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Degger [83]
3 years ago
14

In which quadrant is sec e< 0 and tan e > 0?

Mathematics
1 answer:
iren2701 [21]3 years ago
7 0

Answer:

C

Step-by-step explanation:

Because Tangent is positive in third quadrant

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Distance between (-3,1) (1,5)
Mamont248 [21]

Answer & Step-by-step explanation:

For this problem, we will use the distance formula.

d=\sqrt{(x2-x1)^2+(y2-y1)^2}

For our (x1, y1), we will use (-3, 1). For our (x2, y2), we will use (1, 5). Now let's plug in our numbers and solve the problem.

d=\sqrt{(1-(-3))^2+(5-1)^2}

d=\sqrt{4^2+4^2}

d=\sqrt{16 + 16}

d=\sqrt{32}

d=5.657

So, the distance between (-3,1) and (1,5) is 5.657 units.

5 0
3 years ago
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Step-by-step explanation:

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Answer:

How does y vary with x?

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Step-by-step explanation:

The first table's results would look like:

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