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\(4^{\frac{1}{4}} \cdot 16^{\frac{1}{16}} \cdot 64^{\frac{1}{64}} \cdot 256^{\frac{1}{256}} \dotsm \\ =4^{1\cdot\frac{1}{4}} \cdot 4^{2\cdot\frac{1}{16}} \cdot 4^{3\cdot\frac{1}{64}} \cdot 4^{4\cdot\frac{1}{256}} \dotsm\\ =4^{\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\dotsm}\)

First, let's calculate the infinite series in the exponent; let it equal x. Therefore,

\(x =\frac{1}{4}+\frac{2}{16}+\frac{3}{64}+\frac{4}{256}+\dotsm\\ \frac{1}{4}x=\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\dotsm\\ \text{subtract } \frac{1}{4}x \text{ from }x\text{ to get:}\\ \frac{3}{4}x=\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\dotsm \)

That infinite geometric series on the right-hand side is just equal to \(\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}\) using the infinite geometric sequence formula. Then,

\(\frac{3}{4}x=\frac{1}{3}\\ x=\frac{4}{9}\)

Therefore, the original product is equal to:

\(4^{\frac{4}{9}}=\sqrt[9]{4^4}=\sqrt[9]{256}\)

\(a+b = 9+256=\boxed{265}\)

This is the smallest possible value because the expression \(\sqrt[9]{256}\) cannot be reduced.

if x^2-x-2 is a factor, then both (x-2) and (x+1) will be factors as well.

According to the remainder theorem, f(2) and f(-1) must be equal to zero. That means that:

\(2(2)^3-2(2)^2a+3(2)b-4=0\\16-8a+6b-4=0\\-4a+3b+6=0\)

and

\(2(-1)^3-2(-1)^2a+3(-1)b-4=0\\-2a-3b-6=0\)

both must be true. Add up the 2 system of equations to get:

\(-6a=0\\ a=0\)

Substitute back in a:

\(-4(0)+3b+6=0\\ b=-2\)

Therefore, \(a+b=\boxed{-2}\)

https://web2.0calc.com/questions/help-please_64923

The zeroes of the polynomial are -4 and 7, which means that the bottom 2 solutions are not correct.

Now, look at the x values of the polynomial in between -4 and 7, and evaluate them at each of the 2 possible solutions. The top right one suggests that the graph is negative between -4 and 7, which is not true, so the top-left answer f(x) = (x+4)(x-7)^2 is the correct answer.

Unfortunately, that's incorrect, and instead of actually finding the solutions, there is a way that requires less computation:

First, notice that \(\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}\). Since we already know what a+b is, we just need to compute ab.

To do that, first factor the sum of cubes:

\((a+b)(a^2-ab+b^2)=42\)

Then, replace a+b with 7:

\(a^2-ab+b^2=6\)

Then, square both sides of the first equation:

\(a^2+2ab+b^2=49\)

Then, subtract the first equation from the second one:

\(3ab=43\\ ab=\frac{43}{3}\)

Then, just replace ab with 43/3 and a+b with 7:

\(\frac{7}{\frac{43}{3}} = \frac{21}{43}\)

Let A, B, and C be the angles of the triangle, let a, b, and c be the sides of the triangle, and let R be the circumradius

First, know the sine rule:

\(\frac{a}{\sin{A}} =\frac{b}{\sin{B}}=\frac{c}{\sin{C}} = 2R\)

Rearrange like so:

\(\sin A = \frac{a}{2R}\\\sin B = \frac{b}{2R} \\\sin C = \frac{c}{2R}\)

Also, know the cosine rule (slightly rearranged so I don't have to write as much):

\(\cos{A} = \frac{b^2+c^2-a^2}{2bc}\\\cos{B} = \frac{c^2+a^2-b^2}{2ac}\\cos{C} = \frac{a^2+b^2-c^2}{2ab}\)

Notice that\(\frac{\cot{c}}{\cot{a}+\cot{b}} = \frac{\frac{\cos{c}}{\sin{c}}}{\frac{\cos{a}}{\sin{a}}+\frac{\cos{b}}{\sin{b}}}\)

Now, replace:

\(\frac{\frac{a^2+b^2-c^2}{2ab\cdot\frac{c}{2R}}}{\frac{b^2+c^2-a^2}{2bc\cdot\frac{a}{2R}}+ \frac{a^2+c^2-b^2}{2ac\cdot\frac{b}{2R}}}\)

Multiply the numerator and denominator by \(\frac{2R}{2abc}\):

\(\frac{a^2+b^2-c^2}{b^2+c^2-a^2+a^2+c^2-b^2} = \frac{a^2+b^2-c^2}{2c^2}\)

Rearrange the given equation:

\(a^2+b^2=\frac{19}{9}c^2\)

And then replace:

\(\frac{\frac{19}{9}c^2-c^2}{2c^2} = \frac{\frac{10}{9}c^2}{2c^2} = \boxed{\frac{5}{9}}\)

(this literally took 40 minutes to write so I hope this helps)

Let angle D be x. If 2 sides of a triangle are congruent, then 2 of the angles are also congruent. Therefore, we can label the triangle like this:

For the rightmost triangle, we can set up an equation:

\(5x+5x+90-4x=180\)

Can you take it from there?

Notice that you can form a square in the left corner with side lengths of 5, as shown below. Lines that are tangent to a circle and meet at a point are congruent, so we can label the triangle as shown above. Then, it's just the Pythagorean theorem:

\(18^2+(5+x)^2=(13+x)^2\)

I think you can take it from here.

Since the angles in a triangle add up to 180 degrees, angle ADC is 180-30-90=60. Since BDC = ADC - ADB, BDC = 60-25=35