Suppose an aerial freestyle skier goes off a ramp with her path represented by the equation y=−0.024(x−25)2+40. If the surface o
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<u>Annotation</u>
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equation
d₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation
2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.
Find the velocity of the first car, use the first equation
v₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer
Solve the following system:{12 x = 54 - 6 y | (equation 1)-17 x = -6 y - 62 | (equation 2)
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.
Express the system in standard form:{12 x + 6 y = 54 | (equation 1)-(17 x) + 6 y = -62 | (equation 2)
Swap equation 1 with equation 2:{-(17 x) + 6 y = -62 | (equation 1)12 x + 6 y = 54 | (equation 2)
Add 12/17 × (equation 1) to equation 2:{-(17 x) + 6 y = -62 | (equation 1)0 x+(174 y)/17 = 174/17 | (equation 2)
Multiply equation 2 by 17/174:{-(17 x) + 6 y = -62 | (equation 1)0 x+y = 1 | (equation 2)
Subtract 6 × (equation 2) from equation 1:{-(17 x)+0 y = -68 | (equation 1)0 x+y = 1 | (equation 2)
Divide equation 1 by -17:{x+0 y = 4 | (equation 1)0 x+y = 1 | (equation 2)
Collect results:Answer: {x = 4 {y = 1
Please note the { are supposed to span over both equations but it interfaces doesn't allow it. Please see attachment for clarification.