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3 years ago

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

Mathematics

1 answer:

Otrada [13]3 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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65*h=c

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2 years ago

Justin and his children went into a bakery and where they sell cookies for $0.75 each and brownies for $1.25 each. Justin has $1

enot [183]

Answer:

Justin could buy up to 8 cookies.

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3 years ago

Can't find what c equal's

LiRa [457]

C equals 17.5 you have to cross multiply to find what c is equal to

8 0

2 years ago

Read 2 more answers

Help asap ☺️☺️☺️☺️☺️☺️☺️

never [62]

Hey There!

80+70 is 150

180
-150
is 30 so 30 must be your answer

Hope This Helps!!!

3 0

3 years ago

Read 2 more answers

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago