How do I solve (|x-3|+5|x|) ?

Mathematics

1 answer:

11111nata11111 [884]3 years ago

6 0

Answer:

x≤0 ⇒|x-3|+5|x| = -6x + 3

0≤x≤3 ⇒ |x-3|+5|x| = 4x + 3

x≥3 ⇒ |x-3|+5|x| = 6x - 3

Step-by-step explanation:

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We were never taught this in class, and I understand it, kind of... but i'm just confused with everything.

GaryK [48]

$\bf \cfrac{x^3y}{xy^5}\cdot \cfrac{x^2y^9}{x^8}\implies \cfrac{x^3x^2y^1y^9}{x^1x^8y^5}
\\ \quad \\
\textit{now, using exponent rules, same base, different exponent}
\\ \quad \\
\cfrac{x^{3+2}y^{1+9}}{x^{1+8}y^5}\implies \cfrac{x^5y^{10}}{x^9y^5}\\
----------------------------\\$ $\bf \textit{now, keep in mind that}
\\ \quad \\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad

\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\\ \quad \\
% negative exponential denominator
a^{{ n}} \implies \cfrac{1}{a^{- n}}
\qquad \qquad 

\cfrac{1}{a^{- n}}\implies \cfrac{1}{\frac{1}{a^{ n}}}\implies a^{{ n}}
\\ \quad \\
thus\\
----------------------------\\
$ $\bf \cfrac{x^5y^{10}}{x^9y^5}\implies \cfrac{x^5y^{10}}{1}\cdot \cfrac{1}{x^9}\cdot \cfrac{1}{y^5}\implies x^5y^{10}\cdot x^{\boxed{-9}}\cdot y^{\boxed{-5}}
\\ \quad \\
x^5x^{-9}y^{10}y^{-5}\implies x^{5-9}y^{10-5}\implies x^{-4}y^5
\\ \quad \\
\cfrac{1}{x^{\boxed{4}}}\cdot \cfrac{y^5}{1}\implies \cfrac{y^5}{x^4}$