Assume the triagle is similar, what is the missing side?

Mathematics

2 answers:

solniwko [45]4 years ago

7 0

7 good luck Have a great day

Nina [5.8K]4 years ago

6 0

The missing side would be 7 :)

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HELP PLEASE SIMPLIFY !!!

Lelechka [254]

Answer:

$=x^{\frac{5}{6}}+2x^{\frac{7}{3}}$

Step-by-step explanation:

$x^{\frac{1}{3}}\left(x^{\frac{1}{2}}+2x^2\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\cdot \:2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\\mathrm{Simplify}\:x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}:\quad x^{\frac{5}{6}}+2x^{\frac{7}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=x^{\frac{5}{6}}$

$x^{\frac{1}{3}}x^{\frac{1}{2}}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=\:x^{\frac{1}{3}+\frac{1}{2}}\\=x^{\frac{1}{3}+\frac{1}{2}}\\\mathrm{Join}\:\frac{1}{3}+\frac{1}{2}:\quad \frac{5}{6}\\\frac{1}{3}+\frac{1}{2}\\\mathrm{Least\:Common\:Multiplier\:of\:}3,\:2:\quad 6\\Adjust\:Fractions\:based\:on\:the\:LCM\\=\frac{2}{6}+\frac{3}{6}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{2+3}{6}\\\mathrm{Add\:the\:numbers:}\:2+3=5\\=\frac{5}{6}\\=x^{\frac{5}{6}}\\2x^2x^{\frac{1}{3}}=2x^{\frac{7}{3}}\\=x^{\frac{5}{6}}+2x^{\frac{7}{3}}$