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2 years ago

If Cos A >0, and Csc A < 0, what quadrant does A lie?

Mathematics

1 answer:

Otrada [13]2 years ago

6 0

Answer:

Quadrant IV.

Step-by-step explanation:

You need to remember that :

All trig ratios are positive in the Quadrant I.

Sine and csc are positive in Quadrant II.

Tan and cot are positive in Quadrant III.

Cos and sec are positive in Quadrant IV.

So if cos A > 0 ( ie positive) it is in either Quadrant I or IV.

Csc is <0 (negative) so it is in either Quadrant II or IV.

So if cos A > 0 and csc < 0 , A must be in Quadrant IV.

Hope this helps.

Feel free to ask questions if this is not entirely clear.

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Step-by-step explanation:

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Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x, y) → (0, 0) x4 − 34y2 x2 + 17y2

HACTEHA [7]

Answer:

Step-by-step explanation:

Given the limit of the function $\lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}$ , to find the limit, the following steps must be taken.

Step 1: Substitute the limit at x = 0 and y = 0 into the function

$= \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}\\= \frac{0^4-34(0)^2}{0^2+17(0)^2}\\= \frac{0}{0} (indeterminate)$

Step 2: Substitute y = mx int o the function and simplify

$= \lim_{(x,mx) \to (0,0)} \frac{x^4-34(mx)^2}{x^2+17(mx)^2}\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^4-34m^2x^2}{x^2+17m^2x^2}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2(x^2-34m^2)}{x^2(1+17m^2)}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2-34m^2}{1+17m^2}\\$

$= \frac{0^2-34m^2}{1+17m^2}\\\\= \frac{34m^2}{1+17m^2}\\\\$

<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>

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