3(x+1)^(4/3)=48 how do i solve this?

1 answer:

4 0

$3(x+1)^{\frac{4}{3}}=48\\\\\frac{3(x+1)^{\frac{4}{3}}}{3}=\frac{48}{3}\\\\(x+1)^{\frac{4}{3}}=(x+1)^{(\frac{1}{3}*4)}=\sqrt[3]{(x+1)^{4}}=16\\\\(\sqrt[3]{(x+1)^{4}})^{3}=16^{3}\\\\(x+1)^4=4096\\\\\sqrt[4]{(x+1)^4}=\sqrt[4]{4096}=\pm8\\\\\\x+1=\pm8\\\\x+1=8\\x=7\\\\x+1=-8\\x=-9$

Your answers are x = 7 and x = -9.

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12<br> Using side lengths only, could the triangles be similar?<br> 0.5<br> m05 1 1.5

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Answer:

$\large\boxed{\text{No.}\ \dfrac{0.5}{1}\neq\dfrac{1}{1.5}\neq\dfrac{1.5}{2}}$

Step-by-step explanation:

$\text{Let}\\ a,\ b,\ c\ -\ \text{sides of a triangle ABC, where}\ a\leq b\leq c\\d,\ e,\ f\ -\ \text{sides of a triangle}\ DE F,\ \text{where}\ d\leq e\leq f.\\\\\triangle ABC\sim\triangle DE F\iff\dfrac{a}{d}=\dfrac{b}{e}=\dfrac{c}{f}\\\\/\text{corresponding sides are in proportion}/$

$\bold{WARNING !!!}\\\\\text{No triangle like QRS exists!}\\\\1 + 0.5 = 1.5 !!!\\\\\text{The sum of the lengths of the two shorter sides of the triangle}\\\text{must be greater than the length of the longest side.}$

$\text{Despite this, let's check the ratios}$

$\text{We have:}\\\\\triangle XYZ\to a=1,\ b=1.5,\ c=2\\\triangle QRS\to d=0.5,\ e=1,\ f=1.5$

$\text{Check:}\\\\\dfrac{d}{a}=\dfrac{0.5}{1}=0.5\\\\\dfrac{e}{b}=\dfrac{1}{1.5}=\dfrac{10}{15}=\dfrac{2}{3}\\\\\dfrac{f}{c}=\dfrac{1.5}{2}=\dfrac{15}{20}=\dfrac{3}{4}\\\\\dfrac{d}{a}\neq\dfrac{e}{b}\neq\dfrac{f}{c}\neq\dfrac{d}{a}$