Are these correct?

Question

This is my test: (please tell me if these are correct, and then what would I get in the test according with my answers)
The colon ( : ) means division
The star ( * ) means multiplication
The v ( ^ ) means exponent. 4^5 would be four to the power of five.
Here goes:

8(2-5)^3 : 6^2
I got -6.

(7/8 * 3/5) : (1/5 - 1/2)
I got -7/4.

3/5 * (2-1/3) + 1/6 : 1/2
I got 4/3.

Simplify, don't calculate: (2 + 1/3)^-2 * 3^-2
I got 7-2.

(5/2 - 5/6 + 2/3 * 1/3) : [2 - 1/2 * (1 + 5/3)]
I got 51/18.

Simplify don't calculate: (6^2 * 3^2) : [2^3 * (-3)^2 * 4^2]
I got (3^2) : (2^5).

Thanks in advance :)

Answer 1

You did very well

8(2-5)^3 / 6^2
=8*-3^3 / 6^2
=-6

(7/8 * 3/5) / (1/5 - 1/2)
=21/40 / -3/10
= -21/40 *10/3
=-7/4

3/5 * (2-1/3) + 1/6 : 1/2
= 3/5*5/3 + 1/6*2/1
= 1 + 1/3
= 4/3

(2 + 1/3)^-2 * 3^-2
=(7/3)^-2 * 3^-2
= 1/(7/3)^2 * 1/(3)^2
= 3^2/7^2 * 1/3^2
= 1/7^2
= 7^-2

(5/2 - 5/6 + 2/3 * 1/3) : [2 - 1/2 * (1 + 5/3)]
=(5/2 -5/6 + 2/9)/(2-1/2 * 8/3)
= (34/18*3/2
= 17/6

(6^2 * 3^2) : [2^3 * (-3)^2 * 4^2]
=(3^2*2^2*3^2)/(2^3*3^2*2^4)
= 3^2/2^5

Answer 2

8(2-5)^3 : 6^2
8(-3)^3 : 6^2
(I assume that only the -3 is raised to third power)
8(-1*-3)^3 : (3*2)^2
( -1 raised to an odd power is -1)
8*(-1)*(3^3) : (3^2)*(2^2)

I use 8 = 2^3 = 2*(2^2) and 3^3 = 3*(3^2)
I rearrange the order or the factors (does not change the product):

(-1)*(2)*(3) (3^2)(2^2) : (3^2)(2^2)

The (3^2)(2^2) : (3^2)(2^2) portion is equal to 1 (i.e., it cancels out), leaving
(-1)*(2)*(3)

----
(7/8 * 3/5) : (1/5 - 1/2)

Use (7/8 * 3/5) = 21/40 and
(1/5 - 1/2) = (8/40 - 20/40) = -12/40
This leaves you with:

(21/40) : (-12/40)

The 40s cancel out, and you are left with simplifying
-(21/12)

-----
3/5 * (2-1/3) + 1/6 : 1/2
(3/5)*(5/3) + 1/6 : 1/2

I assume that the entire left side is divided by 1/2.
Dividing by 1/2 is the same as multiplying by 2
2*[ (3/5)*(5/3) + 1/6 ]
2*[ (15/15) + 1/6 ]
2*[ 1 + 1/6 ]
2*(7/6) = 14/6

However, if my assumption is wrong, my answer will also be wrong.

-----
(2 + 1/3)^-2 * 3^-2
(7/3)^-2 * 3^-2

A negative exponent is equivalent to a fraction (e.g., a^-1 = 1/a and a^-2 = (1/a)^2 = 1/(a^2)

(3/7)^2 * (1/3)^2

Same exponent, I can move the bases inside the exponent (the product of two squares is equal to the square of the products -- a^2 * b^2 = (a*b)^2 )

[(3/7)*(1/3)]^2
(1/7)^2
7^-2

-----
(5/2 - 5/6 + 2/3 * 1/3) : [2 - 1/2 * (1 + 5/3)]
(5/2 - 5/6 + 2/9) : (2 - 1/2 - 5/6)
Put everything over 18:
(45/18 - 15/18 +4/18) : (36/18 - 9/18 - 15/18)
(34/18) : (12/18)

The 18s cancel, leaving you with 34/12 = 17/6

(17/6)*(3/3) = (51/18)

51/18 is OK for a value, but if the question asks to simplify, you can still make it simpler (17/6).

-------

Use (6^2) = (2*3)^2 = (2^2)*(3^2)
and (4^2) = (2*2)^2 = (2^2)*(2^2) = 2^4
and (2^3) = 2*(2^2), and
(-3)^2 = 9 = (3^2) (an even power is always positive)

(6^2 * 3^2) : [2^3 * (-3)^2 * 4^2]
(2^2)*(3^2)*(3^2) : 2*(2^2)*(3^2)*(2^4)

The (2^2) and one (3^2) cancel, leaving
(3^2) : 2*(2^4)
(3^2) : (2^5)

Answer 3

we had same ans in 1,2&3
I dont know how to simplify so 4&6 i'm not sure!
#5
(5/2 - 5/6 + 2/3 * 1/3) : [2 - 1/2 * (1 + 5/3)]
(5/2 - 5/6 + 2/6) : [2 - 1/2*(8/3)]
(10/6 - 5/6 + 2/6) : (2 - 4/3)
(7/6) : (2/3)
7/4