anonymous
  • anonymous
There are (3^2)^4 ⋅ 3^0 bacteria in a Petri dish. What is the total number of bacteria in the dish?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@iGreen
iGreen
  • iGreen
Simplify the exponent inside the parenthesis. \(\sf 3^2=~?\)
anonymous
  • anonymous
9

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iGreen
  • iGreen
Correct, now simplify: \(\sf (9)^4\)
anonymous
  • anonymous
6561?
iGreen
  • iGreen
Yep! And what's \(\sf 3^0\)?
anonymous
  • anonymous
1
iGreen
  • iGreen
Correct, that leaves us with: \(\sf 6561\times 1\)
anonymous
  • anonymous
6561
phi
  • phi
they may want the answer as \[ 3^8\] you can find this answer by knowing something to the 4th power means multiply by itself four times, so \[ \left(3^2\right)^4= 3^2\cdot 3^2 \cdot 3^2\cdot 3^2\] you should know \( 3^2= 3 \cdot 3\) so you can also write this as \[ 3^2\cdot 3^2 \cdot 3^2\cdot 3^2= 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \] use the "short way" to show you have 3 multiplied by itself 8 times
phi
  • phi
in other words \[ 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3=3^8\]
anonymous
  • anonymous
okay thank you that is what my question asked for
iGreen
  • iGreen
Yes, @phi is just explaining another way to do it. \(\sf 3^8\) also gives us the answer of 6561
iGreen
  • iGreen
Nice work
anonymous
  • anonymous
ok thanks

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