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\(\sin(55)=\frac{x}{12}\\ x=12\cdot \sin(55) \approx \boxed{9.8 ft}\)

Image for reference:

Question 1:

\(\log_216\) is equal to 4, because \(2^4=16\).

So now, we are trying to find what x satisfies \(x^4=81\), so \(\boxed{x=3}\).

Question 2:

For the biggest value of b, we need to find 2 values in the ranges of f(x) and g(x) that, when multiplied together, results in the biggest possible result. That would be \(-5\cdot-2=\boxed{10}\)

No solutions: the lines are parallel.

If you subtract 7x from both sides of the second equation and compare it to the first, the slopes are the same but the y-intercept is not, which means they will not have any solutions.

I just realized i made a typo, because x^2-x = x(x-1), but the answer remains the same

First, realize that \(x^2-x=x(x+1)\). Also, realize that the prime factorization of 2010 is \(2\cdot3\cdot5\cdot67\). Therefore we can rewrite f(100) and f(101) like this:

\(f(100)=100\cdot99+2\cdot3\cdot5\cdot67 = 10(10\cdot99+3\cdot67) \\f(101)=101\cdot100+2\cdot3\cdot5\cdot67=10(10\cdot101+3\cdot67)\)

Since we cannot factor further, the answer to this question is  \(\boxed{10}\)

First, set this equal to x:

\(x = \frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\)

We can replace the bottom part with x, shown below:

\(x = \frac{1}{1+\frac{1}{2+x}}\)

Now, solve for x:

\(x = \frac{1}{\frac{2+x+1}{2+x}}\\ x = \frac{1}{\frac{3+x}{2+x}}\\ x=\frac{2+x}{3+x}\\ x^2+3x=2+x\\ x^2+2x-2=0 \)

After using the quadratic formula or completing the square, we find that \(x = -1-\sqrt{3}\) or \(x = \sqrt{3}-1\). Since x must be positive, the negative solution is extraneous, and therefore the answer is \(\boxed{\sqrt{3}-1}\)

Faster solution:

The remainder theorem tells us that the remainder is the value of x^3+30x-c when x = 5.

Therefore, 

\((5)^3+30(5)-c=5\\ 125+150-c=5\\ 275-c=5\\ c=270\)

Same answer as guest :)

First, multiply the whole equation by \((x-3)^2(x+5)\):

\(1 = A(x-3)^2+B(x+5)(x-3)+C(x+5)\\ \)

If we set x to be -5, we make B and C equal to zero, which we can use to solve for A:

\(1=A(-5-3)^2\\ 1=A(-8)^2\\ 1=64A\\ \boxed{A=\frac{1}{64}}\)

Your instinct was correct: EG isn't necessarily equal to FG.

Here's a solution:

Notice that DGH and DFE are similar. We can set up an equation:

\(\frac{x-4}{x-7}=\frac{(x-4)+2x}{(x-7)+(x+3)}\\ \frac{x-4}{x-7}=\frac{3x-4}{2x-4}\\ (x-4)(2x-4)=(x-7)(3x-4)\\ 2x^2-12x+16=3x^2-25x+28\\ 0=x^2-13x+12\\ (x-12)(x-1)=0\\ x=1,x=12\)

1 is extraneous because it will be negative for x-4, which is impossible.

Therefore the answer is \(\boxed{24}\)