AdamTaurus

Nice, thanks!

Thanks, man. On d, do you think I need to rationalize the denominator? I know how, but do I need to?

Don't know the first one but.

33 is an answer to question two.

10/33 = 0.303030303030303

\(\frac{1}{4}<\frac{M}{56}<\frac{3}{7}\)

To do this, I would multiply everything by 56.

\((56)(\frac{1}{4})<(56)(\frac{M}{56})<(56)(\frac{3}{7})\)

56*1/4=14

56*M/56=M

56*3/7=24

\(14

All integer values of M between 14 and 24 are \(15, 16, 17, 18, 19, 20, 21, 22, 23\).

Add them all together.

\(15+16+17+18+19+20+21+22+23 = 171​\)

Divide the whole thing by 9, since that is the number of terms present.

\(\frac{171}{9} = 19\)

The average is 19.

Is that\((-2.097+0.1069x+0.2466)*62=32\) or \(-2.097+0.1069x+(0.2466*62)=32\)?

But that would be an arithmetic sequence.

Arithmetic formula: \(t_n=t_1+(n-1)d\)

t_n=n^th term, t₁= first term, d= common difference, n= terms

Geometric Formula: \(t_n=t_1*r^{(n-1)}\)

t_n=n^th term, t₁= first term, r= common ratio, n= terms.

Can you start with the left side instead?

2^(2x+3)=14

The first thing you do is take a log base 2 of both sides, as exponentials and logarithms are inverses. I got the base of 2 from what the 2x+3 is raised to, which is a 2.

\(log_2(2^{2x+3})=log_2(14)\)

Since the base of the logarithm and the base of the exponential are the same, they cancel, leaving only 2x+3.

\(2x+3=log_2(14)\)

Subtract 3 from both sides.

\(2x=log_2(14)-3\)

Divide by 2

\(x=\frac{log_2(14)-3}{2}\)

\(x=0.403677461\)

So, 2^x is equivalent to \(2^{0.403677461}\).

\(2^{0.403677461}=1.322876\)

So, ElectricPavlov is right about the final answer, but I wanted to reiterate the steps in a little more detail, just for clarity's sake.

Thanks Guest!

Thanks CPhill!