So for this question, we're asked to find the quadrant in which the angle of data lies and were given to conditions were given. Sign of data is less than zero, and we're given that tangent of data is also less than zero. Now I have an acronym to remember which Trig functions air positive in each quadrant. . And in the first quadrant we have that all the trig functions are positive. In the second quadrant, we have that sign and co seeking are positive. And the third quadrant we have tangent and co tangent are positive. And in the final quadrant, Fourth Quadrant we have co sign and seeking are positive. So our first condition says the sign of data is less than zero. Of course, that means it's negative, so it cannot be quadrant one or quadrant two. It can't be those because here in Quadrant one, we have that all the trick functions air positive and the second quadrant we have that sign. If data is a positive, so we're between Squadron three and quadrant four now. The second condition says the tangent of data is also less than zero now in Quadrant three. We have that tangent of data is positive, so it cannot be quadrant three, so our r final answer is quadrant four, where co sign and seek in are positive.
<span>You multiply both the numerator and the
denominator by the same number (but not 1 or 0), but don't
simplify. 3/10 is your fraction, and you could multiply the top and
bottom numbers by 2 to get 6/20.
I hope I helped please leave thanks
</span>
You draw a line and then a = and another line on another side
like this but bigger ––=––
Answer:
1. 8% of 50 is 4
2. 95% of 40 is 38
3. 42% of 263 is 110.46
4. 110% of 70 is 77
5. 115% of 20 is 23
6. 130% of 78 is 101.4
7. 6.5% of 50 is 3.25
37. 15% of 50 is 7.5
38. 5% of 19 is 0.95
39. 8% of 275 is 22
I hope this helps! Can I plz have a Brainliest??
Step-by-step explanation:
Answer:
The solution region is x < –5 and x > 2
Step-by-step explanation:
We are given the inequalities, 2x–2 < –12 or 2x+3 > 7
Upon simplifying the inequalities, we get,
A. 
i.e. 
i.e. 
B. 
i.e. 
i.e. 
So, we get the solution is 
or 
and the plotted region can be seen below.
