The answer would be option B "multiply 2 1/2 x 3 and then add 1/4." If you multiply 2 1/2 by three your adding on to the answer. You would have to multiply 1/4 by three then add 2 1/2 to get your answer. It's not option A because your adding the three 1/4's together then multiply it by 2 1/2 which is one of the most common ways to solve this question. It's not option C because you are multiplying 1/4 by 3 which saves all the time adding 1/4 + 1/4 + 1/4. It's also not option D because the / are just in different directions which doesn't matter.
Hope this helps!
Answer:
The answer to your question is below
Step-by-step explanation:
The slope is the coefficient of the x, and the y- intercept is the number after the x.
Slope y-intercept
1. y = 3x - 1 3 -1
2. y = -2/5x + 4 -2/5 4
3. 7 = .5x 0.5 0
a) Order the slopes from least to greatest
-2/5, 0.5, 3
b) Order y-intercept from least to greatest
-1, 0, 4
Answer:
t=0.93 seconds or t=0.07 seconds
Step-by-step explanation:
If ![h=-16t^2+8t-1](https://tex.z-dn.net/?f=h%3D-16t%5E2%2B8t-1)
The time when the projectile lands on the ground is when its height, h=0.
![-16t^2+8t-1=0](https://tex.z-dn.net/?f=-16t%5E2%2B8t-1%3D0)
Using Factoring by Perfect Squares
![-16t^2+8t=1\\\text{Divide all through by the coefficient of t^2}\\\frac{-16t^2}{-16} +\frac{8t}{-16}=\frac{1}{-16}\\t^2-\frac{1}{2}t=-\frac{1}{16}](https://tex.z-dn.net/?f=-16t%5E2%2B8t%3D1%5C%5C%5Ctext%7BDivide%20all%20through%20by%20the%20coefficient%20of%20t%5E2%7D%5C%5C%5Cfrac%7B-16t%5E2%7D%7B-16%7D%20%2B%5Cfrac%7B8t%7D%7B-16%7D%3D%5Cfrac%7B1%7D%7B-16%7D%5C%5Ct%5E2-%5Cfrac%7B1%7D%7B2%7Dt%3D-%5Cfrac%7B1%7D%7B16%7D)
Divide all through by the coefficient of ![t^2](https://tex.z-dn.net/?f=t%5E2)
![\frac{-16t^2}{-16} +\frac{8t}{-16}=\frac{1}{-16}\\t^2-\frac{1}{2}t=-\frac{1}{16}](https://tex.z-dn.net/?f=%5Cfrac%7B-16t%5E2%7D%7B-16%7D%20%2B%5Cfrac%7B8t%7D%7B-16%7D%3D%5Cfrac%7B1%7D%7B-16%7D%5C%5Ct%5E2-%5Cfrac%7B1%7D%7B2%7Dt%3D-%5Cfrac%7B1%7D%7B16%7D)
Next, divide the coefficient of t by 2, square it and add it to both sides.
![t^2-\frac{1}{2}t+(-\frac{1}{2})^2=-\frac{1}{16}+(-\frac{1}{2})^2\\(t-\frac{1}{2})^2=-\frac{1}{16}+\frac{1}{4}\\(t-\frac{1}{2})^2=\frac{3}{16}\\](https://tex.z-dn.net/?f=t%5E2-%5Cfrac%7B1%7D%7B2%7Dt%2B%28-%5Cfrac%7B1%7D%7B2%7D%29%5E2%3D-%5Cfrac%7B1%7D%7B16%7D%2B%28-%5Cfrac%7B1%7D%7B2%7D%29%5E2%5C%5C%28t-%5Cfrac%7B1%7D%7B2%7D%29%5E2%3D-%5Cfrac%7B1%7D%7B16%7D%2B%5Cfrac%7B1%7D%7B4%7D%5C%5C%28t-%5Cfrac%7B1%7D%7B2%7D%29%5E2%3D%5Cfrac%7B3%7D%7B16%7D%5C%5C)
Taking square roots of both sides
![t-\frac{1}{2}=\sqrt{\frac{3}{16}}\\t=\frac{1}{2} \pm \sqrt{\frac{3}{16}}\\t=\frac{1}{2} \pm \frac{\sqrt{3}}{4}](https://tex.z-dn.net/?f=t-%5Cfrac%7B1%7D%7B2%7D%3D%5Csqrt%7B%5Cfrac%7B3%7D%7B16%7D%7D%5C%5Ct%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cpm%20%5Csqrt%7B%5Cfrac%7B3%7D%7B16%7D%7D%5C%5Ct%3D%5Cfrac%7B1%7D%7B2%7D%20%5Cpm%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D)
Therefore:
![t=\frac{1}{2} + \frac{\sqrt{3}}{4} \: OR \: t=\frac{1}{2} - \frac{\sqrt{3}}{4}\\t=\frac{2+\sqrt{3}}{4} \: OR \: \frac{2-\sqrt{3}}{4}\\t=0.93seconds \: OR \: 0.07 seconds](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B1%7D%7B2%7D%20%2B%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%20%5C%3A%20OR%20%5C%3A%20t%3D%5Cfrac%7B1%7D%7B2%7D%20-%20%5Cfrac%7B%5Csqrt%7B3%7D%7D%7B4%7D%5C%5Ct%3D%5Cfrac%7B2%2B%5Csqrt%7B3%7D%7D%7B4%7D%20%5C%3A%20OR%20%5C%3A%20%5Cfrac%7B2-%5Csqrt%7B3%7D%7D%7B4%7D%5C%5Ct%3D0.93seconds%20%5C%3A%20OR%20%5C%3A%200.07%20seconds)
1/8 i think it is good bye