9514 1404 393
Explanation:
<h3>8.</h3>
An exterior angle is equal to the sum of the remote interior angles. Define ∠PQR = 2q, and ∠QPR = 2p. The purpose of this is to let us use a single character to represent the angle, instead of 4 characters.
The above relation tells us ...
∠PRS = ∠PQR +∠QPR = 2q +2p
Then ...
∠TRS = (1/2)∠PRS = (1/2)(2q +2p) = q +p
and
∠TRS = ∠TQR +∠QTR . . . . . exterior is sum of remote interior
q +p = (1/2)(2q) +∠QTR . . . . substitute for ∠TRS and ∠TQR
p = ∠QTR = 1/2(∠QPR) . . . . . subtract q
__
<h3>9.</h3>
For triangle ABC, draw line DE parallel to BC through point A. Put point D on the same side of point A that point B is on the side of the median from vertex A. Then we have congruent alternate interior angles DAB and ABC, as well as EAC and ACB. The angle sum theorem tells you that ...
∠DAB +∠BAC +∠CAE = ∠DAE . . . . a straight angle = 180°
Substituting the congruent angles, this gives ...
∠ABC +∠BAC +∠ACB = 180° . . . . . the desired relation
Answer:
- KEi = 2.256×10^5 J
- KEf = 9.023×10^5 J
- 4 times as much work
Step-by-step explanation:
The kinetic energy for a given mass and velocity is ...
KE = (1/2)mv^2 . . . . . m is mass
At its initial speed, the kinetic energy of the car is ...
KEi = (1/2)(810 kg)(23.6 m/s)^2 ≈ 2.256×10^5 J . . . . . m is meters
At its final speed, the kinetic energy of the car is ...
KEf = (1/2)(810 kg)(47.2 m/s)^2 ≈ 9.023×10^5 J
The ratio of final to initial kinetic energy is ...
(9.023×10^5)/(2.256×10^5) = 4
4 times as much work must be done to stop the car.
_____
You know this without computing the kinetic energy. KE is proportional to the square of speed, so when the speed doubles, the KE is multiplied by 2^2 = 4.
Answer:
I think A
Step-by-step explanation:
Can you mark me brainliest because I am almost to level ace!
Hope you can read this and hope it helps
It goes in 2 times. 6*2=12