Assuming you meant <span>e⁴ thus
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![\bf ln\left( \sqrt[7]{e^4} \right)\implies log_e\left( e^{\frac{4}{7}} \right)\\\\ -----------------------------\\\\ log_{{ a}}{{ a}}^x\implies x=x\impliedby \textit{let's use this cancellation rule}\\\\ -----------------------------\\\\ log_e\left( e^{\frac{4}{7}} \right)\iff \cfrac{4}{7}](https://tex.z-dn.net/?f=%5Cbf%20ln%5Cleft%28%20%5Csqrt%5B7%5D%7Be%5E4%7D%20%5Cright%29%5Cimplies%20log_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_%7B%7B%20%20a%7D%7D%7B%7B%20%20a%7D%7D%5Ex%5Cimplies%20x%3Dx%5Cimpliedby%20%5Ctextit%7Blet%27s%20use%20this%20cancellation%20rule%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Alog_e%5Cleft%28%20e%5E%7B%5Cfrac%7B4%7D%7B7%7D%7D%20%5Cright%29%5Ciff%20%5Ccfrac%7B4%7D%7B7%7D)
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What is the solution for x in the equation? 1/2-x+3/2=x-4
1 answer:
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Answer:
49 units
Step-by-step explanation:
Given the triangles are similar then the ratios of corresponding sides are equal, that is
=
, substitute values
=
( cross- multiply )
22BC = 484 ( divide both sides by 22 )
BC = 22
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Similarly
=
, substitute values
=
( cross- multiply )
22AC = 352 ( divide both sides by 22 )
AC = 16
Thus perimeter of Δ ABC is
perimeter = AB + BC + AC = 11 + 22 + 16 = 49 units