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A Car travels 240 mi. A second car, traveling 12 mph faster than the first,
makes the same trip in 1 hour less time. Find the speed of each car
:
Let s = speed of the slower car
then
(s+12) = speed of the faster
:
Write a time equation; time = dist/speed
:
Slow car time - fast car time = 1 hr
- = 1
:
Multiply by s(s+12), results
240(s+12) - 240s = s(s+12)
:
240s + 2880 - 240s = s^2 + 12s
Combine as a quadratic equation on the right
0 = s^2 + 12s - 2880
Factors to
(s+60)(s-48) = 0
positive solution
s = 48 mph is the speed of the slower car
then
48 + 12 = 60 mph is the faster car
:
Check solution by finding the times of each
- =
5 - 4 = 1 hr; confirms our solutions

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the swimming pool is 12 feet by 20 feet there is a 3 foot wide cement sidewalk all around the pool.
what is the outside perimeter of the sidewalk?
You're right, a 3' sidewalk adds 6' to each dimension

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For problems 5 and 6, solve the system of equations using SUBSTITUTION to eliminate one variable. If the system has no solution, indicate that is is inconsistent. If it has infinite solutions, indicate that it is dependent. If the system has a solution, CHECK YOUR ANSWER.
5.
y = 2x: this means we can substitute 2x for y in the 2nd equation, and find x
3x + y = 20
3x + 2x = 20
5x = 20
x = 4: divided both sides by 5
:
We know that y = 2x, therefore
y = 2(4)
y = 8
:
The solution x=4; y=8,
:
Check this in the 2nd equation
3x + y = 20
Substitute the values for x and y
3(4) + 8 = 20
12 + 8 = 20; equality checks the solutions
:
:
6. 4x + 2y =6
2x + y = 3
Rearrange this equation to use for substitution in the 1st equation
y = -2x + 3; subtracted 2x from both sides
:
Replace y with (-2x+3)in the 1st equation
4x + 2(-2x+3) = 6
4x - 4x + 6 = 6
4x - 4x = 6 - 6
0 = 0, dependent, an infinite number of solutions

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Jim and John drive from point A to point B in separate cars.
Jim leaves at 6 am and arrives at 4 pm. John leaves at 10 am and arrives at 3 pm.
Assume both men drive at constant speeds.
Find when John catches up with Jim.
:
Let d = distance from A to B
Jim's travel time: 10 hrs (6am to 4pm)
Jon's travel time: 5 hrs (10am to 4pm
:
Let t = Jon's travel time when he catches Jim
Then
(t+4) = Jim's travel time when this happens (Jim leave 4 hrs earlier)
and we know
= Jim's speed
and
= Jon's speed
:
When Jon catches Jim, they will have traveled the same distance; Dist = speed * time.
(t+4) = *t
multiply both sides by 10 to get rid of the denominators, results
d(t+4) = 2dt
divide both sides by d
t + 4 = 2t
4 = 2t - t
t = 4 hrs, Jon's travel time
then
4 + 4 = 8 hrs; Jim's travel time
:
10 am + 4 hrs = 2 pm when Jon overtakes Jim
:

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An electric power company charges its consumers $ 0.40 per kilowatt for the
first 100 kilowatts in a month,
$ 0.50 per kilowatt for the second 100 kilowatts,
and $ 0.60 for each additional kilowatt beyond.
Mary pays $ 120 for electricity used in last month.
How much electricity did her family consume in the month?
:
Let x = total no. of kwh used in a month
:
.6(x-200) + .5(100) + .4(100) = 120
.6x - 120 + 50 + 40 = 120
.6x - 30 = 120
.6x = 120 + 30
.6x = 150
x =
x = 250 kwh used
:
:
See if this is true
.4(100) = 40
.5(100) = 50
.6(50) = 30
---------------
total$ = 120
:
:
Did this make sense to you?

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A baseball is hit straight upwards with an initial velocity of 72 feet per
second and leaves the bat at an initial height of 3 feet.
Write a formula, s(t), that models the height of the baseball after t seconds.
:
s(t) = -16t^2 + 72t + 3
Where
s(t) = height after t seconds
-16t^2 = the downward pull of gravity
72t = initial upward velocity
3 = initial height of the bat
:

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A motorist travels 90 miles at a rate of 20 miles per hour.
If he returns the same distance at an average rate of 40 miles per hour,
what is his average speed for the entire trip, in miles per hour?
:
Let a = average speed for the trip
:
Write a time equation, time = dist/speed
:
Go time + return time = total time
+ =
:
Multiply by 40a to clear the denominators, then solve for a

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Marcus and Beth leave college traveling in opposite directions on a straight road.
Beth drives 13 miles faster than Marcus.
After 4 hours they are 452 miles apart.
Find their rates.
Let b = B's driving rate
Let m = M's driving rate
:
Write a distance equation: dist = time * rate
:
M's dist + B's dist = 452
4m + 4b = 452
:
It says,"Beth drives 13 miles faster than Marcus."; the equation for this is:
b = (m+13)
:
Replace b with (m+13) in the 1st equation
4m + 4(m+13) = 452
4m + 4m + 52 = 452
8m = 452 - 52
8m = 400
m =
m = 50 mph,
then
b = 50 + 13
b = 63 mph
:
:
See if that's true
4(50) + 4(63) = 452

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A number has 4 digits.
No digits in the number are repeated.
The digit in the tens place is three times the digit in the thousands place.
The number is odd.
The sum of the digits in the number is 27.
What is the number?
:
Let w = the thousands digit
Let x = the 100's
Let y = the 10's
Let z = the units
:
The digit in the tens place is three times the digit in the thousands place.
y = 3w
Use some an assumption and some logic here
With a total of 27, we have to assume one digit is 9, w has to be less than 4
let y = 9
then
w = 3
:
So we have:
3 + x + 9 + z = 27
12 + x + z = 27
x + z = 27 - 15
x + z = 15
there are only 2 single digit pairs that add up to 15
6 & 9
7 & 8
we have already used 9 (no repeats), it must be 7 & 8
the last digit has to be odd, therefore we have
x = 8
z = 7
:
Our number: 3897

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The road from Tedium to Excitement is uphill for 5 miles, level for 4 miles,
and then downhill for 6 miles.
John Mayer walks from Excitement to Tedium in 4 hours.
Later he walks halfway from Tedium to Excitement and back again in 3 hours and 55 minutes.
Finally he walks all the way to Excitement from Tedium in 3 hours and 52 minutes.
What are his rates going uphill, downhill, and on level ground assuming that
these rates remain constant?
:
Let u = uphill rate
Let f = flat rate
Let d = downhill rate
:
Write a time equation for each statement: Time = dist/rate
We are going to do it in minutes, change it to hrs later
:
"John Mayer walks from Excitement to Tedium in 4 hours."
+ + = 240
:
"he walks halfway from Tedium to Excitement and back again in 3 hours and 55 min."
+ + = 235 (halfway is 7.5 mi)
:
"he walks all the way to Excitement from Tedium in 3 hours and 52 minutes."
+ + = 232
:
What are his rates going uphill, downhill, and on level ground assuming that these rates remain constant?
:
Use elimination on the 1st and 2nd equations
+ + = 240
+ + = 235
-----------------------------------------Subtraction eliminates d
- = 5
:
Multiply the 1st equation by 6 and the 3rd equation by 5
+ + = 1410
+ + = 1160
--------------------------------------------------Subtraction eliminates d again
+ = 250
:
Multiply the 1st 2 unknown equation by 10, add to the above equation
+ = 250
- = 50
------------------addition eliminates f, find u
= 300
u =
u = miles per minute, that's * 60 = 3 mph up hill
:
Use the equation: - = 5 to find f,(were dealing mi/min here)
- = 5
20 - = 5
- = 5 - 20
= -15
f = + mi/min, that's * 60 = 4 mph on the flat area
:
We can use the 1st equation, using hrs, to find d
+ + = 4
2 + 1 + 5/d = 4
5/d = 4 - 3
5/d = 1
d = 5 mph down hill
:
Summarize here
u = 3 mph uphill
f = 4 mph level
d = 5 mph down hill
:
:
See if this works int he original 3rd equation, using hrs
+ + = 3 hr 52 min
1.67 + 1 + 1.2 = 3.87 hrs which is 3 hrs, .87*60 = 52.2 min, close enough

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Joan's age was a factor of her grandfather's age for 6 consecutive years.
What were her grandfather's ages during this time?
:
Looking at 6 consecutive numbers without a prime number over 60, came up with:
gr|Joan
-----------
62| 2
63| 3
64| 4
65| 5
66| 6

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Suppose that the height (in centimeters) of a candle is a linear function of
the amount of time (in hours) it has been burning.
After 11 hours of burning, a candle has a height of 23.4 centimeters.
After 30 hours of burning, its height is 12 centimeters.
What is the height of the candle after 13 hours?
:
Assign the given values as follows:
x1 = 11; y1 = 23.4
x2 = 30; y2 = 12
:
Find the slope using: m =
m = =
:
Find the equation using the point/slope formula: y - y1 = m(x - x1)
y - 23.4 = (x - 11)
y - 23.4 = x +
y = x + + 23.4
y = x + + 23.4
y = x + +
y = x +
y = x + 30, is the equation
:
What is the height of the candle after 13 hours?
x = 13
y = (13) + 30
y = + 30
y = -7.8 + 30
y = 22.2 cm after 13 hrs

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worker throws his wrench downward from a height of 128 feet with an
initial velocity of 32 feet per second.
The height of the wrench above the ground after t seconds is given by
S(t) = -16t^2 – 32t + 128
:
a) What is the height of the wrench after 1 second?
S(t) = the height, t=1; therefore:
S(t) = -16(1^2 - 32(1) + 128
S(t) = -16 - 32 + 128
S(t) = 80 ft after one second
:
b) How long does it take for the wrench to reach the ground?
Replace height [S(t)] with ground level, which is 0
-16t^2 - 32t + 128 = 0
Simplify and change the signs, divide equation by -16
t^2 + 2t - 8 = 0
Factor
(t+4)(t-2) = 0
The positive solution is all we want here
t = 2 seconds for the wrench to reach the ground
:
:
We can prove this:
S(t) = -16(2^2) - 32(2) + 128
S(t) = -16(4) - 64 + 128
S(t) = -64 - 64 + 128
S(t) = 0
:
:
Did this make sense to you now?

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one house can fill a goldfish pond in 84 minutes, and two hoses can fill the
same pond in 35 minutes.
find how long it takes the second hose alone to fill the pond?
:
Let x = time for the 2nd hose to fill the pond by itself
:
Let the completed job = 1 (a full pond)
:
Each hose will do a fraction of the job, the two fractions add up to 1
:
+ = 1
Multiply by 84x, results
35x + 84(35) = 84x
2940 = 84x - 35x
2940 = 49x
x =
x = 60 min for the 2nd hose alone, (but that's without the house getting involved)

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A zebra can run 30 mph.
A tiger can run 50 mph.
If the tiger has run 70 miles further than the zebra, how many minutes have
they been running
:
Let t = no. of hrs they have been running
:
Let z = no. of miles run by the zebra
then
(z+70) = no. of miles run by the tiger
:
Write two distance equations
30t = z
and
50t = (z+70)
Replace z with 30t
50t = 30t + 70
50t - 30t = 70
20t = 70
t =
t = 3.5 hrs
but they want it in minutes: 3.5 * 60 = 210 minutes
:
:
Check solution by finding the distances
3.5(50) = 175 mi
3.5(30) = 105 mi
-----------------
differs: 70 mi

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The function W(d)=0.112d approximates the amount, in centimeters, of water
that results from d cm of snow melting.
Find the amount of water that results from snow melting from depths of 18cm, 28cm and 95 cm.
:
This is not hard, d is the number of cm of snow, W(d) is amt of water it makes
:
W(d) = .112d
Substitute 18 cm for d
W(d) = .112*18
W(d) = 2.016 cm of water
:
Same with d=28cm
W(d) = .112d
Substitute 28 cm for d
W(d) = .112*28
W(d) = 3.136 cm of water
:
You should be able to find it when d=95 cm now

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A suspension bridge with weight uniformly distributed along its length has twin
towers that extend 85 meters above the road surface and are 1200 meters apart.
The cables are parabolic in shape and are suspended from the tops of the towers.
The cables touch the road surface at the center of the bridge.
Find the height of the cables at a point 300 meters from the center. (Assume the road is level.
:
Find the equation for this:
Three coordinates:
0, 85; the vertical suspension point on the left
600, 0; the center point that touches the road (half the length of the bridge)
1200, 85; vertical suspension point on the right
:
Using: ax^2 + bx + c = y
:
0,85, we know c = 85
:
write two equation from coordinates;
x=600, y=0
(600^2)a + 600b + 85 = 0
360000a + 600b + 85 = 0
and
x=1200, y=85
(1200^2)a + 1200b + 85 = 85
2440000a + 1200b + 85 = 85
Multiply the 1st equation by 2, subtract from the 2nd equation
2440000a + 1200b + 85 = 85
720000a + 1200b + 170 = 0
------------------------------subtraction eliminates b, find a
720000a - 85 = 85
720000a = 85 + 85
720000a = 170
a =
a = .0002361

find b, using the 1st equation:
.0002361(600^2) + 600b + 85 = 0
85 + 600b + 85 = 0
600b + 170 = 0
600b = -170
600b =
b = -.2833
:
The equation; y = .0002361x^2 - .2833x + 85
:
Looks something like this:

:
"Find the height of the cable at a point 300 meters from the center"
Now you can answer this question, I'm sure

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A cattle train left Miami and traveled toward New York.
14 hours later a diesel train left traveling at 45 km/h in an effort to catch up to the cattle train.
After traveling for four hours the diesel train finally caught up.
What was the cattle trains average speed?
:
Let s = cattle train speed
:
The cattle train will be traveling 14 + 4 = 18hrs when the diesel catches up
:
When the diesel overtakes the cattle, they will have traveled the same distance
Write a distance equation; dist = speed * time
:
Cattle dist = diesel dist
18s = 4*45
18s = 180
s = 10 kmh is the cattle train
:
Both travel 180 km

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Solve for r in the formula
S = L - rL
:
Factor out L
S = L(1 - r)
:
Divide both sides by L
= 1 - r
Rearrange
r = 1 -

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Let x = the large number
Let y = the small number
:
Write an equation for each statement:
:
Twice the small number is 5 more than 2/3 the larger number.
2y = x + 5
Get rid of that annoying fraction, multiply by 3, results:
6y = 2x + 15
-2x + 6y = 15
:
Three times the larger number is 4 less than 15 times the smaller number.
3x = 15y - 4
3x - 15y = -4
:
Use elimination here, multiply the 1st equation by 3, and the 2nd equation by 2
-6x + 18y = 45
+6x - 20y = -8
-------------------adding eliminates x, find y
-12y = 37
y =
:
Use the 1st equation to find x
-2x + 6() = 15
-2x + ) = 15
Multiply by 2
-4x - 37 = 30
-4x = 30 + 37
-4x = 67
x =
:
A couple nasty fractions, see if they work in the 2nd equation
3x - 15y = -4
Substitute
3() - 15( = -4
+ ( = -4
multiply by 12 to get rid of those annoying fractions
3(201) + 555 = -48
-603 + 555 = -48; it does work, amazing isn't it!

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the formula for dist to horizon: d = ; h in ft. gives d in miles
:
So if you are 6 mi high you are 6(5280) 31680 ft high
d =
d = 251.7 mi so horizon

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A parking lot is 75 feet wide by 60 feet long.
It is being torn up to install a sidewalk of uniform width all the way around.
If the area of the new parking lot is 2/3 of the original one, find the width of the sidewalk.
:
Find the original area; 75 * 60 = 4500 sq/ft
Find 2/3 of that: * 4500 = 3000 sq/ft, is the new parking lot area
:
Let x = the width of the sidewalk
then the dimensions of new parking lot will be: (75-2x) by (60-2x)
:
the area equation
(75-2x)*(60-2x) = 3000
FOIL
4500 - 150x - 120x + 4x^2 = 3000
A quadratic equation
4x^2 - 270x + 4500 - 3000 = 0
4x^2 - 270x + 1500 = 0
simplify, divide by 2
2x^2 - 135x + 750 = 0
Solve for x using the quadratic formula:

In this equation: a=2; b=-135, c=750

:

:

Two solutions

x =
x = 61.4; obviously not a good solution
and

x =
x = 6.1 ft is the width of the side walk
:
:
We can check this by finding the area with this value
(75-2(6.1)) * (60-2(6.1)) =
62.8 * 47.8 = 30001.84 ~ 3000, close enough

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om drives 500 miles on the first day.
On the second day, he drives twice as long and his average speed is 4/5 of
that on the first day.
How long does he drive on the second day?
:
2(*500) = 800 mi

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A carpenter makes bookcases in two sizes, large and small.
let a = no. of large bookcases
Let b = no. of small
:
It takes 6 hours to make a large bookcase and 2 hours to make a small bookcase.
"The carpenter can only spend 24 hours per week making bookcases"
6a + 2b =< 24
:
The profit on a large bookcase is $50 and the profit on a small bookcase is $20.
p = 50a + 20b
:
must make at least two of each size per week.
a => 2
b => 2

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A woman with a basket of eggs finds that if she removes
the eggs from the basket 3 or 5 at a time, there is always 1
egg left. However, if she removes the eggs 7 at a time,
there are no eggs left. If the basket holds up to 100 eggs,
:
We know that the no. of eggs is a multiple of 7
The units digit is 1 or 6, because of the 5
That leaves 21, 56, 91
the 3 tell us it has to be 91 eggs

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Find the measure of angle that the second hand of an accurate clock rotates in
20 seconds.
:
* 360 = 120 degrees

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THEY FLY 900 MILES PER HR WHEN THEY FLY FROM PENSACOLA TO EL CENTRO IN THEIR F/A-18 HORNETS.
ON EVERY TRIP THEIR C130 TRANSPORT LEAVES BEFORE THEM, CARRYING SUPPLIES AND SUPPORT STAFF.
THE C130 TRAVELS AT 370 MILES PER HR.
oN A TRIP FROM PENSACOLA TO EL CENTRO HOW LONG BEFORE THE HORNET LEAVES
SHOULD the C130 LEAVE IF ITS TO ARRIVE 3 HRS BEFORE THE HORNETS?
THE FLYING DISTANCE BETWEEN PENASACOLA AND EL CENTRO IS 1720 MILES.
:
Let t = time for C130 to arrive 3 hrs before F-18's
:
Write a time equation, time = dist/speed
:
t = + 3 -
t = 4.64865 + 3 - 1.91111
t = 5.73754 hrs or about 5 hrs 44 min before the F-18's
:

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a piggy that bank contains only nickels and dimes.
The total value of the coins in Pete's bank is $2.75.
If the nickels were dimes and the dimes were nickels, the total value of the
coins would be $3.25.
Find the number of nickels:
:
Let n = no. of nickels
Let d = no. of dimes
:
Write an equation for each statement:
:
"The total value of the coins in Pete's bank is $2.75."
.05n + .10d = 2.75
:
"If the nickels were dimes and the dimes were nickels, the total value of the
coins would be $3.25."
.10n + .05d = 3.25
:
Multiply the above equation by 2, subtract the 1st equation
.20n + .10d = 6.50
.05n + .10d = 2.75
-----------------------subtraction eliminates d, find n
.15n = 3.75
n =
n = 25 nickels
:
:
Check solution:
find d using the 1st equation
.05(25) + .10d = 2.75
1.25 + .10d = 2.75
.10d = 2.75 - 1.25
.10d = 1.50
d =
d = 15 dimes
:
Find the total using the 2nd equation
.10(25) + .05(15) =
2.50 + .75 = 3.25; confirms our solution

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When the digits of a two-digit number are reversed, the new number is 9 more
than the original number, and the sum of the digits of the original number is 13.
What is the original number?
:
x = the 10s digit, y = the units
then
10x + y = the original number
and
10y + x = the reversed number
:
Rev number = orig number + 9
10y + x = 10x + y + 9
10y - y = 10x - x + 9
9y = 9x + 9
simplify, divide by 9
y = x + 1
:
"the sum of the digits of the original number is 13. "
(kind of silly, the sum of the digits is the same original or reversed, right?"
anyway
x + y = 13
:
Rearrange the y = x + 1 and add to the above equation
-x + y = 1
+x + y = 13
------------addition eliminates x, find y
2y = 14
y = 7
:
I'll let you find x, and check it in the 1st equation

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Maria drove from Chicago, Illinois, to Milwaukee, Wisconsin, a distance of 90 miles, at a mean speed of 65 miles per hour. On her return trip, the traffic was much heavier, and her mean speed was 42 miles per hour. Find Maria's mean speed for the round trip.
:
Let s = average speed for the round trip
:
Write a time equation; time = dist/speed
:
to time + return time = total time
+ =
1.3846 + 2.1429 =
3.5275 =
s =
s = 51.0 mph average speed for the round trip

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i) 3x - 5 > x + 13
Add 5 to both sides, eliminates -5 on the left
3x > x + 13 + 5
Subtract x from both sides
3x - x > 18
2x > 18
divide both sides by 2
x >
x > 9
:
ii) x^2 - 4x - 12 > 0
Factors to
(x-6)(x+2) > 0
x > 6; expression greater than 0 above +6
x < -2; expression is greater than 0 below -2
:
Graphically, this can be seen