TMSCA Practice Tests

[nerualmusic]

hello, does anybody happen to have TMSCA practice tests?

if so, help would greatly appreciated on these questions:

test #6 (december 6, 2008)
09C-18
09C-26
09C-27
09C-46 <- priority, please
09C-57
09C-60

test #7 (january 17, 2009)
09O-26
09O-36
09O-38
09O-46
09O-57
09O-60

yeah...I don't know anything. any answers would help. thank you!

[nsguy1350]

16. Radius of Earth ~ 3960 mi --> Radius of Moon ~ 990 mi
If his eyes are 4 in below the top of his head, and he is 6 feet tall, then his eyes are 5 ft 8 in off the ground.
Set up a right triangle, using the following information:
One leg is the horizon distance, what you're trying to find.
One leg is the radius of the moon.
And the hypotenuse is the radius of the moon plus his height.
You should draw the moon as well, making r vertical, and the radius plus the height as the slanted hypotenuse, with it barely coming off the circle.
Therefore, letting r be the radius of the moon, d be the horizon distance, and h be his height,
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula]
[unparseable or potentially dangerous latex formula] [unparseable or potentially dangerous latex formula].

[nsguy1350]

26. Know that [unparseable or potentially dangerous latex formula]. Faster than 89 conversions, but either way is ok I guess.
Find the volume of the cylinder --- [unparseable or potentially dangerous latex formula].
You have a TI-89, so convert in cubed to ft cubed to cm cubed. Then, you have around 7100 cm cubed worth of water, or 7.1 kg, approximately.
Convert that to pounds to get 15.64 lb, then add the dry weight of the hose, 8.2 lbs, to get [unparseable or potentially dangerous latex formula].

[nsguy1350]

46. A b-low deals with volume. That's cubic. Area is quadratic.

[unparseable or potentially dangerous latex formula].

[nsguy1350]

57. You're maximizing area. Your formula for area is [unparseable or potentially dangerous latex formula]. The 2x is because you're bounded by two sides of a semicircle symmetrical around the y-axis, and x only will give you half of the length of the rectangle (a picture might help).
You have the area formula in one variable. Take the derivative and find the maximum.
The maximum occurs at x = 9.617692. On the TI-89, I used the d(f(x),x) function, then used: solve(ans(1) = 0, x) to find where it equaled zero. One positive result came up.
Plugging that back into the y equation to find height, you see that the height is actually the same as x. Answer: [unparseable or potentially dangerous latex formula].

[nsguy1350]

Test 7 -
26. Use the distance formula and set it equal to two.
[unparseable or potentially dangerous latex formula].
[unparseable or potentially dangerous latex formula]

[nsguy1350]

Test 7 -
38. I'd set it up like this:
[unparseable or potentially dangerous latex formula].
12% concentration unknown solution times its volume plus 35% concentration of 120 mL solution equals 25% concentrated solution with a volume of the unknown volume plus 120.
[unparseable or potentially dangerous latex formula].

[nsguy1350]

46. Hmm... volume is porportional to length cubed.
[unparseable or potentially dangerous latex formula], since the density is 6 times as much in the statue. Convert units, and solve for [unparseable or potentially dangerous latex formula]