$\bf 2log(x-3)+1=5\iff 2log_{10}(x-3)+1=5\\\\\\ 2log_{10}(x-3)=4
\\\\\\
log_{10}(x-3)=\cfrac{4}{2}\implies log_{10}(x-3)=2\\\\
-----------------------------\\\\
{{ a}}^{log_{{ a}}x}=x\impliedby \textit{using this cancellation rule}\\\\
-----------------------------\\\\
10^{\cfrac{}{}log_{10}(x-3)}=10^2\implies x-3=10^2\implies x=100+3$

5 0

3 years ago

17 points if you know. Help please

vovangra [49]

Answer:

Step-by-step explanation:

Angles 153 and QRP are straight angles, and thus Angle QRP is

180 - 153, or 27 degrees.

The interior angles of the triangle must sum up to 180 degrees:

27 + (3y + 5) + (2y - 7) = 180.

combining like terms, we get:

5y - 25 = 180, or 5y = 155, or y = 31

Then Angle Q is 3(31) + 5, or 93

Angle P is 2(31) - 7, or 55, and

Angle QRP is 27 degrees (found earlier).

6 0

2 years ago