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The path of a fired projectile is modeled by the function f(x)=x(2x+8) where x is the height (in feet) and f(x) in the distance(in feet). If the distance is 24 feet, what is the height of the projectile ?

$$f(x)=x(2x+8)\quad f(x) = 24 \\
24 = x(2x+8) \\
24 = 2x^2+8x \quad | \quad :2\\
12 = x^2 + 4x \quad | \quad -12\\
0 = x^2 + 4x -12 \\
x^2 + 4x -12 = 0 \quad \small{\text{factoring}}\\
\underbrace{(x-2)}_{=0}\underbrace{(x+6)}_{=0} =0 \\
x-2=0 \quad \small{\text{we have} }
\quad x=2 \quad \small{\text{okay}} \\
x+6=0 \quad \small{\text{we have}} \quad x=-6 \quad \small{\text{no solution!}}\\$$

the height of the projectile is 2 feet.

Mind providing an example?

A litre of water occupies 1000 cm³. The density of ice is about 0.92 times the density of water so it would occupy about 1000/0.92 ≈ 1087 cm³.

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Everyone does!

I know that feels........not that good of a story

Absolutely correct answer programmer !

you could have done an easy way by just reducing 4 and 16 and then it would just give the answer easily!

but good answer!

Definitely!

If I understood your question correctly the answer should be like this:

1 1/16 is the same as 17/16 so

$${\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{17}} = {\mathtt{68}}$$, $${\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{16}} = {\mathtt{80}}$$

68/80 = 17/20. Right?.

Its ok yours is better anyways :)  

Sorry for the late answer but here goes:

You simply select sin, cos or tan. Then it should say something like $${tan}\left($$

Type the amount of degrees the angle is and close the parenthesis: $$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{50}}^\circ\right)}$$

So let's say I wanted to use tan to figure out a in an right-angled triangle. I have Angle A and side a

The formula is $${\frac{{\mathtt{a}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{A}}\right)}}}$$, so I go: $${\frac{{\mathtt{5}}}{\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{50}}^\circ\right)}}} = {\mathtt{4.195\: \!498\: \!155\: \!887\: \!139\: \!2}}$$

Hope this helped even though it probably didn't...

Thanks everyone! 

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

r = lzl  = √(a^2 + b^2) = √4^2+(-2i)^2 = √4^2+2^2 = √16+4 = √20 = 4,47

b < 0 and a > 0

θ = tan-1(b/a) + pi = tan-1(-2/4) + pi = -23,42 degrees

4,47 and -23,42 degrees

Oooh lala!melody your whatever dinosaur is celebrating becoz our dear Zeg-mammoth has woken Up from his long long hibernation ! And thank you Melody! I'm happy you like it!

tiddle I agree to the fact of you being late! I'm not so of a patient person ! 

Look in the formulary tab at the top of your page. 

You often think of things that don't occur to me.  

One thing I like so much about this forum is that we can share little bits of info like this :)

the sharing makes learning so much easier :)

1l = 1dm3 = 1000cm3

Using x + y = 2   →  y = 2 - x      and xy = 23  ...so....

x(2- x)= 23      expand

2x - x^2 = 23    rearrange

x^2 - 2x + 23 = 0    .....which implies that...

x^2 = 2x -23

And y^2 = (2 - x)^2 = (x -2)^2 = x^2 - 4x + 4

Therefore

x^2 + y^2 =

2x - 23 + x^2 - 4x + 4 =

x^2 - 2x - 19

@@ End of Day Wrap      Thurs 29 /1/15        Sydney, Australia         Time 10:10 pm

Good Evening Everyone,

Well we had lots of answerers today.  It is great when people get involved.  I always feel that people are much more likely to learn if they offer solutions.  (So long as they really do have some idea of what they are doing)  My logic is that if they have given it a go they are much more likely to look and learn if someone else comes in with a different answer.  

Try to use your watch list well to store the questions that you have answered.  They blink at you in a very demanding fashion if someone else posts and you have not seen it yet.  I wrote a post about this a while back.

Rewrite to polar form re ^ iθ, where θ is an angle in radians

$$\small{\text{$\boxed{z=a+bi \quad \to \quad re^{i\phi} }$}}\\\\
\small{\text{
$
\boxed{r=\sqrt{a^2+b^2}
\qquad
\phi= \left\{
\begin{array}{ll}
\arctan{( \frac{b}{a} )}, & a > 0 \\ \\
\arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\
\arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\
 \frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\
-\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\
\end{array}
\right.
}
$
}}$$

I.

   $$\\\small{\text{\ $z = -3$}}\\
\small{\text{ $a=-3, \ b = 0 $}}\\
\small{\text{ $r=\sqrt{(-3)^2+0^2} = 3$}} \\
\small{\text{ $a < 0, \ b=0:\quad$ $\phi=\arctan{( \frac{b}{a} )} + \pi
=\arctan{( \frac{0}{-3} )}+\pi = 0+\pi = \pi
$}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -3 = 3e^{i\pi}}$}}$$

II.

$$\\\small{\text{\ $z = 1-i$}}\\
\small{\text{ $a=1, \ b = -1$}}\\
\small{\text{ $r=\sqrt{(1)^2+(-1)^2} = \sqrt{2}$}} \\
\small{\text{ $a > 0, \ b < 0:\quad$ $\phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-1}{1} )} = -\frac{\pi}{4}
$}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = 1-i = \sqrt{2}e^{-\frac{\pi}{4}
i}}$}}$$

III.

$$\\\small{\text{\ $z = -5i$}}\\
\small{\text{ $a=0, \ b = -5$}}\\
\small{\text{ $r=\sqrt{(0)^2+(-5)^2} = 5$}} \\
\small{\text{ $a = 0, \ b < 0:\quad$ $\phi= -\frac{ \pi }{2} $}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -5i = 5e^{-\frac{ \pi }{2} i}}$}}$$

Ah...that's very clever, Melody.....that thought didn't strike me...!!!

Friday 30/1/15

When in doubt....just look at the answers...b and d would produce positive first terms....but the given problem has a negative first term.....

So...it appears it's a two-horse race....

Expand each remaining one and see what happens....

£280 divide by 6 is £46.66

$$\\\frac{\sqrt{-40}}{\sqrt{4}}\\\\
=\sqrt{-10}\\\\
=\sqrt{10*-1}\\\\
=\sqrt{10}*\sqrt{-1}\\\\
=\sqrt{10}\;i\\\\$$

This is not a real number.

There are no real solutions since no number multiplied by itself = a neg number

This is an 'imaginary number'    

I'm unfamiliar with the notation used in the last one, so I can't comment on that one

On the -5i one....think about where the terminal point of this angle lies.....it's on the y axis, 5 units "down" from the origin in the "imaginary" direction {pardon the non-math terms, here}

So, the correct angle should be -pi.2, not pi/2......