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First one:

The rule of addition of logarithms is: \(\log(x)+\log(y)=log(xy)\). Therefore, we need to calculate \(\displaystyle \prod_{n=1}^{\infty}\frac{n\left(n+2\right)}{\left(n+1\right)^{2}}\) first, and then take the log of that infinite product.

The infinite product would look something like this:

\(\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6 \dotsm}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm}\), and you can rearrange it like this:

\( \frac{1\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm} = \frac{1}{2}\) (since the numerator and denominator both contain \(2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\dotsm\))

Therefore, the answer to the original problem is \(\boxed{\log{\frac{1}{2}}}\), which is approximately -0.301

\(x=\sqrt{420 - \sqrt{420 - \sqrt{420 - \sqrt{420 - \dotsb}}}}\\x^2=420-x\\x^2+x-420=0\\(x-20)(x+21)=0\\\boxed{x=20}\)

lol

The exponent of the prime factorization of the gcd of the 2 numbers will be the biggest power of that prime number that divides into both of the 2 giant numbers. Knowing that here is the prime factorization of the greatest common divisor of the 2 numbers:

\(2^23^35^57^711^{11}13^717^519^323^2\)

Therefore \(\boxed{11}\) is the answer.

1. Almost right; you forgot negative values 

(just as a reminder, the sum \(\displaystyle\sum_{n=1}^{\infty}x^n\) where x is a constant only converges when \(-1  by the ratio test.)

So the answer would be \((-\infty, -2)\bigcup(2, \infty)\)

2. The sum \(\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{y}{n}\right)^a\), where y is a constant, only converges when \(a>1\), so the answer is \((1, \infty)\)

https://web2.0calc.com/questions/please-help_50163

\(4y+7y+6+180-116=180\\11y=110\\y=10\\\text{CDE}=4\cdot10=\boxed{40}\)

Notice that \(x^3-p=x^3-(\sqrt[3]{p})^3=(x-\sqrt[3]{p})(x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2)\), using the rule of difference of cubes. Since we know that \(x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2\)is a factor of \(x^3-p\), we now know that \(\sqrt[3]{p}=b\). Cube both sides of the equation to get \(\boxed{p=b^3}\). 

I genuinely do not understand why the hint would be helpful to solve this problem in any way, but you can show that's true by noticing that \(c=(\sqrt[3]{p})^2=b^2\) 

M can include a 1 as well.

The number of Ms that include a 1 is equal to 4! = 24, so the final answer is 6+24=30

1. It's hard to apply the sine rule here. Instead, notice that the length WY is \(4\sqrt2\) , because triangle WXY is a 45-45-90 triangle, and notice that angle WYZ is a right triangle, so to find A, use the Pythagorean theorem:

\(9^2=(4\sqrt2)^2+a^2\\81=32+a^2\\49=a^2\\a=7\)

2. Let x be the missing side. Using the Pythagorean theorem,

\(7^2=(8-5)^2+x^2\\49=9+x^2\\40=x^2\\x=2\sqrt10\)

your answer is correct; it's just that this problem is very misleading and confusing. 

Notice that if you replace pi with 3.14 you get 16*16-8*8*3.14 = 55.04.