Math Is Fun Forum

No dummy here. Like always, you were right.

However, you said -√3  < x < 3. It cannot be negative, no?

But since none of the sides are of zero length, then we can indeed divide by t.

And Bobby, I THINK (I have to be cautious about being wrong lol) the reason why you cannot have 2t^2 on the left side of the equation is because the angle across of it is not the obtuse angle that we are looking for. Only the angle across from (kt)^2 can be obtuse, that is why I put it on the left side.

What do you mean on the left? I figured the only angle that could be obtuse in this triangle is the one across from side bc.

This is what I saw it as:

http://i32.tinypic.com/33ljxoy.png

A guess here, for all k values greater than √5 and less than 3.

That's assuming that t is just a constant (meaning side ac is half as long as side ab). So I just set t=1.

@Bobby, I used the cosine law too and it somehow worked out. I used k^2 = 1^2+2^2 - 2(1*2)cos(x).

(k^2 - 5)/(-4) > -1 and we get -3 < k < 3 but k cannot be negative.
(k^2 - 5)/(-4) > 0 and we get -√5 < k < √5 but once again k cannot be negative.

So I came up with k values greater than √5 and less than 3.

Not entirely sure if that is correct.

And thank you, now I got only one hard month of full-time summer school left. Speaking of which, I better get ready in a few minutes. Have a fun Tuesday, you two!

We will turn the first two fractions into one fraction by using their lowest common denominator (1827), and do the same with the second set (510).

Take it from there!

It's a simple ratio; where x is the amount produced last month.

2000000/480 = x/300
Multiply both sides by 300 and you get x= 300 * (2000000/480) = 1,250,000 lbs.

And that is the correct answer!

Let me try to comprehend what you did there. http://i46.tinypic.com/2l9gsvl.png

Was that correct analysis?

I forgot that the angle at the bottom is the same as the second angle I pointed (or you did) out in the middle of the diagram.

Oh boy, thanks a lot once more Bobby.

Edit: Didn't see the link you posted. Interesting. And I guess he likely had the same question from the textbook as well.

That makes good sense. Thanks once again Bobbym.

Wish me luck on the exam, it's the last thing to worry about before my long summer.

Yup. I was just putting z=0 for ease but since you mentioned we can put any number, does that mean the plane can have more than one sketch?

With some help from others, I was able to get the answer. In case anyone in the future is interested:

Since the line is perpendicular to the plane and we know that the normal of the plane is also perpendicular to it, we can conclude that the normal of the plane and the direction vector of the line are parallel.

So we can use Ax+By+Cz+D=0 where A=5, B=-4, C=7. x,y,z are simply coordinates of the point in the plane (2,-1,8)
(5)(2)+(-4)(-1)+(7)(8)+D=0
D=-70

Therefore, the Cartesian equation of the plane is 5x-4y+7z-70=0.

As only a SKETCH is being requested, you can factor it as follows:
y= x^2+3x-4
y= (x+4)(x-1)

The zeros are x=-4 and x=+1. The leading co-efficient is positive so it opens up. With that much information, you should be able to sketch the parabola.