Each side is 13cm
the square root of 169 is 13
Answer:
D, Yes, because there is a lot of variability in the science book data
Answer:
![4ln [\frac{x^2 (x^3-1)}{x-5}]](https://tex.z-dn.net/?f=%204ln%20%5B%5Cfrac%7Bx%5E2%20%28x%5E3-1%29%7D%7Bx-5%7D%5D)
Step-by-step explanation:
For this case we have the following expression:
![4[ln(x^3-1) +2ln(x) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%20%2B2ln%28x%29%20-ln%28x-5%29%5D)
For this case we can apply the following property:
![a log_c (b) = log_c (b^a)](https://tex.z-dn.net/?f=%20a%20log_c%20%28b%29%20%3D%20log_c%20%28b%5Ea%29)
And we can rewrite the following expression like this:
![2 ln(x) = ln(x^2)](https://tex.z-dn.net/?f=%202%20ln%28x%29%20%3D%20ln%28x%5E2%29)
And we can rewrite like this our expression:
![4[ln(x^3-1) +ln(x^2) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%20%2Bln%28x%5E2%29%20-ln%28x-5%29%5D)
Now we can use the following property:
![log_c (a) +log_c (b) = log_c (ab)](https://tex.z-dn.net/?f=%20log_c%20%28a%29%20%2Blog_c%20%28b%29%20%3D%20log_c%20%28ab%29)
And we got this:
![4[ln(x^3-1)(x^2) -ln(x-5)]](https://tex.z-dn.net/?f=%204%5Bln%28x%5E3-1%29%28x%5E2%29%20-ln%28x-5%29%5D)
And now we can apply the following property:
![log_c (a) -log_c (b) = log_c (\frac{a}{b})](https://tex.z-dn.net/?f=%20log_c%20%28a%29%20-log_c%20%28b%29%20%3D%20log_c%20%28%5Cfrac%7Ba%7D%7Bb%7D%29)
And we got this:
![4ln [\frac{x^2 (x^3-1)}{x-5}]](https://tex.z-dn.net/?f=%204ln%20%5B%5Cfrac%7Bx%5E2%20%28x%5E3-1%29%7D%7Bx-5%7D%5D)
And that would be our final answer on this case.
Recall Euler's theorem: if
, then
![a^{\phi(n)} \equiv 1 \pmod n](https://tex.z-dn.net/?f=a%5E%7B%5Cphi%28n%29%7D%20%5Cequiv%201%20%5Cpmod%20n)
where
is Euler's totient function.
We have
- in fact,
for any
since
and
share no common divisors - as well as
.
Now,
![37^{32} = (1 + 36)^{32} \\\\ ~~~~~~~~ = 1 + 36c_1 + 36^2c_2 + 36^3c_3+\cdots+36^{32}c_{32} \\\\ ~~~~~~~~ = 1 + 6 \left(6c_1 + 6^3c_2 + 6^5c_3 + \cdots + 6^{63}c_{32}\right) \\\\ \implies 32^{37^{32}} = 32^{1 + 6(\cdots)} = 32\cdot\left(32^{(\cdots)}\right)^6](https://tex.z-dn.net/?f=37%5E%7B32%7D%20%3D%20%281%20%2B%2036%29%5E%7B32%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20%2B%2036c_1%20%2B%2036%5E2c_2%20%2B%2036%5E3c_3%2B%5Ccdots%2B36%5E%7B32%7Dc_%7B32%7D%20%5C%5C%5C%5C%20~~~~~~~~%20%3D%201%20%2B%206%20%5Cleft%286c_1%20%2B%206%5E3c_2%20%2B%206%5E5c_3%20%2B%20%5Ccdots%20%2B%206%5E%7B63%7Dc_%7B32%7D%5Cright%29%20%5C%5C%5C%5C%20%5Cimplies%2032%5E%7B37%5E%7B32%7D%7D%20%3D%2032%5E%7B1%20%2B%206%28%5Ccdots%29%7D%20%3D%20%2032%5Ccdot%5Cleft%2832%5E%7B%28%5Ccdots%29%7D%5Cright%29%5E6)
where the
are positive integer coefficients from the binomial expansion. By Euler's theorem,
![\left(32^{(\cdots)\right)^6 \equiv 1 \pmod9](https://tex.z-dn.net/?f=%5Cleft%2832%5E%7B%28%5Ccdots%29%5Cright%29%5E6%20%5Cequiv%201%20%5Cpmod9)
so that
![32^{37^{32}} \equiv 32\cdot1 \equiv \boxed{5} \pmod9](https://tex.z-dn.net/?f=32%5E%7B37%5E%7B32%7D%7D%20%5Cequiv%2032%5Ccdot1%20%5Cequiv%20%5Cboxed%7B5%7D%20%5Cpmod9)
Answer:
1489.935
Explanation
6.5% of 1399
= 90.935
1399+90.935= 1489.935