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enyata [817]
3 years ago
12

How do you solve the equation:

Mathematics
1 answer:
Molodets [167]3 years ago
5 0
Copy and paste the answer on quizlet
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I need help please help
monitta

Answer:

with wut XD

Step-by-step explanation:

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3 years ago
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(2^8 x 3^−5 x 6^0)−2 x 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 x 2^28
Veronika [31]

(2^8\cdot3^{-5}\cdot6^0)^{-2}\cdot\left(\dfrac{3^{-2}}{2^3}\right)^4\cdot2^{28}\\\\=(2^8)^{-2}\cdot(3^{-5})^{-2}\cdot1^{-2}\cdot\dfrac{(3^{-2})^4}{(2^3)^4}\cdot2^{28}\\\\=2^{-16}\cdot3^{10}\cdot\dfrac{3^{-8}}{2^{12}}\cdot2^{28}\\\\=2^{-16}\cdot2^{28}\cdot2^{-12}\cdot3^{10}\cdot3^{-8}\\\\=2^{-16+28+(-12)}\cdot3^{10+(-8)}\\\\=2^0\cdot3^2=1\cdot9=\boxed{9}\\\\Used:\\\\(a\cdot b)^n=a^n\cdot b^n\\\\(a^n)^m=a^{n\cdot m}\\\\a^n\cdot a^m=a^{n+m}\\\\a^{-n}=\dfrac{1}{a^n}

4 0
4 years ago
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elena55 [62]

Answer:

question 12 is answer B

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Step-by-step explanation:

exponent rule \frac{a^m}{a^n}  = a^(^m^-^n^)

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What is 7.436 rounded to the nearest tenth
Dmitriy789 [7]
It's 7.4 because the 3 doesn't round the 4 up.
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