Answer:
I would say it would also be 64 degrees because if you flip the HGJ to the IJG it would be the same! Hope this helps
:)
<h2>
Hello!</h2>
The answers are:
The possible values for x in the equation, are:
First option, ![5\sqrt[3]{3}](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7B3%7D)
Second option, ![\sqrt[3]{375}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B375%7D)
<h2>
Why?</h2>
To solve the problem, we need to remember the following properties of the exponents and roots:
![a\sqrt[n]{b}=\sqrt[n]{a^{n}*b} \\\\\sqrt[n]{a^{m} }=a^{\frac{m}{n}}\\\\(a^{b})^{c}=a^{b*c}](https://tex.z-dn.net/?f=a%5Csqrt%5Bn%5D%7Bb%7D%3D%5Csqrt%5Bn%5D%7Ba%5E%7Bn%7D%2Ab%7D%20%5C%5C%5C%5C%5Csqrt%5Bn%5D%7Ba%5E%7Bm%7D%20%7D%3Da%5E%7B%5Cfrac%7Bm%7D%7Bn%7D%7D%5C%5C%5C%5C%28a%5E%7Bb%7D%29%5E%7Bc%7D%3Da%5E%7Bb%2Ac%7D)
Then, we are given the expression:

So, finding "x", we have:
![x^{3}=375\\\\(x^{3})^{\frac{1}{3} } =(375)^{\frac{1}{3}}\\\\x=\sqrt[3]{375}=\sqrt[3]{125*3}=\sqrt[3]{125}*\sqrt[3]{3}=5\sqrt[3]{3}](https://tex.z-dn.net/?f=x%5E%7B3%7D%3D375%5C%5C%5C%5C%28x%5E%7B3%7D%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%20%7D%20%3D%28375%29%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%5C%5C%5C%5Cx%3D%5Csqrt%5B3%5D%7B375%7D%3D%5Csqrt%5B3%5D%7B125%2A3%7D%3D%5Csqrt%5B3%5D%7B125%7D%2A%5Csqrt%5B3%5D%7B3%7D%3D5%5Csqrt%5B3%5D%7B3%7D)
Hence, the possible values for x in the equation, are:
First option, ![5\sqrt[3]{3}](https://tex.z-dn.net/?f=5%5Csqrt%5B3%5D%7B3%7D)
Second option, ![\sqrt[3]{375}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B375%7D)
Have a nice day!
Answer:
no one wants to answer this because im hungry
Step-by-step explanation:
Six plus three is nine, thus the lowest answer.
To find if a series is either geometric or arithmetic:
it must satisfy this property:
Arithmetic:
a(n+1) - a(n) = const
Geometric:
a(n+1)/a(n) = const
In your case:
r1 = 7 -4 = 3
r2 = 12 - 7 = 5
r1 != r2 (not arirthmetic)
Geometric check:
r1 = 7/4
r2 = 12/7
r1 != r2 (not Geometric)
so neither.