In AJKL, j = 490 cm, ZJ=128º and ZK=9º. Find the area of AJKL, to the nearest 10th of a square centimeter.

answer: what is the question you're trying to ask?

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation:

ok thats very interesting

i think it’s a i believe