Answer:
8 years approximately
The problem (I'm assuming is):
Solve
.
I put a t in the problem where I suspected it in went.
Step-by-step explanation:

Divide both sides by 100:


Rewrite in logarithm form.

Divide both sides by 0.05:

Put left hand side into calculator:


So about 8 years if I wrote down the equation correctly.
To start, the first thing you need to know is that when you add decimals, you must line the decimal up, meaning, if you were adding 2.7+4.1, you have to make sure 2 is lined up with 4, and 7 is lined up with 1 so that the decimals align. Then, you add straight down from right to left and you get 6.8 as your answer.
Answer:
A
Step-by-step explanation:
We know that the only possibilities with proportionate answers are 6:7 = 12:14, or 7:6 = 14:12. Only 6:7 = 12:14 is an answer choice. Both the numbers in the first pair times two equal the second pair.
Answer:

» Collect like terms, r terms on the left hand side by subtracting r from both sides and adding st to both sides

» On the left hand side, factorise out r

Problem 7)
The answer is choice B. Only graph 2 contains an Euler circuit.
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To have a Euler circuit, each vertex must have an even number of paths connecting to it. This does not happen with graph 1 since vertex A and vertex D have an odd number of vertices (3 each). The odd vertex count makes it impossible to travel back to the starting point, while making sure to only use each edge one time only.
With graph 2, each vertex has exactly two edges attached to it. So an Euler circuit is possible here.
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Problem 8)
The answer is choice B) 5
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Work Shown:
abc base 2 = (a*2^2 + b*2^1 + c*2^0) base 10
101 base 2 = (1*2^2 + 0*2^1 + 1*2^0) base 10
101 base 2 = (1*4 + 0*2 + 1*1) base 10
101 base 2 = (4 + 0 + 1) base 10
101 base 2 = 5 base 10