Sid intended to type a seven-digit number, but the two 3's he meant to type did not appear. what appeared instead was the five-digit number 52115. how many different seven-digit numbers could sid have meant to type? a. 10b. 16c. 21d. 24e. 27

D. 21ways.

Step-by-step explanation:

This is based on selection, fixing two '3' numbers into 5numbers to make it 7 numbers can be done in 7C2 ways.

Generally, nCr = n!/(n-r)!r!

7C2 = 7!/(7-2)!2!

=7!/5!2!

=7×6×5!/5!×2!

=7×6/2

=21ways (D)

109 trust me

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: