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.
Sqr (x2 - x1)^2 + (y2 - y1)^2
Sqr (-4-0)^2 + (2+5)^2
Sqr (16) + 49 = sqr 65
I got around 8.
(I might did something wrong, idk)
The vertex and the t-intercept where t>0, will help solve this problem. Sending the work on the attachment.
You should use the 1 mile because the meter and inch is too short to measure the city block and it’ll take really long this is why you should use 1 mile.
Answer:
bottom of graph will move from (0,0) to point (1,3) after transformation
Step-by-step explanation:
given
original : f(x) = 
transformed; g(x) = 
+ 3
look at this way g(x) = 
+ k
if (x-h), h>0, move h units to the right
if k>0, move k units up
the bottom of the graph will be at point (1,3)
