Please Help!! Im stuck.?

Question

Im trying to get this practice quiz thing to get me ready for this placement this for CSU. Test is coming REAL SOON. And trying to practice as much as I can. The test is Mathematics Placement Test if anyone was interested for Engineering majors.

Here are a few I got stuck on


NOTE- < and > are all underlined, i just didnt know how to put it.
1. |x-2| < 9 is equivalent to which one?
a. -11 < x < 11
b. -7 < x < 11
c. -11 < x < 7
d. x < -7
e. x > 11

Please show steps to this one. Just not letter answer.


2. (2x-1)(x+4) / x-3 = 0 x= ???

3. If f(x) = sin (3x) then f(pie/6) = ???

4. What is the area of this triange? ( link to picture)
http://img380.imageshack.us/img380/1809/1f14ev8.gif
I think it was 0.3

YES, SORRY I MEANT RECTANGLE FOR #4

Answer 1

1.
|x-2| <= 9

implies -9 <= (x-2) <=9

implies -9+2 <= x <= 9+2

implies -7 <= x <= 11

hence b



2. (2x-1)(x+4) / (x-3) = 0

if there is paranthesis in the denominator

(2x-1)(x+4) =0

(2x-1) =0 (x+4) =0

x = (1/2) , x = -4


3. f(pi/6) = sin (3(pi/6)) = sin (pi/2) =1

4. if you mean the rectangle , (and the base line is the x-axis)

l = .8 - .5 = .3

b = .5^2 + 2*.5 + 3

area will be l*b

Answer 2

1. |x-2| < 9 can be rewritten as two parts:

x - 2 > -9 and x-2 < 9 Think about this. The left one is because of the absolute value.

so x>-7 and x<11 and the answer is b.

2. (2x - 1)(x + 4)/x - 3 = 0
2x^2 + 7x - 4 - 3x = 0
2x^2 + 4x -4 = 0
x^2 + 2x - 2 = 0

and x = (-2 +/- SQRT(4 + 8))/2 = -1 +/- SQRT(3)

3. = sin (3*(pi/6)) = sin(pi/2) = 1 x= pi/6 so just substitute it into the equation. and it helps to know that pi/2 is 90 degrees

4. Do you mean rectangle?
y = (.5)^2 + 2(.5) + 3 = 4.25 This is the height of the rectangle

Area = height times width
so the area of the rectangle is (.8 - .5)*(4.25) = 1.275

Answer 3

1. when x < 2 => 2-x < 9 => -x<9-2 => x > -7
when x > 2 => x-2<9 => x< 11
from those 2 we have -7
3. f(pie/6)= sin3*pie/6= sin(pie/2) = 1

2. (2x^2+8x-x-4)/x-3 =0 ax^2+bx+c=0
2x^2+7x-4 = 0 delta= b^2-4ac
delta= 49+32=81 x1,2= (-b+/-radical delta)/2a
x1= (-7+9)/4=> x= 1/2
x2= (-7-9)/4=> x= -4

4. ? sorry

Answer 4

first answer is B.

-11-2= negative 13, but if you take the absolete value, which is 13, thats not less than or equal to 9 A IS WRONG

B. x is greater than or equal to -7 and less than or equal to 11

ok well negative 7-2 =negative nine and the absolute value is 9 so ok, and 11-2 is =9 so the range is good

c. we know -11 is no good as per a

d. we know is wrong

e. we know is wrong.


SORRY ONLY HAVE TIME FOR ONE.


ATESHOBE's math should not be followed!!