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Help Needed with Algebra
- Thread starter JohnRobbo
- Start date
Couple of tips.Really struggling with these questions would really appreciate some help,
answers need to be given with positive indices.
Question 1:
(0.2a^-2/3) ^3
Question 2:
3x^-3 (x-4x^2)
Queston 3:
(9x^8) ^-1/2
Thanks
.
(1) Please show your work even if you know it is wrong. It allows us to see where you are having trouble.
.
(2) In the future, please post one question per thread.
.
(3) Give us a clue about your situation. The answer appropriate for an 8th grader studying algebra for the first time will differ from the answer appropriate for a college student taking a review course.
.
(4) When posting, be very careful to follow PEMDAS so we can be sure we understand your problem. What your first problem involves as you have written it is:
.
\(\displaystyle \left(\dfrac{2a^{-2}}{3}\right)^3.\) But I suspect you mean:
.
\(\displaystyle \left(2a^{(-2/3)}\right)^3.\)
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Laws of exponents: \(\displaystyle a^xa^y= a^{x+ y}\) and \(\displaystyle (a^x)^y= a^{xy}\).
Couple of tips.
.
(1) Please show your work even if you know it is wrong. It allows us to see where you are having trouble.
.
(2) In the future, please post one question per thread.
.
(3) Give us a clue about your situation. The answer appropriate for an 8th grader studying algebra for the first time will differ from the answer appropriate for a college student taking a review course.
.
(4) When posting, be very careful to follow PEMDAS so we can be sure we understand your problem. What your first problem involves as you have written it is:
.
\(\displaystyle \left(\dfrac{2a^{-2}}{3}\right)^3.\) But I suspect you mean:
.
\(\displaystyle \left(2a^{(-2/3)}\right)^3.\)
had a go at each
Q1
1/125a^-2/27 and with positive indices = 1/125 / a^2/27.
Q2
3x^-2 - 12x^-1 and with positive indices = 3/x^2 - 12/x
Q3
1/3x^-4 and with positive indices = 1/3 / x^4
Dont know if they are correct could you help me if any are wrong
D
Deleted member 4993
Guest
Really struggling with these questions would really appreciate some help,
answers need to be given with positive indices.
Question 1:
(0.2a^-2/3) ^3
Is this \(\displaystyle \left ( 0.2a^{-\frac{2}{3}}\right )^3\)
Or \(\displaystyle \left ( \dfrac{0.2a^{-2}}{3}\right )^3\)
Question 2:
3x^-3 (x-4x^2)
Queston 3:
(9x^8) ^-1/2
Thanks
.
D
Deleted member 4993
Guest
\(\displaystyle \left [0.2a^{-\frac{2}{3}}\right ]^3\)
= \(\displaystyle (0.2)^3 \ * \ \left [a^{-\frac{2}{3}}\right ]^3\)
= \(\displaystyle (0.008) \ * \ \left [a^{-\frac{2}{3} \ * \ 3}\right ]\)
= \(\displaystyle (0.008) \ * \ a^{-2}\)
= \(\displaystyle \dfrac{1}{125a^2}\)
= \(\displaystyle (0.2)^3 \ * \ \left [a^{-\frac{2}{3}}\right ]^3\)
= \(\displaystyle (0.008) \ * \ \left [a^{-\frac{2}{3} \ * \ 3}\right ]\)
= \(\displaystyle (0.008) \ * \ a^{-2}\)
= \(\displaystyle \dfrac{1}{125a^2}\)
Last edited by a moderator:
.had a go at each
Q1
1/125a^-2/27 and with positive indices = 1/125 / a^2/27.
Q2
3x^-2 - 12x^-1 and with positive indices = 3/x^2 - 12/x No one has a clue what the second question is so please repost it correctly.
Q3
1/3x^-4 and with positive indices = 1/3 / x^4 This answer is correct AS FAR AS IT GOES \(\displaystyle \dfrac{1}{3} \div x^4 = \dfrac{1}{3} * \dfrac{1}{x^4} = \dfrac{1}{3x^4}.\)
Dont know if they are correct could you help me if any are wrong