you are correct only on #1, 4, 6, 11
yikes
#2. anything to the zero power is 1. Consider
(-8.3)^2 / (-8.3)^2 = (-8.3)^(2-2) = (-8.3)^0 = 1
#3. moving powers from top to bottom (or bottom to top) changes the sign of the exponent.
1/a^2 = a^-2
1/a^-2 = 1/(1/a^2)) = a^2
So, -1/a^-2 = -1 * a^2 = -a^2
#5. (3)^10 (3)^-9 = 3^10 / 3^9 = 3
you're getting confused with 3^2 * 2^2 = (3*2)^2 = 6^2
if the bases are the same add exponents
if the powers are the same, combine bases
#7. (6q^6)^-4
= 6^-4 * (q^6)^-4 = 1/6^4 & q^-24 = 1/(1296 q^24)
NOT as you wrote it, which is (1/1296) q^24
#8. 11x^-3 y^-1 for x= -1 and y=2
= 11/(x^3 y) = 11/((-1^3)*2) = 11/-2 = -11/2
#9. (t^-4)^-9 t^2 = t^(-4 * -9) t^2 = t^36 t^2 = t^38
how ever did you get -26?
#10. Not sure on this one. As written, you have
(4/7y^5)^2 = ((4/7)(y^5))^2 = (16/49) y^10
But, as you probably meant,
(4/(7y^5))^2 = 4^2/(7y^5)^2 = 16/(49y^10)
Sometimes they look complicated, but they're just a set of easy steps.
i need help. not saying these are right, i just tried my best, could someone check them for me and just tell me the letters for the correct ones? (NOT ALL ARE ANSWERED)
1. (-5)^-1
A. - 1/5 **********
B. -5
C. 5
D. 1/5
2. (-8.3) ^0
A. 1
B. -1
C. 0
D. -8.3 **********
3. -1/a^-2
A. -a^2
B. -2/a
C.-2a
D. a^2 **********
4. 4^6 * 4^3
A. 4^9 **********
B. 4^18
C. 12^9
D. 64^9
5. (3)^10 (3)^-9
A. 3
B. 1/3
C. 9^-90 **********
D. 9
6. (y^6)^2
A. 2y^12
B. y^12 **********
C. y^8
D. y^36
7. (6q^6)^-4
A. 6q^1296
B. 1296q^2
C. 1/1296 q^24
D. 6q^-24 **********
8. What is the value of 11x^-3 y^-1 for x= -1 and y=2
A. -2/11
B. -22
C. -66
D. -11/2
(yikes i dont know this one...)
9. (t^-4)^-9 t^2
A. t^26
B. t^-26 **********
C. t^38
D. t^-38
10. (4/7y^5) ^2
A. 8 / 14y^10
B. 16/ 49 y^25
C. 16/9^10
D. 8/14 y^25
11. Determine if the number 3.14*10^-6 is written in scientific notation. If not, explain.
A. Yes; the number is written in scientific notation **********
B. No; the first number is not a number between 1 and 10
C. No; it is not written as a number times a power of 10
There is 23 questions overall but I will submit in the rest once I try to finish and see if anyone is able to help me here.
1 answer