For each of the following, find the transition matrix from B, to B, and the transition matrix from B2 to Bi. Be sure to label wh
ich is which. Then use the transition matrices to write the given vector relative to the other basis. You may use a calculator for row reduction (just show the original matrix and it's reduced row echelon form). You may also use a calculator for finding matrix inverses. (a) V =R2, B = {(-4, 3), (6,2)), B2 = {(2, -3),(-1,4)}, TB = (2,-5). = {(0,4,-1),(1,2,-6), (2, -2,0)}, (b) V =R3, B1 = {(3,2,1),(0,3, -2), (0, -2, 1)}, B 7 B2 = (4, -5,3). (e) V = P2, B, = {rº, 1, 1), B2 = {4.x2 – 2x + 5, 2x2 – 3.2 +1, 4.r? – 2}, (6)b = (-4,3,1).
1 answer:
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Answer:
Final answer is .
Step-by-step explanation:
Given expression is
Now we need to find about what is product of both factors.
We see that both factors are in exponent form having equal base "x".
So we can apply formula:
Hence final answer is .
Answer:
2 hours, 150 miles
Step-by-step explanation:
The relation between time, speed, and distance can be used to solve this problem. It can work well to consider just the distance between the drivers, and the speed at which that is changing.
<h3>Separation distance</h3>Jason got a head start of 20 miles, so that is the initial separation between the two drivers.
<h3>Closure speed</h3>Jason is driving 10 mph faster than Britton, so is closing the initial separation gap at that rate.
<h3>Closure time</h3>The relevant relation is ...
time = distance/speed
Then the time it takes to reduce the separation distance to zero is ...
closure time = separation distance / closure speed = 20 mi / (10 mi/h)
closure time = 2 h
Britton will catch up to Jason after 2 hours. In that time, Britton will have driven (2 h)(75 mi/h) = 150 miles.
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<em>Additional comment</em>
The attached graph shows the distance driven as a function of time from when Britton started. The distances will be equal after 2 hours, meaning the drivers are in the same place, 150 miles from their starting spot.
