The line segments between (4,-6) and (-6,-16) is 1.
slope = y2 -y1 /x2-x1 = -16+6/-6-4=-10/-10=1
It’s an isosceles triangle so it’s 4
11 hundredths , or 11/100
We take the equation <span>d = -16t^2+12t</span> and subtract d from both sides to get
0<span> = -16t^2+12t - d
We apply the quadratic formula to solve for t. With a = -16, b = 12, c = -d, we have
t = [ -(12) </span><span>± √( 12^2 - 4(-16)(-d) ) ] / [2 * -16]</span>
= [- 12 ± √(144-64d) ] / (-32)
= [- 12 ± √16(9-4d)] / (-32)
= [- 12 ± 4√(9-4d)] / (-32)
= 3/8 ±√(9-4d) / 8
The answer to your question is t = 3/8 ±√(9-4d) / 8
The little lines on each side of the rhombus mean that all the sides are the same length.
We can set line LM and MN equal to solve for X, then we can solve the length of a side.
3x-3 = x+7
Add 3 to each side:
3x = x +10
Subtract x from each side:
2x = 10
Divide both sides by 2:
x = 10/2
x = 5
Now we have the value for x, replace x in one of the side formulas:
x +7 = 5+7 = 12
Each side = 12 units.
The perimeter would be 12 + 12 + 12 + 12 = 48 units.
