1. ## Solving Problems Involving Quadratic Relations.

its on quadratic relations and different forms ie. standard factored vertex exc all word problems (there are 3 questions)
Looking for someone to do this out of generosity
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I dont know if these answers are right, feel free to correct me on them.

Is this something upon which you are being assessed?

No this is a test review and i need help understand these questions because the test is on monday (tommorow)

Well we have:

$\displaystyle h=-0.04(d-19)^2+14.4$
1) Is fine except that I would advise you to label the points of intersection with the axis.
Your factors are correct. The distance is correct.You're missing a minus sign on the $\displaystyle ax^2+bx+c=0$ form.

Moving onto 2: The rocket will hit the ground when $\displaystyle h=0$

Can you solve
$\displaystyle -4.9t^2+44.1t=0$? Take out a common factor of $\displaystyle t$ and then solve.

For the max height, why not put the equation into vertex form? Alternatively, it will be the point halfway between the x-intercepts, so you could find this point and then substitute it into the formula.

I don't feel like the COMM part is something I can answer without knowing what exactly the examiners are looking for in an answer - for that, I'd consult a teacher.

I have fixed number 1, but i am still confused for #2 on how to remove the common factor then solve. If you can further explain 2 than i should be able to turn it into factored form. As for communications mark that does not matter on this, it will matter on the actual test.

$\displaystyle -4.9t^2+44.1t=0$

Divide both sides by $\displaystyle 4.9$:

$\displaystyle -t^2+9t=0$

$\displaystyle t(9-t)=0$

Once i do that do i FOIL so it will equal $9t+t^2$ ?

The equation marks the flight path of a rocket. We already know the equation of the quadratic, which is the height, h(t) relates to the time, t, in seconds after launch. The equation is $\displaystyle h(t)=-4.9t^2+44.1t$. The height will be $\displaystyle 0$ when $\displaystyle h(t)=0$ so you are being asked to solve

$\displaystyle 0=-4.9t^2+44.1t$
I noticed a common factor of $\displaystyle 4.9$ (knowing my $\displaystyle 4.9$ times table has helped immeasurably in life) so divided through both sides by $\displaystyle 4.9$. This gave:

$\displaystyle 0=-t^2+9t$
There is now a common factor of t, which can be factored:
$\displaystyle t(-t+9)=0$
By the null factor law, there are only two ways this can be possible: if $\displaystyle t=0$ or if $\displaystyle -t+9=0$

Ok, but once i get $t(-t+9)=0$ What is the next step? because if $t=0$ or $-t+9=0$ than i dont know how to get there.

If you cannot solve:
$\displaystyle -t+9=0$, then I am incredibly concerned about your test tomorrow! Can you not see that this means $\displaystyle t=9$?

No i did not see this because i have not done it this exact way before, the teacher does not explain these kinds of things well which is why i have to come online to get the help required.

Ok. What about part b? Can you put the quadratic into vertex form? (Which is the form it was in at the start of q1)

Im not sure if i did it right but i got 793.8m on for b.

I have to go to bed now so i wont be able to correct it right now if its wrong.