Posted by mysterychicken on Saturday, June 1, 2013 at 3:00pm.
Can I please get some help on these questions:
1. If , find the value of 2 square root of 17x = 2x10, find the value of x/4.
2. If absolute value of 3x+6 < 9, then which of the following is a possible value of x ?
My answer choices are 7, 1, 5, 1, 10. When I solve for x, I get x<1, so can't it be 7, 5, OR 1?
3. If 5<absolute value of 2x+1<11, then which of the following is a possible value of x ?
I get 2<x<5, My answer choices are 6, 4, 2, 3, and 5.
4. If y is inversely proportional to x and y = 8 when x = 1.55, what is the value of x when y = 0.62?
I'm not sure on this
5. Simplify the following expression: 41abc(1 + z)41abcz
...
Thanks to anyone who can help me!

SAT Math  mysterychicken, Saturday, June 1, 2013 at 3:02pm
#1 is worded wrong, if 2 square root of 17x = 2x10, find the value of x/4.

SAT Math  Steve, Saturday, June 1, 2013 at 3:16pm
#1 Can't quite parse the problem, but if
2√(17x) = 2x10
√(17x) = x  5
17x = x^2  10x + 25
x^2  9x + 8 = 0
(x1)(x8) = 0
x = 1 or 8
but x=1 does not fit the original equation, so x=8, and x/4 = 2
#2 Recall that n = n if n>=0 and n if n<0
3x+6<9
if 3x+6>=0, then
3x+6 < 9
x < 1
If 3x+6 < 0,
(3x+6) < 9
3x  6 < 9
3x < 15
x > 5
so 5 < x < 1
Looks like 1 is the only choice.
Think of the graph of x. It is a Vshape. So, if x < n you get the pointy part of the v below the line y=n.
#3
5 < 2x+1 < 11
This is just a sneaky way of getting you to solve two inequalities in one problem. You end up with
6 < x < 3 or 2 < x < 5
Looks like 4 is the only choice.
#4
y = k/x
8 = k/1.55
k = 12.4
so, when y = .62
.62 = 12.4/x
x = 20
#5
41abc(1 + z)41abcz
ever heard of the distributive property?
41abc(1) + 41abc(z)  41abcz
41abc + 41abcz  41abcz
41abc 
SAT Math  mysterychicken, Saturday, June 1, 2013 at 3:46pm
Thank you!