For number 1, theat is not a right triangle as a squared plus b squared does not equal c squared
The parabolic motion is an illustration of a quadratic function
The equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31
<h3>How to model the function?</h3>
Given that:
x stands for time and y stands for height in feet
So, we have the following coordinate points
(x,y) = (5,0), (11,0) and (10,80)
A parabolic motion is represented as:
y =ax^2 + bx + c
At (5,0), we have:
25a + 5b + c = 0
c= -25a - 5b
At (11,0), we have:
121a + 11b + c = 0
Substitute c= -25a - 5b
121a + 11b -25a - 5b = 0
Simpify
96a + 6b = 0
At (10,80), we have:
100a + 10b + c = 80
Substitute c= -25a - 5b
100a + 10b - 25a -5b = 80
75a -5b = 80
Divide through by 5
15a -b = 16
Make b the subject
b = 15a + 16
Substitute b = 15a + 16 in 96a + 6b = 0
96a + 6(15a + 16) = 0
Expand
96a + 90a + 96 = 0
This gives
186a = -96
Solve for a
a = -16/31
Recall that:
b = 15a + 16
So, we have:
b = -15 * 16/31 + 16
b =-240/31 + 16
Take LCM
b =(-240 + 31 * 16)/31
[tex]b =256/31
Also, we have:
c= -25a - 5b
This gives
c= 25*16/31 - 5 * 256/31
Take LCM
c= -880/31
Recall that:
y =ax^2 + bx + c
This gives
y = -16/31x^2 + 256/31x - 880/31
Hence, the equation that models that path of the rocket is y = -16/31x^2 + 256/31x - 880/31
Read more about parabolic motion at:
brainly.com/question/1130127
Since the slope is -2, then m=-2. Since the y-intercept is 5, b=5. Then, we can create the equation: y=-2x+5
Since the slope is 
, then m=. Since the y-intercept is -3, b=-3.
Then, we can create the equation: y=x-3
Answer:
6u
Step-by-step explanation:
The two terms 10u and 4u are like terms because they both carry the same variable (u) with the same degree (1). So, we can just subtract them as if they were normal numbers 10 - 4:
10u - 4u = 6u
Thus, the answer is 6u.
Hope this helps!
