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15390..15419 , 15420..15449 , 15450..15479 , 15480..15509 , 15510..15539 , 15540..15569 , 15570..15599 , 15600..15629, >>NextGraphs/95328: Help me please i dont know how to graph this! please explain??!
Graph
-y <_ 3x
1 solutions
Answer 69477 by ankor@dixie-net.com(15624)   on 2007-08-31 14:20:32 (Show Source):
You can put this solution on YOUR website!Help me please i dont know how to graph this! please explain??!
Graph
-y <= 3x
:
You always want a positive y for this graphing equation
Multiply equation by -1
y >= -3x
:
A sign change like this always reverses the inequality sign
:
Plot a line using the coordinates:
x = -3, y = +9
x = 0, y = 0
x = +3, y = -9
:
The graph would look like this:
:
Area should be shaded at or above the line
:
Did this help?
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Systems-of-equations/95323: Lucia's aunt is now three times as old as Lucia. Fifteen years ago, her aunt was five times as old as Lucia. How old are Lucia and her aunt?
1 solutions
Answer 69474 by ankor@dixie-net.com(15624) on 2007-08-31 14:07:45 (Show Source):
You can put this solution on YOUR website!Lucia's aunt is now three times as old as Lucia. Fifteen years ago, her aunt was five times as old as Lucia. How old are Lucia and her aunt?
:
Let x = Lucia's age now
Then
3x = Auntie's age now
:
15 years ago they would be:
(x-15) = Lucia
and
(3x-15) = Auntie
:
It says,"15 yrs ago aunt was five times as old as Lucia."
(3x-15) = 5(x-15)
3x - 15 = 5x - 75
3x - 5x = -75 + 15
-2x = -60
x = -60/-2
x = +30 yrs is Lucia's age now
Then
3(30) = 90 yrs is Auntie's age now.
:
Check solutions by looking at their ages 15 years ago
Lucia = 15 years
Auntie = 75 years which is 5 times 15 years
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Travel_Word_Problems/95270: John walked from point P to point Q and back again, a total distance of 2 miles. If he averaged 4 miles per hour on the trip from P to Q and 5 miles per hour on the return trip, what was his average walking speed for the entire trip?
1 solutions
Answer 69462 by ankor@dixie-net.com(15624) on 2007-08-31 08:25:34 (Show Source):
You can put this solution on YOUR website!John walked from point P to point Q and back again, a total distance of 2 miles. If he averaged 4 miles per hour on the trip from P to Q and 5 miles per hour on the return trip, what was his average walking speed for the entire trip?
:
We want to find out the total time it took him to walk the total distance, then
it is an easy calculation to find the average speed
:
Time = Dist/speed
:
One way distance = 1 mile
:
To time + return time = Total time
 =  =  = .45 hrs to walk the two miles
:
Speed = dist/time
Av speed =  = 4.44 mph
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Quadratic_Equations/95407: Rita rows 12 km upstream and 12 km downstream in 3 hours. The speed of her boat in still water is 9 km/hr. Find the speed of the stream.
*note* I really would like some help as to how to setup this question. Thank you!
1 solutions
Answer 69438 by ankor@dixie-net.com(15624) on 2007-08-30 21:49:59 (Show Source):
You can put this solution on YOUR website!Rita rows 12 km upstream and 12 km downstream in 3 hours. The speed of her boat in still water is 9 km/hr. Find the speed of the stream.
:
Let x = speed of the stream
then
(9 - x) = Boat speed up-stream
and
(9 + x) = Boat speed down-stream
:
Write a time equation: Time = dist/speed
:
Time upstream + Time downstream = 3 hrs
 +  = 3
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Quadratic_Equations/95406: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
*note* I really would like some help as to how to setup this question. Thank you!
1 solutions
Answer 69436 by ankor@dixie-net.com(15624) on 2007-08-30 21:42:39 (Show Source):
You can put this solution on YOUR website!Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
:
Let s = the "certain speed"
:
The faster speed = (s+20)
:
Write a time equation: Time = Dist/speed
:
The problems states;
Certain speed time - Faster speed time = 1 hr
 -  = 1
:
Is this set up enough for you?
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Quadratic_Equations/95298: A garden area is 30ft long and 20ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400ft2, what is the width of the path?
1 solutions
Answer 69430 by ankor@dixie-net.com(15624) on 2007-08-30 21:08:22 (Show Source):
You can put this solution on YOUR website!A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft^2, what is the width of the path?
:
Draw diagram of this; label the outside dimensions of the rectangle 30 by 20.
Label the width of the path as x, it will be apparent that the dimensions of
the garden (inside the path), will be (30-2x) by (20-2x)
:
The area of the garden is given as 400 sq/ft
:
A simple area equation:
:
length times width = 400 sq/ft
(30-2x) * (20-2x) = 400
:
FOIL:
600 - 60x - 40x + 4x^2 = 400
4x^2 - 100x + 600 = 400
4x^2 - 100x + 600 - 400 = 0
4x^2 - 100x + 200 = 0; a quadratic equation
:
Simplify, divide by 4 and you have:
x^2 - 25x + 50 = 0
:
We need to use the quadratic formula to solve this: a=1; b=-25;; c=50
:
:
:
Two solutions:
x = 22.8, not a possible solution, obviously
and
x = 2.19 ft is the width of the path
:
:
Check our solution by finding the area of the garden
We have to subtract 2x from the outside dimensions: 2*2.19 = 4.38
:
(30-4.38) * (20-4.38) =
25.62 * 15.62 = 400.2 ~ 400 sq/ft
:
How about this, did it make sense to you, any questions?
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Quadratic_Equations/95266: Could you show me how to solve this equation, I know the answer , thanks to my calculator , but I do not know how to get to that stage
x+1 6
_____ + ______ = 0
x 2x-3
1 solutions
Answer 69429 by ankor@dixie-net.com(15624) on 2007-08-30 21:01:57 (Show Source):
You can put this solution on YOUR website!:
 +  = 0
:
Multiply thru by x(2x-3) and you have:
:
(2x-3)(x+1) + 6x = 0
:
FOIL
2x^2 + 2x - 3x - 3 + 6x = 0
:
2x^2 + 5x - 3 = 0
:
Factor:
(2x - 1)(x + 3) = 0
:
2x = +1
x = 1/2
and
x = -3
:
Substitute both solutions to confirm our solutions
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Expressions-with-variables/95335: I don't know how to set this problem up and i can solve it on my own if you could just show me how to set the problem.
"Kara has driven 75 miles. She averages 55 mi/h. How many more hours must kara drive to travel a total of 350 miles?"
This is a variable problem that is in word problem form.
If you could just please get back to me soon it would be much appreciated..
Thanks!
Katie
1 solutions
Answer 69400 by ankor@dixie-net.com(15624) on 2007-08-30 19:48:47 (Show Source):
You can put this solution on YOUR website!Kara has driven 75 miles. She averages 55 mi/h. How many more hours must kara drive to travel a total of 350 miles?"
:
Time = Dist/speed
:
Let t = time in hrs to complete the 350 mi trip
:
t = 
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Quadratic_Equations/95257:  =14
1 solutions
Answer 69391 by ankor@dixie-net.com(15624) on 2007-08-30 19:10:41 (Show Source):
You can put this solution on YOUR website! = 14
:
Subtract 11 from both sides:
 = 14 - 11
:
= 3
:
Cube both sides:
 = 
:
That gives us;
 = 27
:
Square both sides and you have:
7x - 9 = 729
7x = 729 + 9
7x = 738
x = 738/7, a rather nasty fraction, but that is the solution
:
:
Check our solution:
Substitute for x in 
 =  = 27
Cube rt of 27 = 3, so we have 3 + 11 = 14
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Linear-systems/95142: 5/6x+1/3= 3/4x-1/2
1 solutions
Answer 69389 by ankor@dixie-net.com(15624) on 2007-08-30 18:52:37 (Show Source):
You can put this solution on YOUR website!Assume you mean (x is not in the denominator)
:
Find a common multiple of all these denominators and multiply
12 is good choice:
:
After some easy canceling we have:
2(5)x + 4(1) = 3(3)x - 6(1)
:
10x + 4 = 9x - 6
:
Some basic algebra operations:
10x - 9x = -6 - 4
x = -10
:
You can substitute -10 for x in the original equation and check for equality
|
Rate-of-work-word-problems/95345: This question is from textbook prentice hall mathematics algebra 1
At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up?
I don't understand how to solve the problem. I dont even know where to begin. Could you please help with steps and tips of how to solve this problem?
1 solutions
Answer 69387 by ankor@dixie-net.com(15624) on 2007-08-30 18:39:49 (Show Source):
You can put this solution on YOUR website!At 1:30 pm, Tom leaves in his boat from a dock and heads south. He travels at a rate 0f 25 miles per hour. Ten minutes later, Mary leaves the same dock in her speedboat and heads after Tom. If she travels at a rate of 30 miles per hour, when will she catch up?
:
Let t = T's travel time to when M catches up
:
Since M left 10 min later we can say: (change 10 min to hrs 10/60 = 1/6)
(t-(1/6) = M's travel time
:
One thing we do know is that when M catches up with T they will have traveled
the same distance. (This applies to all these "catch-up" problems)
:
That means we can make a distance equation:
Dist = speed * time
:
M's dist = T's dist
30(t-(1/6)) = 25t
30t - (30/6) = 25t
30t - 5 = 25t
30t - 25t = +5
5t = 5
t = 1 hr + 1:30 = 2:30 Pm when M catches up with T
:
we can check this to see if they do, indeed, travel the same distance
in the time we found. M's time = (5/6) hr and T's time = 1 hr
30*(5/6) = 25 mi
25 * 1 = 25 mi
:
Did this make sense to you? Any questions?
|
Trigonometry-basics/95228: This question is from textbook Algebra and Trigonometry
The amount of a radioactive tracer after t days is given by A=A lower case o e^-0.18t, where A is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
1 solutions
Answer 69355 by ankor@dixie-net.com(15624) on 2007-08-30 16:32:47 (Show Source):
You can put this solution on YOUR website!The amount of a radioactive tracer after t days is given by A =A lower case o e^-0.18t, where A is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
:
Actually, I think they mean the starting amt is Ao, and
A is the amt after t days.
:
 = A
:
Substitute 40 for A and 3 for t in the above equation:
 = 40
:
 = 40
:
Find  on a good calc
.58275Ao = 40
Ao = 
Ao = 68.64 grams, is the initial amt to have 40 grams in 3 days
:
:
Check on a calc: enter: 68.64 * e^-.54 = 39.9998 ~ 40
|
Geometry_Word_Problems/95253: If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches. If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches. What is the area of the original rectangle?
1 solutions
Answer 69340 by ankor@dixie-net.com(15624) on 2007-08-30 15:26:25 (Show Source):
You can put this solution on YOUR website!If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches. If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches. What is the area of the original rectangle?
:
Let x = length, and y = width of the original rectangle
x*y = area of original rectangle
:
1st rectangle equation from the statement:
"If the length of a rectangle is decreased by 2 inches and its width is increased by 2 inches, the area of the rectangle increases by 16 square inches.
(x-2)*(y+2) = xy + 16
;
xy + 2x - 2y - 4 = xy + 16; FOILed (x-2)*(y+2)
xy - xy + 2x - 2y = 16 + 4; arrange the unknowns on the left, numbers on right
2x - 2y = 20
x - y = 10; simplified, divided by 2
x = (y + 10)
:
1nd rectangle equation from the statement:
" If the length is increased by 5 inches and the width is decreased by 3 inches, the area of the rectangle increases by 15 square inches."
(x+5)*(y-3) = xy + 15
:
xy - 3x + 5y - 15 = xy + 15
xy - xy - 3x + 5y = 15 + 15
-3x + 5y = 30
:
Substitute (y+10) for x from the 1st equation:
-3(y+10) + 5y = 30
-3y - 30 + 5y = 30
-3y + 5y = 30 + 30
2y = 60
y = 30 inches, width of the original square
:
Find x:
x = y + 10
x = 40 inches, length of the original square
:
What is the area of the original rectangle?
40 * 30 = 1200 sq inches
:
:
Check solutions using the changed rectangles
1st rectangle: 38 * 32 = 1216 sq inches
2nt rectangle: 45 * 27 = 1215 sq inches
:
Did this make sense to you? Any questions?
|
Expressions-with-variables/95120: This question is from textbook Algebra 1
I'm in 8th grade taking algebra so I can get a high school credit and so far we have had four teachers since school started in July. We have the smart board system but it has been down for several days and the teacher realy didn't explain the work that we have been doing and it really confuses me can you please help me undetstand this more?
1 solutions
Answer 69307 by ankor@dixie-net.com(15624) on 2007-08-30 08:26:32 (Show Source):
You can put this solution on YOUR website!On this website, you should state the actual problem or problems clearly, and ask for an explanation of it. someone will explain it to you
|
Rectangles/95169: I have been trying this problem for three days now and still can not come up with anything. Please Help show me the light. I was asked to find two different rectangles with the same perimeter and the same area. Tried all kinds of different combos of numbers. I know sides of 0 will work, but that isn't a rectangle. Thank you much, Rich
1 solutions
Answer 69246 by ankor@dixie-net.com(15624) on 2007-08-29 20:57:27 (Show Source):
You can put this solution on YOUR website! find two different rectangles with the same perimeter and the same area. Tried all kinds of different combos of numbers. I know sides of 0 will work, but that isn't a rectangle.
:
Let x = length and y = width
;
Perimeter = area
2x + 2y = x*y
:
2y - xy = -2x
:
y(2-x) = -2x
:
y =  ; plot this equation
:
Here is a graph which plots the length as x and the width as y, which has
perimeter = area (for x > 2).
:
I should have added here the statement,
"Any value greater than 2, that you plug in for x in the above equation, will yield a value for y which will satisfy the problem."
;
:
A couple integer values 6 by 3, 4 by 4, non-integers are unlimited,
10 by 2.5, 12 by 2.4, 22 by 2.2, etc.
:
Did this help you?
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Graphs/95066: The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial heigth in feet.
If a rock is throne upward with an initial velocity of 64 ft per sec. from the top of a 25 ft building, write the height (s) equation using this information.
a) how high is the rock after 1 sec?
b)after how many sec will the graph reach max heigth?
c) what is the max. heigth?
1 solutions
Answer 69173 by ankor@dixie-net.com(15624) on 2007-08-29 15:33:26 (Show Source):
You can put this solution on YOUR website!The path of a falling object is given by the function s=16t^2+v0t+s0 where Vo represents the initial velocity in ft/sec and s0 represents the initial height in feet.
If a rock is throne upward with an initial velocity of 64 ft per sec. from the top of a 25 ft building, write the height (s) equation using this information.
:
s = -16t2 + 64t + 25
:
Reviewing the Three parts of this equation
-16t^2 is the force of gravity, hence it is negative
64t is the initial upward velocity, it is positive
25 is the initial height of the building where this starts
:
:
a) how high is the rock after 1 sec?
Substitute 1 for t in the above equation and find s:
s = -16(1^2) + 64(1) + 25
s = -16 + 64 + 25
s = 73 ft above ground after 1 second
:
:
b)after how many sec will the graph reach max height?
:
Max height will occur at the axis of symmetry of this equation:
:
Formula for that is: x = -b/(2a)
:
In our equation this would be:
s = -64/(2*-16)
s = -64/-32
s = +2 seconds to reach max height
:
:
c) what is the max. height?
Find the this by substituting 2 for t in the equation:
s = -16(2^2) + 64(2) + 25
s = -16(4) + 128 + 25
s = -64 + 128 + 25
s = 89 ft is the max height occurring after 2 second
:
A graphical presentation would show this clearly.
x axis in seconds and y axis in feet
:
Note after 1 sec it looks like it is about 70 ft and after 2 sec it's over 88 ft
:
Did I succeed in making this understandable to you?
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Miscellaneous_Word_Problems/94986: If the sides of a square are decreased by 2 cm, the area is decreased by 36 cm2. What were the dimensions of the original square?
1 solutions
Answer 69161 by ankor@dixie-net.com(15624) on 2007-08-29 13:20:40 (Show Source):
You can put this solution on YOUR website!If the sides of a square are decreased by 2 cm, the area is decreased by 36 cm2. What were the dimensions of the original square?
:
Let x = side of original square
Then
(x-2) = side of the new square
:
Original square area - new square area = 36
x^2 - (x-2)^2 = 36
:
x^2 - (x^2 - 4x + 4) = 36: FOILed (x-2)(x-2)
:
x^2 - x^2 + 4x - 4 = 36; removing bracket changes the signs
:
4x = 36 + 4; x^2's cancel out
:
4x = 40
:
x = 40/4
:
x = 10 cm, side of original square:
:
:
Check: 10^2 - 8^2 = 36
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Equations/95028: Business and finance. If the total bill at a restaurant, including a 15% tip, is
$65.32, what was the cost of the meal alone?
1 solutions
Answer 69160 by ankor@dixie-net.com(15624) on 2007-08-29 13:14:20 (Show Source):
You can put this solution on YOUR website!If the total bill at a restaurant, including a 15% tip, is
$65.32, what was the cost of the meal alone?
;
Let x = cost of the meal; tip would be .15x
x + .15x = 65.32
1.15x = 65.32
x = 65.32/1.15
x = $56.80
:
Check solution:
56.80 + .15(56.80) = 65.32
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Polynomials-and-rational-expressions/94748: This question is from textbook Beginning Algebra
Good evening tutor,
I do not understand multiplying polynomials. Could someone please help me with this problem?
Multiply the following:
(x/3 + 3/4)(3x/4 - 3/5)
Thank you in advance for your help I am going crazy trying to understand this.
1 solutions
Answer 69083 by ankor@dixie-net.com(15624) on 2007-08-28 18:28:53 (Show Source):
You can put this solution on YOUR website!Multiply the following:
 * 
:
We need to put each pair of fractions over a single denominator, using a common
denominator much as you do in numerical fractions:
:
Find the common denominator of the 1st pair: It would be 12, right?
=  = 
:
Do the same with the 2nd pair of fractions, the common denominator would be 20:
=  = 
:
Now we have a single fraction times a single fraction
*
:
FOIL the numerators:
 = 
:
Factor out 3 in the numerator, then cancel it into the denominator:
 =  ; that's about all you can do with this
:
Did this help you understand this?
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Expressions-with-variables/94727: find a value for y so that the value of 10y^3 is the same as the value for
(10y)^3 ?
1 solutions
Answer 69051 by ankor@dixie-net.com(15624) on 2007-08-28 16:06:21 (Show Source):
You can put this solution on YOUR website!find a value for y so that the value of 10y^3 is the same as the value for
(10y)^3 ?
:
 = 
:
I inadvertently sent this in, but worse than that I have not come up with a solution. I will keep trying but perhaps you can resubmit it, someone smarter than me may come up with it. Sorry. A
:
On further effort on this, I am going to say only y = 0 will satisfy this
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Percentage-and-ratio-word-problems/94722: Step right up anad guess the nimber of candies in the jar?
Three people have already guessed, and their guesses are 315, 350, and 327. Now here's the big clue: One of the guesses is off by 26, one is off by 14, and one is close, but off by 9 candies?
How many candies are in the jar?
1 solutions
Answer 69044 by ankor@dixie-net.com(15624) on 2007-08-28 15:55:37 (Show Source):
You can put this solution on YOUR website!Three people have already guessed, and their guesses are 315, 350, and 327. Now here's the big clue: One of the guesses is off by 26, one is off by 14, and one is close, but off by 9 candies?
How many candies are in the jar?
:
The "Brute Force" method
:
x +/- 14 = 315; x = 301 or 329
x +/- 26 = 350; x = 376 or 324
x +/- 9 = 327; x = 336 or 318
No matches here:
:
x +/- 14 = 327; x = 341 or 313
x +/- 26 = 315; x = 341 or 289
x +/- 9 = 350; x = 359 or 341
:
No need to go any further: the number is 341
:
There is probably a more elegant method, but have to admit, I can't come up with it.
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Complex_Numbers/94817: I have a worded problem that is giving me trouble and I would really appreciate if someone could assist me in making sense of it all. It goes like this.
Four brothers are born in successive intervals of three years. The eldest is twice the age of the youngest. What is the age of their father, who is eighteen years older than the eldest son?
I have given this a go, but all I am able to come up with is this
B1= Brother 1 =9
B2= Brother 2 =12
B3=Brother 3 =15
B4=Brother 4 =18
F=Father =36
B1+B2+3+B3+6B4+9
4B=+18
F=2B4+9
My problem is even though I think I can get an answer to this problem in my head. I am unable to sort it into an algebraic equation, which is what I desperately need. Any help would be greatly appreciated. Thank you so much.
1 solutions
Answer 69038 by ankor@dixie-net.com(15624) on 2007-08-28 15:28:17 (Show Source):
You can put this solution on YOUR website!Four brothers are born in successive intervals of three years. The eldest is twice the age of the youngest. What is the age of their father, who is eighteen years older than the eldest son?
:
Let x = age of the youngest
Then the remaining brothers' ages would be: x+3, x+6, x+9
:
It says,"The eldest is twice the age of the youngest."
(x+9) = 2x; you can do this in your head, but continuing in algebra
9 = 2x - x
x = 9 age of the youngest
:
Oldest, obviously is x + 9 = 18, + 18 = 36 is father
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Graphs/94818: The equation of the line with y-intercept of 3 and perpendicular to the graph of 2x+y=5 is?
This is an sample test questions for my math assessment test. I will be taking in a few hours and this is my last question that I don't know how it works. I know the answer but I don't know how to get to that answer. Thank you very much!
1 solutions
Answer 69036 by ankor@dixie-net.com(15624) on 2007-08-28 15:20:35 (Show Source):
You can put this solution on YOUR website!The equation of the line with y-intercept of 3 and perpendicular to the graph of 2x+y=5 is?
:
y intercept occurs when x = 0, therefore the equation in the form y = mx + b
would be y = mx + 3
:
Now we want to find the slope (m) of the given equation:
Put equation in the slope/intercept (y = mx + b) form:
2x + y = 5
y = -2x + 5
The slope m = -2
:
Slope relationship between perpendicular lines is given as: m1*m2 = -1
Assume the m1 = -2 from the above equation,
find m2 (the slope of the perpendicular line) y = mx + 3
:
-2 * m2 = -1
Divide both sides by -1 and you have:
m2 = -1/-2
m2 = +1/2
:
The equation of the perpendicular line: y =  x + 3
:
It will help you to memorize the perpendicular lines slope relationship
m1 * m2 = 1, and that y intercept occurs when x = 0
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Graphs/94773: 1/9x-1/5y=20/9
-10x+18=-200
Solve the system
I can't get this? Please I need help, thanks
1 solutions
Answer 69032 by ankor@dixie-net.com(15624) on 2007-08-28 13:43:30 (Show Source):
You can put this solution on YOUR website!
The first thing we want to do here is get rid of those annoying denominators.
Multiplying each term in the equation by 9*5 or 45, would do this, leaving us with:
5x - 9y = 5(20)
5x - 9y = 100
:
On this equation I assume you mean:
-10x + 18y = -200
:
Now it becomes apparent that this system cannot be solved. See what happens if
we multiply the 1st equation by 2, and add to the 2nd equation:
:
10x - 18y = 200
-10x + 18y = -200
------------------- adding eliminates everything!
0x + 0y = 0
:
This is just a form of "math student harassment" don't you think?
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Radicals/94797: the equation d=1.2 square root h, gives the distance, D, in miles that a person can see to the horizon from a height ,h, in feet.
A. solve equation for h,
B. a national park peak is 14,255 feet in elevation. How far can you see to the horizon from the top of peak.
I am lost on how to solve the equation and then figure out the steps to take to come up with the answer for B.
1 solutions
Answer 69027 by ankor@dixie-net.com(15624) on 2007-08-28 13:27:26 (Show Source):
You can put this solution on YOUR website!the equation d=1.2 square root h, gives the distance, D, in miles that a person can see to the horizon from a height ,h, in feet.
:
A. solve equation for h,
1.2  = d
Divide both sides by 1.2
= 
Square both sides:
h = 
:
B. a national park peak is 14,255 feet in elevation. How far can you see to the horizon from the top of peak.
Given the formula: d = 1.2*
d = 1.2*  ; find Square Root on a calculator
d = 1.2 * 119.3943
d = 143.273 miles to the horizon from an elevation of 14,255 ft
:
:
Did this help you understand this?
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Probability-and-statistics/94739: This question is from textbook Introductory algebra
A triangular region has a perimeter of 66 meters. The first side is two-thirds of the second side. The third side is 14 meters shorter than the second side. What are the lengths of the three sides of the triangular region?
1 solutions
Answer 69021 by ankor@dixie-net.com(15624) on 2007-08-28 12:55:36 (Show Source):
You can put this solution on YOUR website!Let x = length of the 2nd side
:
It says," The first side is two-thirds of the second side." Therefore:
1st side = 2x/3
:
It also says,"The third side is 14 meters shorter than the second side." Therefore:
3rd side = (x-14)
:
What are the lengths of the three sides of the triangular region?
It says "A triangular region has a perimeter of 66 meters." Therefore:
:
side 1 + side 2 + side 3 = 66
(2x/3) + x + (x-14) = 66
:
Multiply equation by 3 to get rid of the denominator:
2x + 3x + 3(x-14) = 3(66)
2x + 3x + 3x - 42 = 198
8x = 198 + 42
8x = 240
x = 240/8
x = 30 m, the length of side 2
:
2/3(30) = 20 m, the length of side 1
:
30 - 14 = 16m, the length of side 3
:
:
Check solution, using the perimeter:
20 + 16 + 30 = 66
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Polynomials-and-rational-expressions/94584: Science and Nutrition. A passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi faster than the speed of the freight train, find the speed of each.
1 solutions
Answer 68986 by ankor@dixie-net.com(15624) on 2007-08-27 21:51:48 (Show Source):
You can put this solution on YOUR website!A passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. If the speed of the passenger train is 25 mi faster than the speed of the freight train, find the speed of each.
:
Let s = speed of the freight
Then
(s+25) = speed of the pass train
:
They are giving the time as the same for both trains, make a time equation from this fact:
Time = Dist/speed
:
 = 
:
Cross multiply and you have:
:
325s = 200(s+25)
:
325s = 200s + 5000
:
325s - 200s = 5000
:
125s = 5000
:
s = 5000/125
:
s = 40 mph, speed of the freight
and
(s+25) = 65 mph, speed of the pass train
:
:
Check solutions, is the time the same?
200/40 = 325/65
5 = 5
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