Step-by-step explanation:
A is in Quadrant I
D is in Quadrant III
Main street is vertical to park entrance
You know that BC is congruent to x so you need to solve for x using the ratio:

So then we need to find BC.
We know:


Therefore BC =8
Then:

Average=(total number)/(number of items)
given that the final exam counts as two test, let the final exam be x. The weight of the final exams on the average is 2, thus the final exam can be written as 2x because any score Shureka gets will be doubled before the averaging.
Hence our inequality will be as follows:
(67+68+76+63+2x)/6≥71
(274+2x)/6≥71
solving the above we get:
274+2x≥71×6
274+2x≥426
2x≥426-274
2x≥152
x≥76
b] The above answer is x≥76, the mean of this is that if Shureka is aiming at getting an average of 71 or above, then she should be able to get a minimum score of 76 or above. Anything less than 76 will drop her average lower than 71.
Answer:
the slope is 4
Step-by-step explanation:
y+3=4(x-5)
This is in point slope form
y-y1=m(x-x1)
where (x1,y1) is a point on the line and m is the slope
The slope is 4 and a point on the line is (5,-3)