Hey everyone, can someone please show me the steps i'd need to take in order to get the answers to the following questions. Thanks! I already have the answers, but can you shoe me the steps.
Simplify
a) (-27x^3)1/3 = -9x^3 = -3x
b) (-16x^4)-1/2 = no solution
c) (27a3^b^3)2/3 = (9a3b3)2 = 9a2b2
d) (49c^6)-3/2 = (1/343c^9)
Simplify
a) (-27x^3)1/3 = -9x^3 = -3x
b) (-16x^4)-1/2 = no solution
c) (27a3^b^3)2/3 = (9a3b3)2 = 9a2b2
d) (49c^6)-3/2 = (1/343c^9)
-
a)
(-27x³)⅓ = (-27)⅓ (x³)⅓ = ((-3)³)⅓ (x³)⅓
= (-3)³ˣ⅓ (x)³ˣ⅓ = (-3)¹ (x)¹ = -3x
b)
16x⁴ is always positive. So -16x⁴ is always negative and √(-16x⁴) can not be simplified.
c)
(27a³b³)⅔ = (27)⅔ (a³)⅔ (b³)⅔ = (3³)⅔ (a³)⅔ (b³)⅔
= 3²a²b² = 9a²b²
d)
(49c⁶)ˉ³/² = (49)ˉ³/²(c⁶)ˉ³/² = (7²)ˉ³/² (c⁶)ˉ³/² = 1/[(7²)³/² (c⁶)³/²]
= 1/[7³ c⁹] = 1/343c⁹
(-27x³)⅓ = (-27)⅓ (x³)⅓ = ((-3)³)⅓ (x³)⅓
= (-3)³ˣ⅓ (x)³ˣ⅓ = (-3)¹ (x)¹ = -3x
b)
16x⁴ is always positive. So -16x⁴ is always negative and √(-16x⁴) can not be simplified.
c)
(27a³b³)⅔ = (27)⅔ (a³)⅔ (b³)⅔ = (3³)⅔ (a³)⅔ (b³)⅔
= 3²a²b² = 9a²b²
d)
(49c⁶)ˉ³/² = (49)ˉ³/²(c⁶)ˉ³/² = (7²)ˉ³/² (c⁶)ˉ³/² = 1/[(7²)³/² (c⁶)³/²]
= 1/[7³ c⁹] = 1/343c⁹
-
OK.
a). Because of the laws of roots you can take the cube root of -27 and that is -3 and take the cube root of x^3 which is x and multiply them together to get -3x. Your intermediate step above is not correct.
b. this is the same as 1/sqrt(-16*x^4). This can be broken down into (1/sqrt(-16))*1/sqrt(x^4).
Now, sqrt of -16 is +/-4i and sqrt of x^4 is x^2 so your answer would be = 1/(x^2*4i). This involves complex number i which you may not have had yet. i = sqrt(-1).
c. Ok--first, break each part down and take the cube root to get 3ab. Now square this quantity to
get 9a^2b^2.
d. Ok---first take and put everything in the denominator so the exponent will now be +.
Then take the sqrt of each part to get 7c^3. Now cube this result to get 1/343*c^9
a). Because of the laws of roots you can take the cube root of -27 and that is -3 and take the cube root of x^3 which is x and multiply them together to get -3x. Your intermediate step above is not correct.
b. this is the same as 1/sqrt(-16*x^4). This can be broken down into (1/sqrt(-16))*1/sqrt(x^4).
Now, sqrt of -16 is +/-4i and sqrt of x^4 is x^2 so your answer would be = 1/(x^2*4i). This involves complex number i which you may not have had yet. i = sqrt(-1).
c. Ok--first, break each part down and take the cube root to get 3ab. Now square this quantity to
get 9a^2b^2.
d. Ok---first take and put everything in the denominator so the exponent will now be +.
Then take the sqrt of each part to get 7c^3. Now cube this result to get 1/343*c^9
-
a) (-27x^3)^(1/3) = -3x
b) (-16x^4)^(-1/2) = cant take the square root of a negative number = no solution
c) (27a3^b^3)2/3 = (9ab)2 = 9a2b2
d) (49c^6)^(-3/2) = ( 7c^3)^(-3) = (343c^9)^(-1) = 1 / (343c^9)
If you care to skype and ask questions on these steps I am a fantastic math tutor.
If you have questions on this or any other math problem feel free
to let me know how to contact you.......Read Below.......
I tutor free online since moving to France from Florida in June 2011.
Hope this is most helpful and you will continue to allow me to tutor you.
If you dont fully understand or have a question PLEASE let me know
how to contact you so you can then contact me directly with any math questions.
If you need me for future tutoring I am not allowed to give out my
info but what you tell me on ***how to contact you*** is up to you
Been a pleasure to serve you Please call again
Robert Jones.............f
"Teacher/Tutor of Fine Students"
b) (-16x^4)^(-1/2) = cant take the square root of a negative number = no solution
c) (27a3^b^3)2/3 = (9ab)2 = 9a2b2
d) (49c^6)^(-3/2) = ( 7c^3)^(-3) = (343c^9)^(-1) = 1 / (343c^9)
If you care to skype and ask questions on these steps I am a fantastic math tutor.
If you have questions on this or any other math problem feel free
to let me know how to contact you.......Read Below.......
I tutor free online since moving to France from Florida in June 2011.
Hope this is most helpful and you will continue to allow me to tutor you.
If you dont fully understand or have a question PLEASE let me know
how to contact you so you can then contact me directly with any math questions.
If you need me for future tutoring I am not allowed to give out my
info but what you tell me on ***how to contact you*** is up to you
Been a pleasure to serve you Please call again
Robert Jones.............f
"Teacher/Tutor of Fine Students"