Find the midpoint:
m= x1+x2/2; y1+y2/2
m= 9+-1/2; 8+-2/2
m= 8/2; 6/2
m= (4,3)
(4,3) is your answer.
I hope this helps!
~kaikers
Answer:
There are 23 dimes in the box
Step-by-step explanation:
Let nickels=n
Let dimes=(35-n)
Equation: 0.05n+0.10(35-n)=2.90
Multiply both sides by 100: 5n+10(35-n)=290
-5n+350=290
-5n=-60
n=12
35-12=23
Therefore, there are 23 dimes in the box
The braking distance is the distance the car travels before coming to a stop after the brakes are applied
a. The braking distances are as follows;
- The braking distance at 25 mph, is approximately <u>63.7 ft.</u>
- The braking distance at 55 mph, is approximately <u>298.35 ft.</u>
- The braking distance at 85 mph, is approximately <u>708.92 ft.</u>
b. If the car takes 450 feet to brake, it was traveling with a speed of 98.211 ft./s
Reason:
The given function for the braking distance is D = 2.6 + v²/22
a. The braking distance if the car is going 25 mph is therefore;
25 mph = 36.66339 ft./s

At 25 mph, the braking distance is approximately <u>63.7 ft.</u>
At 55 mph, the braking distance is given as follows;
55 mph = 80.65945 ft.s

At 55 mph, the braking distance is approximately <u>298.35 ft.</u>
At 85 mph, the braking distance is given as follows;
85 mph = 124.6555 ft.s

At 85 mph, the braking distance is approximately <u>708.92 ft.</u>
b. The speed of the car when the braking distance is 450 feet is given as follows;

v² = (450 - 2.6) × 22 = 9842.8
v = √(9842.2) ≈ 98.211 ft./s
The car was moving at v ≈ <u>98.211 ft./s</u>
Learn more here:
brainly.com/question/18591940
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Answer:
a) average rate = (total distance)/(total time)
b) Rave = 2·R1·R2/(R1 +R2)
c) cheetah's average rate ≈ 50.91 mph
Step-by-step explanation:
a) Let AB represent the distance from A to B. Let t1 and t2 represent the travel times (in hours) on leg1 and leg2 of the trip, respectively. Then the distances traveled are...
First leg distance: AB = 70·t1 ⇒ t1 = AB/70
Second leg distance: AB = 40·t2 ⇒ t2 = AB/40
The average rate is the ratio of total distance to total time:
average rate = (AB +AB)/(t1 +t2)
average rate = 2AB/(AB/70 +AB/40) = 2/(1/70 +1/40) = 2(40)(70)/(70+40)
average rate = 560/11 = 50 10/11 . . . mph
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No equations are given, so we cannot compare what we wrote with the given equations. In each step of the solution, we have used the rules of algebra and equality.
b) For two rates over the same distance (as above), the average is their harmonic mean:
average rate = 2r1·r2/(r1+r2)
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c) The cheetah's average rate was 50 10/11 mph ≈ 50.91 mph.
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