The two diagonals intersects at right-angle for a kite
the long diagonal also bisects the short diagonal so QP = PS = QS/2 = 6/2 = 3
QPR is a right-angle triangle with QR at 5 and QR at 3
RP = sqrt (5^2 - 3^) = 4m
This is a right angle triangle problem
drawing a vertical line at from the point where the ramp touches the car park leaves a right angle triangle with the
opposite being 2m
hypothenus being 10m
adjacent unknown
we could use sine
SineO equal to opposite over hypothenus
SineO equal to 2/10
SineO equal to 0.2
O equal to Sine^1(0.2)
O equal to 11 .5
The angle between the ramp and the horizontal is 11.5 degrees
Question 1:
We are going to add.
1/5 + 1/2
2/10 + 5/10
7/10
Note that we are going DOWN so the number will be negative.
Therefore, the answer is A.) -7/10
Question 2:
We are going to subtract.
4 1/2 - 1 2/3
9/2 - 5/3
27/6 - 10/6
17/6 or 2 5/6
Note that we are still BELOW sea level so the number will be negative.
Therefore, the answer is C.) -2 5/6
Question 3:
We are going to subtract.
2 2/3 - 1 1/5
8/3 - 6/5
40/15 - 18/15
22/15 or 1 7/15
Therefore, the answer is B.) 1 7/15
Best of Luck!
-9 because 2+7=9 so the negative part oh then must be -9
