H=0
0=-16t^2+40
-40=-16t^2
-40/-16=2.5=t^2
t= sqrt2.5
t=1.58seconds
a
Well, let's take a peek at the graph of that line, hmmm let's pick two points, heck, those tow on the extremes anyway, and those are (-5,2) and (5,-1), alrite.. now, let's do some checking on that.
![\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2=-\cfrac{3}{10}[x-(-5)] \\\\\\ y-2=-\cfrac{3}{10}(x+5)\implies y-2=-\cfrac{3}{10}x-\cfrac{3}{2}\implies y=-\cfrac{3}{10}x-\cfrac{3}{2}+2 \\\\\\ y=-\cfrac{3}{10}x+\cfrac{1}{2}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bpoint-slope%20form%7D%7D%7By-%7B%7B%20y_1%7D%7D%3D%7B%7B%20m%7D%7D%28x-%7B%7B%20x_1%7D%7D%29%7D%5Cimplies%20y-2%3D-%5Ccfrac%7B3%7D%7B10%7D%5Bx-%28-5%29%5D%0A%5C%5C%5C%5C%5C%5C%0Ay-2%3D-%5Ccfrac%7B3%7D%7B10%7D%28x%2B5%29%5Cimplies%20y-2%3D-%5Ccfrac%7B3%7D%7B10%7Dx-%5Ccfrac%7B3%7D%7B2%7D%5Cimplies%20y%3D-%5Ccfrac%7B3%7D%7B10%7Dx-%5Ccfrac%7B3%7D%7B2%7D%2B2%0A%5C%5C%5C%5C%5C%5C%0Ay%3D-%5Ccfrac%7B3%7D%7B10%7Dx%2B%5Ccfrac%7B1%7D%7B2%7D)
Answer:
<h2>
y = -⁵/₂x - 12
</h2>
Step-by-step explanation:
The point-slope form of the equation is y - y₀ = m(x - x₀), where (x₀, y₀) is any point the line passes through and m is the slope:
m = -⁵/₂
(-4, -2) ⇒ x₀ = -4, y₀ = -2
The point-slope form of the equation:
y + 2 = -⁵/₂(x + 4)
So:
y + 2 = -⁵/₂x - 10 {subtract 2 from both sides}
y = -⁵/₂x - 12 ← the slope-intercept form of the equation
Multiply length times width to find the area. 6.5x11= 71.5
