Adi3
  • Adi3
pls help (64m^4)^3/2 and (25b^6)^-1.5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Adi3
  • Adi3
@mathstudent55
dayakar
  • dayakar
do u know exponential laws
Adi3
  • Adi3
ye, |dw:1446960219665:dw|

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More answers

dayakar
  • dayakar
change the terms in the square form
Adi3
  • Adi3
64m^4*3/2
Adi3
  • Adi3
qwerty, say something
dayakar
  • dayakar
|dw:1446960447168:dw|
Adi3
  • Adi3
what is that
dayakar
  • dayakar
8^2 =64
campbell_st
  • campbell_st
well the power of 3/2 applies to both parts \[(64m^4)^{\frac{3}{2}} = (8^2m^4)^{\frac{3}{2}}\] the power of a pwer index laws says to multiply the powers \[(a^xb^y)^z = x^{x \times z}b^{y \times z}\] apply this to your problem
Adi3
  • Adi3
|dw:1446960686962:dw|
Adi3
  • Adi3
@campbell_st i dont get it
campbell_st
  • campbell_st
ok... if you can't multiply the powers, perhaps you need to review your index laws
campbell_st
  • campbell_st
I'm not into giving answers... I was looking to help you understand...
campbell_st
  • campbell_st
the 2nd question... uses the same index law power of a power.... so multiply \[(25b^6)^{-1.5} = (5^2b^6)^{-1.5} = 5^{2 \times -1.5} b^{6 \times -1.5}\] just simplify
Adi3
  • Adi3
@campbell_st wait for a while
Adi3
  • Adi3
ohhh the answer is 0.008b^-9
campbell_st
  • campbell_st
or you can rewrite it with positive indices what does the negative indicate...?
campbell_st
  • campbell_st
I'd expect you would be asked to rewrite the problem with positive indices
Adi3
  • Adi3
the answer is 1/125b^9
Adi3
  • Adi3
is the wrong answer
mathstudent55
  • mathstudent55
\(\Large (64m^4)^{\frac{3}{2} }\) First, write 64 as a power. 64 = 2^6 since 2 * 2 * 2 * 2 * 2 * 2 = 64 So the problem is now: \(\Large =(2^6m^4)^{\frac{3}{2} }\) Now to raise a product to a power, raise each factor to the power. \(\Large =(2^6) ^{\frac{3}{2} } ( m^4)^{\frac{3}{2} }\) Now for each factor, to raise a power to a power, multiply powers: \(\Large =(2^{ 6 \times \frac{3}{2} }) ( m^{4 \times \frac{3}{2} })\) Now multiply out each exponent.
Adi3
  • Adi3
i get the part 1 but the part 2 answer is wrong
Adi3
  • Adi3
@imqwerty
Adi3
  • Adi3
|dw:1446961484278:dw|
imqwerty
  • imqwerty
ok can u write -1.5 in fraction form? like this -> p/q
Adi3
  • Adi3
3/2
imqwerty
  • imqwerty
3/2 is positive :) -1.5 is negative
Adi3
  • Adi3
-
Adi3
  • Adi3
3/2
imqwerty
  • imqwerty
yeah -3/2
Adi3
  • Adi3
aggee kuch bol
dayakar
  • dayakar
\[5^{2*(-3/2)} *b ^{6*(-3/2)}\]
Adi3
  • Adi3
i got the answer
dayakar
  • dayakar
\[5^{-3}*b ^{-9}\]
campbell_st
  • campbell_st
well if you know a negative power idicates a fraction then you have \[\frac{1}{(5^2)^{1.5}(b^6)^{1.5}}\] now multiply the powers... and you'll get your answer actually the answer you had was correct, but not in the form that the marker wanted
dayakar
  • dayakar
\[\frac{ 1 }{ 125b^{9} }\]
dayakar
  • dayakar
can u understand
campbell_st
  • campbell_st
there you go... someone always gives an answer
mathstudent55
  • mathstudent55
\(\Large (25b^6)^{-1.5} \) We can write 1/2 as a fraction: \(\Large =(5^2b^6)^{-\frac{3}{2}} \) Now we raise each factor to the power: \(\Large =(5^2)^{-\frac{3}{2}} ( b^6)^{-\frac{3}{2}} \) To raise a power to a power we multiply powers: \(\Large =[5^{2 \times (-\frac{3}{2})}] [ b^{6 \times (-\frac{3}{2})} ]\) \(\Large =(5^{-3}) ( b^{-9 } )\) Now we deal with the negative exponents: \(\Large = \dfrac{1}{5^3} \times \dfrac{1}{b^9} \) Finally, we simplify: \(\Large = \dfrac{1}{125b^9} \)

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