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n200080 [17]
2 years ago
14

Please help meeee!!!!The domain (input values) of the sine function is:

Mathematics
2 answers:
bogdanovich [222]2 years ago
3 0

Answer:

the domain of the function y=sin(x) ) is all real numbers (sine is defined for any angle measure), the range is −1≤y≤1 .

Step-by-step explanation:

Alexeev081 [22]2 years ago
3 0
Your answer is choice C
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Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol
koban [17]

Answer:

\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}

Step-by-step explanation:

\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228

=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}

\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}

4 0
2 years ago
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Answer:

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Butoxors [25]
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3 years ago
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