Answer:
BD
AC
SN (might not be one)
AB
Step-by-step explanation:
Scientific notation is als<span>o </span>called power-of-10 notation, is a method of writing extremely large and small numbers.
when a number written in scientific notation includes a positive exponent, the magnitude of the number is <span>greater than one</span>
<span>Here let the quadratic equation be ax^2 + bx + c
We know that a=5 from the question.
Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2).
However, let's assume that a2=1 and b2=6,
Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10
So, the equation is (5x-10)(x-6)</span>
The first question is c i believe, I dont have time to answer the others
Answer:
Step-by-step explanation:
The expression to transform is:
![(\sqrt[6]{x^5})^7](https://tex.z-dn.net/?f=%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7)
Let's work first on the inside of the parenthesis.
Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:
![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Therefore ![\sqrt[6]{a} = a^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now let's replace 
with 
which is the algebraic form we are given inside the 6th root:
![\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E5%7D%20%3D%20%28x%5E5%29%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now use the property that tells us how to proceed when we have "exponent of an exponent":
Therefore we get: 
Finally remember that this expression was raised to the power 7, therefore:
^7=(x^\frac{5}{6})^7=x^\frac{35}{6}](https://tex.z-dn.net/?f=%5Btex%5D%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7%3D%28x%5E%5Cfrac%7B5%7D%7B6%7D%29%5E7%3Dx%5E%5Cfrac%7B35%7D%7B6%7D)
[/tex]
An use again the property for the exponent of a exponent:
