These techniques for elimination are preferred for 3rd order systems and higher. They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b. Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method. As the system of equations increases, the "condition" of a matrix becomes extremely important. Some of this may sound completely alien to you. Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n') become larger than 2.
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Answer:
- northbound: 2.5 hours
- southbound: 1.5 hours
Step-by-step explanation:
When the second train leaves the station, the trains are already 63 miles apart. The rate of increase of their separation is (57 +63) = 120 miles per hour. The remaining 243-63 = 180 miles of separation will be achieved in ...
(180 mi)/(120 mi/h) = 1.5 h
The northbound train will have traveled 2.5 hours; the southbound train will have traveled 1.5 hours when the trains are 243 miles apart.
Answer:
<em>The slope of the line</em> 3.
Step-by-step explanation:
The line is going through 3. Although<em> it is not -3,</em> because if the line is <em>sloping upward from left to right,</em> then the slope is <em>positive (+)</em>. If the line is <em>sloping downward from left to right,</em> the slope is <em>negative (-).</em>
Problem One
Call the radius of the second can = r
Call the height of the second can = h
Then the radius of the first can = 1/3 r
The height of the first can = 3*h
A1 / A2 = (2*pi*(1/3r)*(3h)] / [2*pi * r * h]
Here's what will cancel. The twos on the right will cancel. The 3 and 1/3 will multiply to one. The 2 r's will cancel. The h's will cancel. Finally, the pis will cancel
Result A1 / A2 = 1/1
The labels will be shaped differently, but they will occupy the same area.
Problem Two
It seems like the writer of the problem put some lids on the new solid that were not implied by the question.
If I understand the problem correctly, looking at it from the top you are sweeping out a circle for the lid on top and bottom, plus the center core of the cylinder.
One lid would be pi r^2 = pi w^2 and so 2 of them would be 2 pi w^2
The region between the lids would be 2 pi r h for the surface area which is 2pi w h
Put the 2 regions together and you get
Area = 2 pi w^2 + 2 pi w h
Answer: Upper left corner <<<<< Answer
