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Let = Carole's age now
Let = Carl's age now
= Carole's age years ago
= Carl's age years ago
= Carole's age years ago
= Carl's age years ago
--------------
given:
(1)
(2)
---------------
(1)
(1)
(1)
and
(2)
(2)
(2)
Subtract (2) from (1)
(3)
(3)
Substitute this back in (1) or (2)
(1)
(1)
(1)
Carole is 23 and Carl is 11

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Sometimes I do it this way:


The point (3,2) is given, so




So the equation is:

-----------------------------
The other way is:





And, of course,

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For each value of , would be
greater. The greater the value of ,
the greater the increase in .
That would move the graph closer to the
y-axis. (c) is the answer

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This is a minimum when and

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The roots occur where


(note that )
There is just 1 root, since

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Let = the percentage of global energy used by China
Let = the percentage of global energy used by Russia
Let = the percentage of global energy used by United States
given:
(1)
(2)
(3)
---------------------
Substitute (1) and (2) in (3)




And, from (1)
(1)
That's all I have time for

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Let = Jill's speed
For Jack:

For Jill:

given:
hr
km/hr
hr
km
----------------------
Rewriting equations:
(1)
(1)
(1)
(2)
(2)
(2)
(2) km/hr answer

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Let = the rate of the wind
Let = Jim's rate of walking without wind
given:
mi/hr


--------------------

(1)
and

(2)
Add (1) and (2)
(3)
(3)
and from (1)
(1)
(1)
(1)
(1) mi/hr
check:
(2)
(2)
(2)
(2)
OK

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Is this the formula:
? If so,
t is in seconds and s is in meters.
How long will it take her to free fall 750 meters?



sec
It will take her 12.37 sec to fall 750 m
------------------
check answer:


square both sides

Multiply both sides by

Divide both sides by

m
OK

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Let = the first amount invested @ 7%
given:






The amounts invested were:
$129,714 and $428,143

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The minute hand and the hour hand are each going in
a circle at a certain speed in degrees/hr
Between 2 and 3 o'clock, the hour hand covers 5 minutes
of the 60 minutes on the clock face
, so if the whole face is 360 degrees
degrees
When the hour hand moves 30 degrees, the minute and
goes all the way around, or 360 degrees. That's 12 times
faster than the hour hand
-------------------
To solve the problem, I ask the question:
If the hour hand moves a certain number of degrees be
tween 2 and 3, what time is it when the minute hand is
at the hour hand's position + 180 degrees? That would put
them directly opposite eachother.
-------------------
Let = the elapsed time in hours after 2 o'clock when the hands
are opposite eachother
= the rate of travel of the minute hand in degrees/ hr
= the rate of travel of the hour hand in degrees/hr
--------------------
Reading from 12 and going clockwise, the hour hand is at the 2
which is of the 360 degrees
degrees
So, when the hands are opposite the hour hand is at

I want the minute hand to be this position + 180 degrees.
I also know that { is the minute's hand rate of travel)
so, I can write



hrs
min
min
and
sec
The hands are opposite eachother at 43 min and 38 sec past 2 o'clock
---------------------
check answer:
In degrees the hour hand has moved
degrees
In degrees, the minute hand has moved
degrees
degrees
If I subtract the 60 degree headstart the hour hand had,

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Subtract from both sides

Add to both sides

Use complete the square method



Take the square root of both sides

and



check the answers:




and




OK

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The more increases, the smaller becomes
(a) She will sell less if she raises the price
(b) Every time she raises the price $1, the 2nd term
on the right increases by , so decreses
by , so the nuber of cans she sells drops by

is the vertical axis
is the horizontal axis

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Let = Bob's weight
given:

This is a solvable equation
Multiply both sides by


Bob weighs 200 pounds

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Multiply both sides by

Use complete the square method







check answers:





OK






OK

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With powers or what is the same thing, exponents,
this is the picture to have in your mind:
2^5 = 2*2*2*2*2
2^4 = 2*2*2*2
2^3 = 2*2*2
2^2 = 2*2
2^1 = 2
2^0 = 1 (any number to the power is
2^-1 = 1/2
2^-2 = 1/(2*2)
2^-3 = 1/(2*2*2)
2^-4 = 1/(2*2*2*2)
2^-5 = 1/(2*2*2*2*2)
---------------------

As with any fraction, if there is factor in the numerator
and the same factor in the denominator, they can be
cancelled, so
(I cancelled a 2 in the
numerator with a 2 in the denominator
I did the same thing again
I did the same thing again
If I follow this back to the beginning, I have:
answer

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Let = the speed of the current
Let = the speed of the boat in still water
For upstream:

For downstream:

given:
mi/hr
mi
mi
(I'll call them both )
Now I can rewrite the equations
(1)
(2)
This is 2 equations and 2 unknowns,
so it's solvable
(1)
(1)
(2)
(2)
I'll add (1) and (2)
(1)
(2)
(3)
hr
Plug this back into (1) or (2)
(2)
(2)
(2) mi/hr
The speed of the current is 3 mi/hr
check answer:
(1)
(1)
(1)
(2)
(2)
(2)
OK

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(1)
Divide both sides by
(1)
(1)
(2)
Divide both sides by
(2)
Substitute (1) in (2)
(2)
(2)
(2)
(2)
Substitute this back into (1)
(1)
(1)
(1)
(1)
The answer is (6,-2)
check answer:
(1)
(1)
(1)

(2)
(2)
(2)
(2)
OK

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For the 30 mi/hr car:
(1)
For the 46 mi/hr car:
(2)
given:



------------------

The faster car goes 72 mi further than the slower car
I can rewrite (1) and (2)
(1)
(2)
(2)
Note that since they travel different
distances but in the same amount of time
Since both left sides are , I can set the
right sides equal to eachother


hr
and
hr also
In 4.5 hrs they will be 72 mi apart
Now check the answer:
(1)
(1)
(1) mi
and
(2)
(2)
(2)
OK

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I'm not sure what your main difficulty is, but I'll
go through it in detail
First you need the distributive law which is

is the same as


Note that +(-5) = -5 (or any number, I just picked 5)
So far, I have:

I'll use the same law again on the right side


Now subtract from both sides

Simplify the right side

Subtract from both sides

Divide both sides by

or, write it like

Now check the answer:



Add the number inside the ( ) on right side



OK

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This is advanced algebra and you're not used to
seeing the infinity sign. I think you need lots
of basic reading, maybe tutoring. Infinity
should be like an old friend if you're going
to tackle this stuff - /: |

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Just add these together


Then substitute this back into
either equation


Add to both sides

Subtract from both sides



The answer is (3,-4)

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With current:

Against current:

Math excercises are just pushups for the brain.
They can be about boats in water or alligators in mud
What good are pushups? Maybe no good at all.
You decide.

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Let the smallest number be
The next larger number will be
The next larger number will be
given:






The numbers are 654, 655, and 656
check:




OK

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Jogging:

Walking:

given:
mi
mi
mi/hr
hr
hr
--------------------------
Jogging:



(1)
Walking:


Substitute this in (1)
(1)
multiply both sides by
(1)
(1)
Use quadratic formula to solve for
Then find , then find

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This is because

This is becaue

This is because


This is because

This is because
Now divide top & bottom by
answer
-----------------------------------------------------
There is a quick & easy way to test answer
I'll say and (I could use any numbers)




and



OK

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Let = number of nickels
Let = number of dimes
Let = number of quarters
given:
(1)
(2)
(3)
-----------------------
There are 3 unknowns and 3 equations, so it's solvable
(2)
(2)
Subtract (1) from (2)
(2)
(1)
(4)
(4)
(4)
And, from (3),
(3)
(3)
From(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
and
(4)
(4)
and
(3)
(3)
(3)
(3)
Harry has 15 nickels, 10 dimes, and 2 quarters
check:
(1)
(1)
(1)
(1)
OK

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Multiply both sides by


Use quadratic formula










The sum of 3 and it's reciprocal is 10/3
also, the sum of 1/3 and it's reciprocal is 10/3

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Let = the smaller
Let = the larger
given:
(1)
(2)
(2)
-------------------------
Substitute (2) in (1)
(1)
Add to both sides
(1)
Add to both sides
(1)
Divide both sides by
(1)
The smaller number is
check:
(1)
(1)
(2)
(2)
OK

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Multiply both sides by

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Let the total distance =
given:

---------------
Multiply both sides by



The total distance is 120 miles
check answer:




OK