Word Problems
Here is a motion word problem:
The height of a rock thrown from a walkway over a lagoon can be approximated by the formula h = -5t^2 + 20 t+ 60, where t is the time in seconds, and h is the height in meters.
a) Write the above formula in factored form
h=-5t^2 +20t +60
h= -5(t^2 -4 -12)
h= -5(t-6) (t+2)
b) When will the rock hit the water?
h= -5(t-6)(t+2)
The rock will hit the water at 6 seconds because that is the only positive time. You can't have a negative time it is impossible. Therefore the rock hits the water at 6 seconds.
Here is my second word problem on geometry:
A) Write and simplify an expression to represent the area of the given composite figure.
The height of a rock thrown from a walkway over a lagoon can be approximated by the formula h = -5t^2 + 20 t+ 60, where t is the time in seconds, and h is the height in meters.
a) Write the above formula in factored form
h=-5t^2 +20t +60
h= -5(t^2 -4 -12)
h= -5(t-6) (t+2)
b) When will the rock hit the water?
h= -5(t-6)(t+2)
The rock will hit the water at 6 seconds because that is the only positive time. You can't have a negative time it is impossible. Therefore the rock hits the water at 6 seconds.
Here is my second word problem on geometry:
A) Write and simplify an expression to represent the area of the given composite figure.
a0 a)
A1=3(x) A2= (x=2)(x-2)
=3x and =x^2-4
A=x^2+3x-4
B) If the area of the shape is 36cm^2, determine the value of x.
36=x^2+3x-4
0=x^2+3x-40
0=(x+8)(x-5)
x+8=0 or x-5=0
x=-8 or x=5
x can not be -8 because you can not have a negative length.
Therefore x = 5
Here is my third word problem and it is on Revenue:
A ferry service has about 240000 riders per day for $2.The port wants to increase the fare to help with increasing operational costs. Research has shown for every $0.10 increase in the fare the number of riders will drop by 10000.
a) What increase in the fare will maximize the revenue?
let x represent the number of increases by $0.10.
Revenue = cost * number of rides
R=(2+0.10x)(240000-10000x)
To find the max revenue you need the axis of symmetry. For that you need the X-intercepts of the equation.
0=(2+0.10x)(240000-10000x)
0=2+0.10x and 0= 240000-10000x
-2= 0.10x and -240000=10000x
-20 and 24 are the X-intercepts.
AOS = -20 + 24 / 2
=4/2
2
There fore you will need two increases for it to make the max revenue.
b) What is the new fare?
To find the new fare you take the cost from the equation which is (2+0.10x). From here you sub in the axis of symmetry which is a x value.
2+0.10(2)
2+0.20
2.20
There fore $2.20 will be the new fare for the max revenue.
c) What is the revenue that will be received from the new fare?
To find the revenue you use this equation R=(2+0.10x)(240000-10000x). Then you sub in the AOS.
R=(2+0.10x(2))(240000-10000(2))
R=(2+0.2)(240000-20000)
R=(2.2)(220000)
R=484000
There fore the revenue for the new fare is $484000.
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A1=3(x) A2= (x=2)(x-2)
=3x and =x^2-4
A=x^2+3x-4
B) If the area of the shape is 36cm^2, determine the value of x.
36=x^2+3x-4
0=x^2+3x-40
0=(x+8)(x-5)
x+8=0 or x-5=0
x=-8 or x=5
x can not be -8 because you can not have a negative length.
Therefore x = 5
Here is my third word problem and it is on Revenue:
A ferry service has about 240000 riders per day for $2.The port wants to increase the fare to help with increasing operational costs. Research has shown for every $0.10 increase in the fare the number of riders will drop by 10000.
a) What increase in the fare will maximize the revenue?
let x represent the number of increases by $0.10.
Revenue = cost * number of rides
R=(2+0.10x)(240000-10000x)
To find the max revenue you need the axis of symmetry. For that you need the X-intercepts of the equation.
0=(2+0.10x)(240000-10000x)
0=2+0.10x and 0= 240000-10000x
-2= 0.10x and -240000=10000x
-20 and 24 are the X-intercepts.
AOS = -20 + 24 / 2
=4/2
2
There fore you will need two increases for it to make the max revenue.
b) What is the new fare?
To find the new fare you take the cost from the equation which is (2+0.10x). From here you sub in the axis of symmetry which is a x value.
2+0.10(2)
2+0.20
2.20
There fore $2.20 will be the new fare for the max revenue.
c) What is the revenue that will be received from the new fare?
To find the revenue you use this equation R=(2+0.10x)(240000-10000x). Then you sub in the AOS.
R=(2+0.10x(2))(240000-10000(2))
R=(2+0.2)(240000-20000)
R=(2.2)(220000)
R=484000
There fore the revenue for the new fare is $484000.
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