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Tyler
06-04-02, 03:32 PM
Ahhhhhh! My math exam is tomorrow and I found out today!!!!!

Anyway, I have last years exam here in front of me and I need a little bit o' help. I'm using '{sq}' to represent the squaring symbol. I would use '{sqrt}' for the square root symbol! Anyone with a min, could you please help! Thank you!

Problem one:

"Solve: x{sq} 7x -3
------- - ----- = ----
2.......... 4.........2 "

and it apaprently wants the two possible answers as it is a quadratic function.

Problem two:

"For the equation y = -3x{sq} - 2x + 4 find the vertex of the graph."

Problem three:

"Solve: 2x{sq} + 1 = 5x"

Thanks again!!!!

Joeman
06-04-02, 03:56 PM
Originally posted by Tyler

Problem one:

"Solve: x{sq} 7x -3
------- - ----- = ----
2.......... 4.........2 "

and it apaprently wants the two possible answers as it is a quadratic function.

multiply both sides by 4
2x^2 - 7x + 6 = 0 Factor: => (x-2)(2x-3)=0
x = 2 and 3/2

Problem two:

"For the equation y = -3x{sq} - 2x + 4 find the vertex of the graph."

The trick is to realize the derivative is zero at vertex of the graph.

dy/dx = -6x -2 Set it to zero. -6x - 2 = 0 x = -1/3

Problem three:

"Solve: 2x{sq} + 1 = 5x"

2x^2 -5x + 1 = 0

You have to use quadratic formula for this. I don't remember quadratic formula anymore. It has been awhile.

I think it is probably something like (-b +- sqrt(b^2 - 4ac))/(2a)

Joeman
06-04-02, 05:02 PM
Originally posted by Xev
Sorry Ty, I can't get the second one. Working on the third now, Joeman, thanks, that should help.

Xev,

Since you are taking SAT soon, I hope you can do the second one. The derivative means the (microscopic) slope or change at that point. At vertex the slope is zero. The slope is delta_dy/delta_dx = 0. So you take derivative for y, set it to zero, and solve for x. Wait, is it first or second directive? First directive is the slope, and second is the rate of change. I think it is the first. Man it has been at least 6 years since I see this stuff. I hope I don't screw up :D Maybe wait for James R to show up just to be on the safe side :D

Edit to add. I can't believe I actually got quadratic formula right :p

Tyler
06-04-02, 05:12 PM
Thank you very much guys!!!!!!!! I owe you both!!!!

Here's another one I got stuck with. I got an answer but it fits under 'None of the above' so I need to check if it's right!

"Louis stands at a point P as shown in the diagram below. He wants to find the height of the tallest building in his city. He is standing 166 m away from the building. There is a tree 31 m in front of him, which he knows is 20 m tall. How tall is the building, to the nearest metre?"

.................................................. ..B
.................................................. ..B
....................T............................. .B
P..................T.............................. B

(There should be a line showing that P seeing the building {B} looks directly above the tree {T})

Joeman
06-04-02, 05:24 PM
Tyler,

Now you have two similar triangles right?

tangent(angle) = height of tree / distance of tree

tan(theta) = 20/31 => theta = tan-1 (20/31) = 33.69 degrees

Tan(33.69 degrees) = height of buildings / distance of buildings.

Height of buildings = Tan(33.69) * 166 = 110 m

hope that helps.

Tyler
06-04-02, 06:19 PM
Hmmmmm, okay I see where I missed up. For some reason I figured out 20/31 first.

Thanks again!