Solve and type in a different form by using the given theorems of logarithms

Mathematics

1 answer:

scoray [572]3 years ago

4 0

$\log\left(\frac{10}{y}\right) = \log(10) - \log(y)$

The general rule is

$\log\left(\frac{A}{B}\right) = \log(A) - \log(B)$

Other log rules are

$\log(AB) = \log(A) + \log(B)$

$B\log(A) = \log(A^B)$

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Is 126 divisible by 3

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Answer:

yes

Step-by-step explanation:

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Reduce the following fraction: -36x^4y^4z^5/-12x^6y^3z

Evgesh-ka [11]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\$

$\bf \cfrac{-36x^4y^4z^5}{-12x^6y^3z}\quad 
\begin{cases}
36=2\cdot 2\cdot 3\cdot 3\\
12=2\cdot 2\cdot 3
\end{cases}\implies \cfrac{-36}{-12}\cdot \cfrac{x^4y^4z^5}{x^6y^3z^1}
\\\\\\
\cfrac{\underline{-2\cdot 2\cdot 3}\cdot 3}{\underline{-2\cdot 2\cdot 3}}\cdot \cfrac{x^4y^4z^5x^{-6}y^{-3}z^{-1}}{1}\implies \cfrac{3}{1}\cdot 
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$\bf \cfrac{3}{1}\cdot \cfrac{x^{4-6}y^{4-3}z^{5-1}}{1}\implies 
\cfrac{3}{1}\cdot\cfrac{x^{-2}y^{1}z^4}{1}\implies 
\cfrac{3x^{-2}yz^4}{1}\implies \cfrac{3yz^4}{x^2}$

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In equilateral △ABC, y = ?

Nina [5.8K]

Well since it’s equilateral then we know that x+3x+2x=180/3=6
So 6x=6 x=6
Angle ACB is gonna be 60 degrees (bc it’s equilateral)
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