ElectricPavlov

This will help:

He gave away 1/8     so he has 7/8ths left    7/8 * 250 = 218.75 ac

66² = 4356₁₀      =  2630₁₂

Note that QCA is a right triangle....area = 1/2 *  base * height

1/2 base * height = area

1/2 * 3 * (12-p) = area

18 - 3/2 p = area

When the ball hits the ground, height = y   will be ZERO

0 = - 16t^2 - 80t + 54         Use quadratic equation    a = -16      b = -80    and   c = 54   to solve for t

                                                     throe out the negative 't' value

\(t = {80 \pm \sqrt{(-80)^2-4(-16)(54)} \over 2(-16)}\)

You need to know how to calculate using the Quadratic Formula......if you already do, then here is a quick and easy Quadratic Formula calculator that I often use :

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

-2(s + 7) > 4s + 8      expand L side

-2s-14 > 4x+8            add 2s and 12 to both sides of the equation

0 > 6x+ 22                 subtract 22 from both sides

-22 > 6x                     divide both sides by 6

-22/6 >x                     or...re-wrote as

x< - 22/6                      (-inf, - 22/6)

This is not possible in a right triangle....   the shorter side + the 20 side do not add up to more than 99 ....necessary to meet the triangle side length rule. .

Normally, you cannot have the sqrt of a negative number....so the domain  is

[0,+inf)

3x^2 + 12x + 4 - 2x^2 + 11x - 7      simplify

x^2 + 23x -3                                  complete the square for 'x'

(x + 11.5)²  - 135.25      You can finish !    (hint: a =1)

t^2 +2t + 9       in simplified form     is a bowl shaped parabola   the minumum     

                              will occur at 't' value = - b/2a    where b = 2   a = 1        

                                     Use this value of 't' in the equation to calculate the minumum value fo the quadratic