Given m with arrow = 3 î + 2 ĵ − 5 k and n with arrow = 6 î − 2 ĵ − 4 k, calculate the vector product m with arrow ✕ n with arro
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By definition, the average rate of change is given by:
We evaluate each of the functions in the given interval.
We have then:
For f (x) = x ^ 2 + 3x:
Evaluating for x = -2:
Evaluating for x = 3:
Then, the AVR is:
For f (x) = 3x - 8:
Evaluating for x =4:
Evaluating for x = 5:
Then, the AVR is:
For f (x) = x ^ 2 - 2x:
Evaluating for x = -3:
Evaluating for x = 4:
Then, the AVR is:
For f (x) = x ^ 2 - 5:
Evaluating for x = -1:
Evaluating for x = 1:
Then, the AVR is:
Answer:
from the greatest to the least value based on the average rate of change in the specified interval:
f(x) = x^2 + 3x interval: [-2, 3]
f(x) = 3x - 8 interval: [4, 5]
f(x) = x^2 - 5 interval: [-1, 1]
f(x) = x^2 - 2x interval: [-3, 4]
Two figures are similar if one is the scaled version of the other.
This is always the case for circles, because their geometry is fixed, and you can't modify it in anyway, otherwise it wouldn't be a circle anymore.
To be more precise, you only need two steps to prove that every two circles are similar:
- Translate one of the two circles so that they have the same center
- Scale the inner circle (for example) unit it has the same radius of the outer one. You can obviously shrink the outer one as well
Now the two circles have the same center and the same radius, and thus they are the same. We just proved that any two circles can be reduced to be the same circle using only translations and scaling, which generate similar shapes.
Recapping, we have:
- Start with circle X and radius r
- Translate it so that it has the same center as circle Y. This new circle, say X', is similar to the first one, because you only translated it.
- Scale the radius of circle X' until it becomes
. This new circle, say X'', is similar to X' because you only scaled it
So, we passed from X to X' to X'', and they are all similar to each other, and in the end we have X''=Y, which ends the proof.