Select the correct answer.
What is the approximate solution to this equation?
[tex]$15(3)^{2 x}=90$[/tex]

A. [tex]$x=\frac{2 \log 6}{\log 3}$[/tex]
B. [tex]$x=\frac{2 \log 3}{\log 6}$[/tex]
C. [tex]$x=\frac{\log 3}{2 \log 6}$[/tex]
D. [tex]$x=\frac{\log 6}{2 \log 3}$[/tex]

3m + 4c +7d = 5.45 => m = (5.45 - 4c - 7d)/3;
One solution is c = -1, d = 0, m =3.15; and so on .........

Answered by crisforp | 2024-06-10

i dont know but good luck ;

Answered by xmantaylor09 | 2024-06-12

To solve for m , c , and d in the equation 3 m + 4 c + 7 d = 5.45 , we can express one variable in terms of the others since we have more variables than equations. The expressions become m = 3 5.45 − 4 c − 7 d ​ , c = 4 5.45 − 3 m − 7 d ​ , and d = 7 5.45 − 3 m − 4 c ​ . This allows for multiple solutions based on chosen values for two of the three variables.
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Answered by Anonymous | 2024-12-20