10) What is the base of the triangle?From this triangle we know:THE AREA:[tex]A=80cm^2[/tex]THE HEIGHT:[tex]H=B+12[/tex]We know that the area, height, and base of a triangle are related according to the following formula:[tex]A=\frac{BH}{2} \\ \\ Substituting \ H: \\ \\ A=\frac{B(B+12)}{2} \\ \\ Substituting \ A: \\ \\ 80=\frac{B(B+12)}{2} \\ \\ 2(80)=B(B+12) \\ \\ 160=B^2+12B \\ \\ B^2+12B-160=0[/tex]Solving B by quadratic formula:[tex]B_{12}=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ B_{12}=\frac{-12 \pm \sqrt{12^2-4(1)(-160)}}{2(1)} \\ \\ B_{12}=\frac{-12 \pm \sqrt{12^2-4(1)(-160)}}{2(1)} \\ \\ B_{1}=8 \ and \ B_{2}=-20[/tex]Since we can’t have a negative value of the base, the correct option is:
[tex]B=8cm[/tex]
11) How long will it take the water balloon to hit the ground?We must use the formula:[tex]h(t)=-5.2t^2+v_{0}t+h_{0}[/tex]Since James drops the balloon from a height of 45m, then this is the initial height, so [tex]h_{0}=45m[/tex]. Moreover, at the very instant he drops the balloon the initial velocity is zero, so [tex]v_{0}=0[/tex]. When the ballon hit the ground [tex]h(t)=0[/tex]. Therefore:[tex]0=-5.2t^2+45[/tex]Solving this equation:[tex]5.2t^2=45 \\ \\ t^2=\frac{45}{5.2} \\ \\ t^2=8.653 \\ \\ t=\sqrt{8.653} \\ \\ t=2.941s[/tex]Rounding to the nearest tenth:[tex]\boxed{t=2.9s}[/tex]Finally, the water balloon will hit the ground after 2.9 seconds.12) Height and initial velocityWe have the equation:[tex]h(t)=-16t^2+32t+12[/tex]But we know that this equation follows the form:[tex]h(t)=-16t^2+v_{0}t+h_{0}[/tex]According to this:The height of the platform is [tex]h_{0}=12ft[/tex]The initial velocity of the ball is [tex]v_{0}=32ft/s[/tex]