WEBVTT

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>> Welcome to the Cypress
College Math Review

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on Applications of
Quadratic Equations.

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Objective one, perimeter
and area problems.

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Here's some formulas
that you need to know.

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For a square, the perimeter is
4 times the length of a side.

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The area is side squared.

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For a rectangle, the
perimeter or distance around,

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is 2 times the length
plus 2 times the width.

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The area is length times width.

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For a triangle, the perimeter
is the sum of the three sides,

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A plus B plus C. The area
is 1/2 base times height

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where the height and base are
perpendicular to each other.

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For a circle, the distance
around is called circumference.

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And there's two different
formulas you can use,

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2 pi R or pi times
D. And the reason

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for that is the diameter
is twice the radius.

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The area of a circle is pi
times the radius squared.

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The area of a triangle
is 112 square meters.

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The height is 2 meters
longer than the base.

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What are the dimensions
of the triangle?

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So we're told that the area

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of the triangle is
112 square meters.

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The formula for area

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of a triangle is 1/2
base times height.

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So we have to figure
out what to put

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in for the base and
for the height.

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The height is 2 meters
longer than the base.

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So height equals 2 plus
base, longer than, plus base.

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So we would let X be the base.

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Then the height would be 2 plus
X. Substitute in X for base

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and 2 plus X or X plus 2 for the
height and we have our equation.

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Pause the video if you'd
like to go through the steps

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to solve this equation.

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Here's the answer.

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The area of a circle
is 12 square inches.

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What is the radius
of the circle?

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So area of a circle equals 12.

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Our formula is area
equals pi R squared

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or pi times the radius squared.

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So our unknown is the radius.

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We'll let R be the radius.

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We substitute in 12 for the
area and here's the equation

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that we need to solve.

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Pause the video if
you want to go

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through the steps to solve it.

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And then here's our answer.

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Objective two, work
rate problems.

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It takes Guiana 2 hours longer

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to paint a room than
it does Bryson.

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It takes 7 hours for the two
of them to paint the room

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if they are working together.

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How long does it
take each one of them

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if they did the job
all by themselves?

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So we have a work rate problem.

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And the set up is the part
of the job done by Guiana

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in 1 hour plus the part
of the job done by Bryson

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in 1 hour equals the part of
the job done together in 1 hour.

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The formula is 1 over time.

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Now, Guiana takes 2 hours
longer to paint the room

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than Bryson does, so we should
let X be the time it would take

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Bryson to paint the room if
he was working by himself.

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Then X plus 2, 2 hours longer,
is the time it would take Guiana

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to paint the room working alone.

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So the part of the job done
by Guiana in 1 hour is 1

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over how long it would
take her to do it

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by herself, which is X plus 2.

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The part of the job done
by Bryson in 1 hour is 1

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over the time it would take
him to do it by himself,

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so that's 1 over X. And the
part of the job done together

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in 1 hour is 1 over 7, since 7
hours is how long it would take

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them if they were
working together.

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So this is the equation
that models that problem.

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Pause the video if you want
to go through the steps

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to solve this equation.

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And here's our answer.

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Pause the video and
try these problems.

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Objective three, Pythagorean
Theorem applications.

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If you have a right triangle,
then the sum of the squares

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of the legs is equal to the
square of the hypotenuse.

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Example, the diagonal
of a square is 26 feet.

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What is the length
of one of the sides?

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So we have a square.

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Here's its diagonal.

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So the diagonal is 26 feet.

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Since this is a square,
all the sides are the same.

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We'll let X be the length
of the side of the square

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and we fill that in our diagram.

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We'll use the Pythagorean
Theorem to solve for X,

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substituting in X
for A and X for B.

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And the hypotenuse is
26, that's the diagonal.

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Here's the equation
that models our problem.

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Pause the video if you want to
go through the steps to solve.

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And here's our answer.

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A guy wire for an
antenna is 27 yards long.

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The height of the
antenna is 18 yards.

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How far from the base

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of the antenna should the guy
wire be anchored to the ground?

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Here's an antenna and here's
the guy wire that anchors it.

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So a guy wire is a
diagonal from the top

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of the antenna down
to the ground.

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That forms a right triangle.

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The length of the guy
wire was 27 yards long.

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The height of the
antenna is 18 yards.

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We're trying to find out
how far from the base

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of the antenna the guy
wire should be anchored

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to the ground.

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Well let X be the distance
from the base of the antenna

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to where the guy wire
will be anchored.

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Now, we have a right triangle
with the three sides filled in.

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Here's the Pythagorean Theorem
that we're going to use.

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We'll let X be A and 18 be
B. Those are interchangeable,

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but C has to be the 27.

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Here's the equation
that models our problem.

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Pause the video if you want
to go through the steps

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to solve this equation.

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And here's our answer.

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Objective four, consecutive
integer problems.

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Here are some consecutive
integers, 3, 4, 5.

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If we let X be the
smallest integer, 3,

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how do we get from 3 to 4?

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Well, we would add 1.

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How do we get from 4 to 5?

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We would add 1 again.

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So if the consecutive integers
were unknown, we could set

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up the problem using X for the
smallest, X plus 1 for the next,

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and X plus 2 for the one after
that and so on and so forth.

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What about consecutive even
integers, like 4, 6, 8?

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If X was the smallest,
how do I get from 4 to 6?

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Well, you would add 2.

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Then you would add 2
again to get from 6 to 8.

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So if the first one is X, the
next one would be X plus 2.

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The next one would be X plus
4 and so on and so forth.

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Consecutive odd integers.

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This is the one that
messes most people up.

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Here's my example, 7, 9, and 11.

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How do we get from 7 to 9?

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We add 2. How do we
get from 9 to 11?

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We add 2. We don't
add 1, we don't add 3.

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A lot of people think it's going

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to be something different
for odd integers.

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But to get to the next odd
integer, you have to skip

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over the even integer
that's in between them.

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So you're going to be adding 2
no matter whether you're working

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with consecutive even
or consecutive odd.

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So for consecutive off
integers, you're going to set it

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up where X is the smallest,
X plus 2 is the next one,

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and then X plus 4 and
so on and so forth.

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The product of two consecutive
odd integers is 23 more

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than 5 times their sum.

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Find the integers.

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Since we have consecutive
odd integers,

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I'll represent them
as X and X plus 2.

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Now let's read the
sentence again.

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The product, so product, is,
that's the equals, 23 more than,

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so that would be
23 plus, 5 times,

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so it would be 5 times
something, their sum.

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Sum. So that sentence
translates into this equation.

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Product equals 23
plus 5 times sum.

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It's important for you
to be able to translate

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in between a sentence
and an equation.

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Now what do we write
down for the product

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since our numbers
are X and X plus 2?

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Well, we would multiply them.

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So multiply X times
the quantity X plus 2.

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That's equal to 23
plus 5 times their sum.

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What do we put in for sum?

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That means add them.

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So we're going to
add X and X plus 2.

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This is the equation
that models our problem.

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Pause the video if you want
to go through the steps

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to solve the equation.

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Here's our answer.

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Be sure you have both
pairs of solutions.

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It's a good idea to check
your answer when you can.

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Here's the check
for this problem.

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Pause the video and
try these problems.

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Objective five, projectile
problems.

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The height in feet of a ball
thrown upward from a height

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of 18 feet is given by H equals
negative 16 T squared plus 32 T

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plus 18 after T seconds.

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So H is the height of the
ball and T is the time.

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How many seconds will
it take for the ball

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to reach a height of 30 feet?

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We want the height
to equal 30 feet.

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What's the height equal to?

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Well, the height's equal to
that quadratic expression.

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We're going to let T be the
time since the ball was thrown.

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We're going to replace H with
30 because we want to know

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when is the height equal to 30.

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Here's the equation
that models our problem.

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Pause the video if
you want to go

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through the steps to solve it.

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Notice on our answer that
we have two different times.

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So on the way up,
after a half a second,

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the ball reached 30 feet.

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And then on the way back down,
after 1 and a half seconds,

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the ball was at 30
feet once again.

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The height in feet of a ball
thrown upwards from a height

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of a6 feet is given by H equals
negative 16 T squared plus 32 T

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plus 6 after T seconds.

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So the height is H and the time
is T. How many seconds will it

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take for the ball
to hit the ground?

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Well, if it's on the
ground what's the height?

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The height would be zero.

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So we're going to let
the height equal zero.

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Well, the height is equal to
this quadratic expression.

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We're going to let T be the
time since the ball was thrown.

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We're going to let H be zero.

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And solve for T. So
here's the equation

00:12:46.516 --> 00:12:49.556 A:middle
that will give us the time
when this ball hits the ground.

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Pause the video if
you'd like to go

00:12:53.266 --> 00:12:55.676 A:middle
through the steps to solve it.

00:12:55.816 --> 00:12:57.000 A:middle
And here's our answer.

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Objective six, compound
interest problems.

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For interest that is
compounded annually,

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annually means one
time per year,

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use the formula A equals P times
the quantity 1 plus R raised

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to the T power.

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Where A is the amount in the
account at the end of the term,

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P is the amount invested or
the principal, T is the number

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of years, and R is
the interest rate.

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Find the rate R at which $500
compounded annually grows

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to $525 in two years.

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Round your answer to the
nearest tenth of a percent.

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Since the money's being
compounded annually,

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we use this formula and we
don't know the interest rate,

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so let R be the unknown
interest rate.

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$500 is growing to
$525, which tells us

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that 500 is our principal and
$525 is the amount we have

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at the end, which is A. Two
years, so T is equal to 2.

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Substitute these into our
formula and here's the equation

00:14:16.466 --> 00:14:17.756 A:middle
that models that problem.

00:14:18.356 --> 00:14:20.956 A:middle
The first step to
solve this is to get

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that quantity squared by itself.

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We want to get rid of the 500.

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It's being multiplied
times that parentheses,

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so we would divide
both sides by 500.

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Then you take the
square root of both sides

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and we have 1 plus R. Then we
subtract 1 from both sides.

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One of the answers
will be negative.

00:14:43.306 --> 00:14:44.426 A:middle
We discard that one.

00:14:44.796 --> 00:14:46.656 A:middle
And the other answer
is listed here.

00:14:49.876 --> 00:14:52.066 A:middle
The question asked us
to round our answer

00:14:52.066 --> 00:14:53.926 A:middle
to the nearest tenth
of a percent.

00:14:54.416 --> 00:14:55.836 A:middle
That can be tricky for people.

00:14:56.126 --> 00:15:00.066 A:middle
I recommend you first change it
into a percent, like I did here,

00:15:01.526 --> 00:15:03.886 A:middle
then round to the nearest
tenth of a percent.

00:15:03.886 --> 00:15:07.000 A:middle
So here's our answer.

00:15:12.496 --> 00:15:15.776 A:middle
Objective seven, distance,
rate, time problems.

00:15:16.956 --> 00:15:20.046 A:middle
Sharon and Jill each drove
from their home in St. George,

00:15:20.046 --> 00:15:24.156 A:middle
Utah to Salt Lake City,
a distance of 330 miles.

00:15:24.946 --> 00:15:27.996 A:middle
Sharon drove 11 miles an
hour faster than Jill did.

00:15:28.916 --> 00:15:31.246 A:middle
Sharon arrived 1
hour before Jill did.

00:15:31.946 --> 00:15:33.816 A:middle
How fast did each of them drive?

00:15:34.626 --> 00:15:35.116 A:middle
All right.

00:15:35.116 --> 00:15:35.956 A:middle
What are we given here?

00:15:36.606 --> 00:15:38.006 A:middle
Well, we're given the distance.

00:15:38.296 --> 00:15:40.736 A:middle
Each of them drove 330 miles.

00:15:40.736 --> 00:15:42.616 A:middle
It's only listed once,
but that's the distance

00:15:42.616 --> 00:15:43.656 A:middle
that each of them drove.

00:15:44.016 --> 00:15:45.966 A:middle
So we are given both
of their distances.

00:15:46.616 --> 00:15:49.866 A:middle
Any time you're given both of
the distances, the best way

00:15:49.866 --> 00:15:52.126 A:middle
to set it up is using the time.

00:15:52.126 --> 00:15:53.906 A:middle
So what relationship do we know

00:15:53.906 --> 00:15:56.066 A:middle
about the times for
Sharon and Jill?

00:15:56.836 --> 00:16:00.286 A:middle
Sharon arrived 1
hour before Jill did.

00:16:00.636 --> 00:16:03.076 A:middle
So her time was 1 hour less.

00:16:03.456 --> 00:16:07.596 A:middle
So the time for Sharon is 1 hour
less than the time for Jill.

00:16:08.106 --> 00:16:10.566 A:middle
One hour less than,
that reverses the order,

00:16:10.566 --> 00:16:13.236 A:middle
so it would be time
for Jill minus 1.

00:16:13.516 --> 00:16:16.976 A:middle
So the time for Sharon is 1 hour
less than the time for Jill.

00:16:18.256 --> 00:16:19.626 A:middle
Formula for distance.

00:16:19.886 --> 00:16:21.466 A:middle
Distance equals rate times time.

00:16:22.066 --> 00:16:25.446 A:middle
So time, which is what we
need, is distance over rate.

00:16:26.776 --> 00:16:30.256 A:middle
So we substitute distance
over rate for time for Sharon

00:16:30.776 --> 00:16:31.986 A:middle
and for the time for Jill.

00:16:33.816 --> 00:16:35.466 A:middle
What are our unknowns?

00:16:36.906 --> 00:16:38.326 A:middle
We don't know their rates.

00:16:39.326 --> 00:16:42.166 A:middle
We know that Sharon drove
11 miles an hour faster

00:16:42.166 --> 00:16:42.906 A:middle
than Jill did.

00:16:43.936 --> 00:16:47.076 A:middle
So we have Sharon's rate
in terms of Jill's rate,

00:16:47.566 --> 00:16:49.466 A:middle
which means X should
be Jill's rate.

00:16:49.466 --> 00:16:51.276 A:middle
If we let X be Jill's rate,

00:16:52.006 --> 00:16:54.586 A:middle
then Sharon's rate
would be X plus 11.

00:16:56.106 --> 00:17:00.056 A:middle
Now we're ready to
substitute in to our equation.

00:17:00.406 --> 00:17:03.376 A:middle
The distances are
both 330 miles.

00:17:04.396 --> 00:17:06.956 A:middle
Sharon's rate was X plus 11.

00:17:08.336 --> 00:17:12.516 A:middle
Jill's rate was X.
Here's the equation

00:17:12.516 --> 00:17:13.766 A:middle
that models our problem.

00:17:15.336 --> 00:17:16.886 A:middle
Pause the video if
you want to go

00:17:16.886 --> 00:17:18.596 A:middle
through the steps to solve it.

00:17:20.506 --> 00:17:22.000 A:middle
And then here's our answer.

00:17:27.186 --> 00:17:28.976 A:middle
Pause the video and
try these problems.