When the height is 0 that is when it hits the ground
solve
0=-16t²+25x+3
cant factor so use quadratic formula
for
0=ax²+bx+c
![x=\frac{-b+/-\sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%2B%2F-%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
so for
0=-16t²+25t+3
a=-16
b=25
c=3
so
![x=\frac{-25+/-\sqrt{25^2-4(-16)(3)}}{2(-16)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-25%2B%2F-%5Csqrt%7B25%5E2-4%28-16%29%283%29%7D%7D%7B2%28-16%29%7D)
![x=\frac{-25+/-\frac{625+192}}{-32}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-25%2B%2F-%5Cfrac%7B625%2B192%7D%7D%7B-32%7D)
![x=\frac{-25+/-\frac{817}}{-32}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-25%2B%2F-%5Cfrac%7B817%7D%7D%7B-32%7D)
it has to be a positive time
so it will hit the ground after
![\frac{25+\sqrt{817}}{32}](https://tex.z-dn.net/?f=%5Cfrac%7B25%2B%5Csqrt%7B817%7D%7D%7B32%7D)
seconds
Answer:
-19
Step-by-step explanation:
~ Distribute:
- 11 = 1 / 2 x + -3 / 2
~ Flip the equation:
1 / 2 x + -3 / 2 = -11
~ Use transposition method:
(take -3 / 2 to the other side)
1 / 2 x = -11 - (-3 / 2)
= -19 / 2
x = -12 / 2 * 2
= -19
Hope this helped
Wishing u a gr8 day
:D
Answer:191/94 i hope this helps!!(✿◠‿◠)
<u>Answer:</u>
○ ![y = \frac{1}{4}x - \frac{1}{2}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B4%7Dx%20-%20%5Cfrac%7B1%7D%7B2%7D)
<u>Step-by-step explanation:</u>
To find the equation of the line, let's first consider the points whose coordinates we have been given:
• (6, 1)
• (2, 0).
The point (2, 0) is what is called the x-intercept, which is the point where the line crosses the x-axis. This means that at this point, the y-coordinate of the line is 0.
Next, let's calculate the slope (gradient) of the line using the formula:
![m = \frac{y_2 - y_1}{x_2 - x_1}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D)
where:
m = gradient,
and
= points on the line.
Using the formula:
![m = \frac{1 - 0}{6 - 2}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B1%20-%200%7D%7B6%20-%202%7D)
⇒ ![m =\bf \frac{1}{4}](https://tex.z-dn.net/?f=m%20%3D%5Cbf%20%5Cfrac%7B1%7D%7B4%7D)
Finally, now that we have two points on the line as well as the line's slope, we can use the following formula to find the equation of the line:
![\boxed{y - y_1 = m(x - x_1)}](https://tex.z-dn.net/?f=%5Cboxed%7By%20-%20y_1%20%3D%20m%28x%20-%20x_1%29%7D)
You can use any of the points on the line as
and
.
Using (2, 0):
![y - 0 = \frac{1}{4}( x - 2)](https://tex.z-dn.net/?f=y%20-%200%20%3D%20%5Cfrac%7B1%7D%7B4%7D%28%20x%20-%202%29)
⇒ ![y = \frac{1}{4}x - \frac{1}{2}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B1%7D%7B4%7Dx%20-%20%5Cfrac%7B1%7D%7B2%7D)
Therefore the equation of the line is
.
Learn more about point-slope form at:
brainly.com/question/15143525