Answer:
dont know
Step-by-step explanation:
Answer and step-by-step explanation:
We learn that (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
So in this case, it would be:
= x^3 + (3*x^2*2) + (3*x*2^2) + 2^3
= x^3 + 6x^2 + 3x * 4 + 8
= x^3 + 6x^2 + 12x + 8
This is the standard form of the equation
Hope it help you :3
Abc is the answer hope that helps
Answer:
D
Step-by-step explanation:
We assume the rotation R is <em>counterclockwise</em> 60°.
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The exponent on R is the number of times it is applied. That is, R² = R(R(figure)). So, the composition is equivalent to R^(2-4) = R^-2.
When the exponent of R is negative, it is essentially the inverse function. That is, applying the function R to the result will give the figure you started with. Equivalently, it is rotation in the other direction.
![(\mathcal{R}^2\circ\mathcal{R}^{-4})(\text{figure})=\mathcal{R}^{-2}(\text{figure})=\text{figure rotated $120^{\circ}$ CW}](https://tex.z-dn.net/?f=%28%5Cmathcal%7BR%7D%5E2%5Ccirc%5Cmathcal%7BR%7D%5E%7B-4%7D%29%28%5Ctext%7Bfigure%7D%29%3D%5Cmathcal%7BR%7D%5E%7B-2%7D%28%5Ctext%7Bfigure%7D%29%3D%5Ctext%7Bfigure%20rotated%20%24120%5E%7B%5Ccirc%7D%24%20CW%7D)
The point 120° clockwise from B is D.
The desired image point is D.
13 : 7
(as eli has seven
and his dister has 6 more therefore 6+7 =13)
so this must be the correct answer
hope it will help you