![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-10})\qquad B(\stackrel{x_2}{x}~,~\stackrel{y_2}{-4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ 10=\sqrt{[x-(-6)]^2+[-4-(-10)]^2}\implies 10=\sqrt{(x+6)^2+(-4+10)^2} \\\\\\ 10^2=(x+6)^2+(6)^2\implies 100=x^2+12x+36+36 \\\\\\ 100=x^2+12x+72\implies 0=x^2+12x-28 \\\\\\ 0=(x+14)(x-2)\implies x= \begin{cases} -14\\ \boxed{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20A%28%5Cstackrel%7Bx_1%7D%7B-6%7D~%2C~%5Cstackrel%7By_1%7D%7B-10%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7Bx%7D~%2C~%5Cstackrel%7By_2%7D%7B-4%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%2010%3D%5Csqrt%7B%5Bx-%28-6%29%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%2010%3D%5Csqrt%7B%28x%2B6%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%2010%5E2%3D%28x%2B6%29%5E2%2B%286%29%5E2%5Cimplies%20100%3Dx%5E2%2B12x%2B36%2B36%20%5C%5C%5C%5C%5C%5C%20100%3Dx%5E2%2B12x%2B72%5Cimplies%200%3Dx%5E2%2B12x-28%20%5C%5C%5C%5C%5C%5C%200%3D%28x%2B14%29%28x-2%29%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%20-14%5C%5C%20%5Cboxed%7B2%7D%20%5Cend%7Bcases%7D)
because B is on the IV Quadrant, the x-coordinate must be positive.
Ground: y = 0
- 16t² + 128t + 50 = 0
Apply quadratic equation.
t = 0, 8 (rounded to the nearest second)
8 seconds
Answer:
246
Step-by-step explanation:
He should increase the length and width by a factor = sqrt 3.
so new length should be 78* sqrt3 = 135.1 - that is increase of 58.1 ft
and new width should be 65 * sqrt3 = 112,58 = increase of 47.58 ft
Answer:
(14,3)
Step-by-step explanation:
2x + y = 19
x - y = 11
-2(x - y) = -2(11)
-2x + 2y = -22
-2x + 2y = -22 + 2x + y = 19
3y = -3
y = 3
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x - y = 11
x - 3 = 11
x = 11 + 3
x = 14
