Mathematics
10 Online
OpenStudy (anonymous):
Identify the Exponent Law shown in the example below.
(ab)3 = a3b3
Power of a Product
Product of Powers
Power of a Power
None of the Above
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jimthompson5910 (jim_thompson5910):
(...)^3 is a power
a*b is a product
jimthompson5910 (jim_thompson5910):
combine the two, in the right order, to say that (a*b)^3 is a power of a product
OpenStudy (anonymous):
thank you so much :)
jimthompson5910 (jim_thompson5910):
you're welcome
OpenStudy (anonymous):
can you help me with more or do I have to close this one and put another question up?
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jimthompson5910 (jim_thompson5910):
I don't mind either way
jimthompson5910 (jim_thompson5910):
do what works best for you
OpenStudy (anonymous):
(5x4y2z2)(3x4y3z5).
15x16y6z10
15x8y5z7
8x8y5z7
8x16y6z10
wouldn't it be c?
jimthompson5910 (jim_thompson5910):
5*3 isn't 8, it's 15
jimthompson5910 (jim_thompson5910):
only the exponents add, the coefficients multiply
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OpenStudy (anonymous):
so B
OpenStudy (anonymous):
No its A
jimthompson5910 (jim_thompson5910):
exponent for x: 4+4 = 8
exponent for y: 2+3 = 5
exponent for z: 2+5 = 7
jimthompson5910 (jim_thompson5910):
hmm decisions decisions...A or B
OpenStudy (anonymous):
its B
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jimthompson5910 (jim_thompson5910):
good
OpenStudy (anonymous):
do you do the same thing for this?
Choose the correct simplification of the expression (2ab5)3.
8ab15
8a3b8
5a3b15
8a3b15
jimthompson5910 (jim_thompson5910):
you would use the rule given at the top of the page
jimthompson5910 (jim_thompson5910):
(a*b)^c = a^c * b^c
jimthompson5910 (jim_thompson5910):
so this means that
(2ab^5)^3 = 2^3 * a^3 * (b^5)^3 = ???
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OpenStudy (anonymous):
so D...?
jimthompson5910 (jim_thompson5910):
perfect
jimthompson5910 (jim_thompson5910):
you're a pro at this already
OpenStudy (anonymous):
thank you :)
jimthompson5910 (jim_thompson5910):
sure thing
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OpenStudy (anonymous):
im not sure about this one
Choose the correct simplification of the expression (3xy4)2(y2)3.
6x2y14
9x2y14
9x3y11
6x3y11
jimthompson5910 (jim_thompson5910):
simplify (3xy^4)^2 first and tell me what you get
OpenStudy (anonymous):
7xy8?
jimthompson5910 (jim_thompson5910):
hmm no unfortunately
jimthompson5910 (jim_thompson5910):
(3xy^4)^2 = 3^2 * x^2 * (y^4)^2
(3xy^4)^2 = 9 * x^2 * y^8
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jimthompson5910 (jim_thompson5910):
what do you get when you simplify (y^2)^3
OpenStudy (anonymous):
y16
jimthompson5910 (jim_thompson5910):
y^6 so you are close
jimthompson5910 (jim_thompson5910):
(3xy^4)^2(y2)^3 becomes 9x^2*y^8*y^6
now simplify 9x^2*y^8*y^6
OpenStudy (anonymous):
18xy16y12.. I'm doing this wrong.. :/
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jimthompson5910 (jim_thompson5910):
all you do at this point is add the y exponents
jimthompson5910 (jim_thompson5910):
8+6 = 14
OpenStudy (anonymous):
so 18xy14
jimthompson5910 (jim_thompson5910):
so the final answer is \[\large 9x^{2}y^{14}\]
jimthompson5910 (jim_thompson5910):
idk how you're getting 18
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OpenStudy (anonymous):
idk either lol
jimthompson5910 (jim_thompson5910):
hmm
jimthompson5910 (jim_thompson5910):
well that 2 is an exponent and you're not multiplying it by 9 to get 18
OpenStudy (anonymous):
so what do you do with it?
jimthompson5910 (jim_thompson5910):
you just leave the 9 as it is
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OpenStudy (anonymous):
okay thank you
jimthompson5910 (jim_thompson5910):
np
OpenStudy (anonymous):
what about this? is it D or B?
Choose the correct simplification of the expression (−13g2h5k2)0.
1
0
13g2h5k2
−13
jimthompson5910 (jim_thompson5910):
0 is the exponent right?
OpenStudy (anonymous):
yes
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jimthompson5910 (jim_thompson5910):
anything to the 0th power (except 0 itself) is always 1
jimthompson5910 (jim_thompson5910):
ex: 2^0 = 1
9^0 = 1
115^0 = 1
you get the idea
jimthompson5910 (jim_thompson5910):
so if x is not zero, then x^0 = 1
OpenStudy (anonymous):
ohh okay
jimthompson5910 (jim_thompson5910):
so you could have some really long and really ridiculous expression raised to the 0th power....and it would all boil down to 1
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OpenStudy (anonymous):
I don't feel like doing that lol
jimthompson5910 (jim_thompson5910):
yeah just know that (−13g^2h^5k^2)^0 = 1 and you're good to go
OpenStudy (anonymous):
thank you!
jimthompson5910 (jim_thompson5910):
yw glad to be of help