Answer:
However, that's the length of "c" and you're looking to drag it in. So,o you drag in 10^2+7^2=a^2.
Then, as shown by the first picture attatched, you can drag in the one with "c" and "a" with the five shown.
Step-by-step explanation:
To find "c" the diagonal, as explained, you need to use the theorem twice. You can first use it by finding "a", as you already have "b". To find a, you do 10^2+7^2=a^2, the hypotenuse.
100+49=149.
So, a is root 149.
a^2+b^2=c^2 so
149+25=174
Step-by-step explanation:
Answer: t=d/r
Your welcome
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
