You know where the glacier is now, and how far it moves in
one year. The question is asking how close to the sea it will be
after many years.
Step-1 ... you have to find out how many years
Step-2 ... you have to figure out how far it moves in that many years
Step-3 ... you have to figure out where it is after it moves that far
The first time I worked this problem, I left out the most important
step ... READ the problem carefully and make SURE you know
the real question. The first time I worked the problem, I thought
I was done after Step-2.
============================
Step-1: How many years is it from 2010 to 2030 ?
(2030 - 2010) = 20 years .
Step-2: How far will the glacier move in 20 years ?
It moves 0.004 mile in 1 year.
In 20 years, it moves 0.004 mile 20 times
0.004 x 20 = 0.08 mile
Step-3: How far will it be from the sea after all those years ?
In 2010, when we started watching it, it was 6.9 miles
from the sea.
The glacier moves toward the sea.
In 20 years, it will be 0.08 mile closer to the sea.
How close will it be ?
6.9 miles - 0.08 mile = 6.82 miles (if it doesn't melt)
Answer:
The volume of the solid is the volume of the prism minus the volume of the cylinder.
For the cylinder, diameter = d = 4 cm
radius = d/2 = (4 cm)/2 = 2 cm
V = volume of prism - volume of cylinder
The volume of a prism is length times width times height.
The volume of a cylinder is pi times the square of the radius times the height.
V = LWH - (pi)r^2h
V = 6 cm * 6 cm * 15 cm - (pi)(2 cm)^2(15 cm)
V = 540 cm^3 - 60pi cm^3
V = (540 - 60pi) cm^3
The graph would intercept at (0,-4) on the y-axis & at (5.3,0) on the x-axis. And the shaded region would be on the right side of the line on the graph.
Assuming this is a system of equations, here is how to find x.
2x + 3y = 45
x + y = 10
Multiply x + y = 10 by 2 so you are able to use the elimination method.
2x + 3y = 45
2x + 2y = 20
Subtract.
y = 25
Now that we've found y, we can plug it in to find x.
x + 25 = 10
Subtract 25 from both sides.
x = -15
