + √7.
x-1 = 2 + √7.
So it would take the new machine 2 + √7
or around 4.65 hours to do the job alone.


3. You are looking for the vertex of the parabola
s = -16t² + 48t.
Useful trick: If y = ax²+bx+c is the equation of
any parabola, the x coordinate of the vertex
always occurs at x = -b/2a.
So the t coordinate of maximum height is
t = 48/32 = 1.5
and the maximum height is -16(2.25) + 48(1.5) = 72-36 =
36 feet.
Another way to find the t coordinate of the
vertex:
s(t) = 0 when t= 0 and t = 3. By symmetry,
the t coordinate of the vertex is halfway in between
these 2, or it is at t= 1.5.

Lots and lots of good luck on your final tomorrow!

Answer 3

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Answer 4

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Answer 5

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