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sattari [20]
1 year ago
6

which of the following represent s of the graph of f (x) equals x exponent 3 think about the reflectional drive using Desmos.

Mathematics
1 answer:
ololo11 [35]1 year ago
6 0

The correct answer is A.

lets discard the other options:

For B:

B is always positive, but f(x) is not, so B graph is not the answers

For C:

C is always positive too, so it cannot be the desire graph.

For D:

D is always negative, but we know that x exponent 3 is positive for all positive numbers,

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Answer: We are given there are 49 cars

Also these 49 cars are in a line stretched 528 feet.

Now the average length of the car is:

Average length = \frac{(length-covered-by-49-cars)}{Number-of-cars}

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In this exercise, T: R2 → R2 is a function. For each of the following parts, state why T is not linear. (a) T(a1, a2)= (1, a2) (
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Answer:

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  • c) T is not homogeneous
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Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

  • a) T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0). These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).
  • b) T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5). Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).
  • c) T(\frac{1}{2}(\pi,0))=T((\frac{\pi}{2},0))=(\sin(\frac{\pi}{2}),0)=(1,0); \frac{1}{2}T((\pi,0))=\frac{1}{2}(0,0)=(0,0) so T is not homogeneous.
  • d) T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0) then T is not additive.
  • e) T((0,0)+(1,0))=T(1,0)=(2,0); T((0,0))+T((1,0))=(1,0)+(2,0)=(3,0) then T is not additive.
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