Help for my spec test sac

[ejf8888]

Hi,
I have my first spec test/SAC today and i just have a few last desperate questions that I'm not sure how to answer.

1. Solve 3x^2 - 4x - 5= cosx

2. How many intercepts will the graph of y=x(x +9)^2(x^2 - 9) have?

3. Find the equations of the asymptotes for the hyperbola with rule x^2 - ((y-2)^2)/9 = 15.

4. The triangle base ABC of a tetrahedron has side lengths AB = 15cm, BC = 12cm and AC = 9cm. If the apex D is 9cm vertically above C, then find the angles that the sloping edges make with the horizontal.

I know it is EXTREMELY early, so any help with any questions will be VERY greatly appreciated.

Thankyou so much in advance!

[pHysiX]

Qu1:
Hmmm....I'm lost with this one....I'd say just graph it on your gfx calc and find the intersects and hence the solutions

I got x=-0.86 and x=2.06

Qu2:
let y=0

so x=0 or (x+9)^2 =0 or (x^2)-9=0
so x=0 or x=-9 or x=3 or x=-3

Also, when x=0, y=0

Hence, 4 intercepts including the y-intercept =]

Qu3:
Divide both sides by 15 so that u get a hyperbola
(x^2)/15 - ((y-2)^2)/135 = 1

equation for the asymptotes is (y-k)=+ or - (b/a)(x-h)
where in this case h=0, k=2, a=SQRT(15), b=SQRT(135)

so after all the working, it's y=3x+2 and y=-3x+2

Qu4:
I'm not sure about this one coz i really hate these ones. But here's my guess:

From recognition, we know that triangle ABC is a right-angled triangle with the right angle at C.
So if you think about it, let the line AC and triangle ABC lie on the horizontal plane. (Note that BC is also a horizontal for later).
Now, D is directly above C, so when you draw it out, we get a right-angled isosceles triangle ACD and a right-angled triangle BCD.

The slope is line AD and the angle is Angle DAC in essence; because remember, the line AC was the horizontal and the point D is directly above C, so that is a vertical/perpendicular plane, if that makes any sense at all. Try getting some sticks and make a rough model. You'll understand more.

Anyway, since the sides AC=CD=9cm, and Angle ACD is 90 degrees, this means that Angle DAC is 45 degrees.

The other slope is line BD and the angle is Angle DBC. We have dimensions BC=12cm and CD=9cm. Applying trig, the angle is tan^-1(9/12)=29.76 degrees

Therefore, the angles made by the sloping edges are 29.76 and 45 degrees to the horizontal. =]