<span>1. (f+g)(x) = f(x) +g(x)
.. = (</span>x^2-36) +(<span>x^3+2x^2-10)
.. = x^3 +3x^2 -46
2. </span>(f•g)(x) = f(x)•g(x)
.. = (x^4-9)•(x^3+9)
.. = x^7 +9x^4 -9x^3 -81
<span>3. (f-g)(x) = f(x) -g(x)
.. = (x^3-2x^2+12x-6) -(4x^2-6x+4)
.. = x^3 -6x^2 +18x -10</span>
Answer: ∠A = 70°
because ABC is an isosceles triangle and AB = AC
=> ∠B = ∠C = 55°
=> ∠A = 180-(∠B + ∠C) = 180° - 55°.2 = 70°
Step-by-step explanation:
Separate the vectors into their <em>x</em>- and <em>y</em>-components. Let <em>u</em> be the vector on the right and <em>v</em> the vector on the left, so that
<em>u</em> = 4 cos(45°) <em>x</em> + 4 sin(45°) <em>y</em>
<em>v</em> = 2 cos(135°) <em>x</em> + 2 sin(135°) <em>y</em>
where <em>x</em> and <em>y</em> denote the unit vectors in the <em>x</em> and <em>y</em> directions.
Then the sum is
<em>u</em> + <em>v</em> = (4 cos(45°) + 2 cos(135°)) <em>x</em> + (4 sin(45°) + 2 sin(135°)) <em>y</em>
and its magnitude is
||<em>u</em> + <em>v</em>|| = √((4 cos(45°) + 2 cos(135°))² + (4 sin(45°) + 2 sin(135°))²)
… = √(16 cos²(45°) + 16 cos(45°) cos(135°) + 4 cos²(135°) + 16 sin²(45°) + 16 sin(45°) sin(135°) + 4 sin²(135°))
… = √(16 (cos²(45°) + sin²(45°)) + 16 (cos(45°) cos(135°) + sin(45°) sin(135°)) + 4 (cos²(135°) + sin²(135°)))
… = √(16 + 16 cos(135° - 45°) + 4)
… = √(20 + 16 cos(90°))
… = √20 = 2√5
Answer:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Step-by-step explanation:
So we have the expression:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
Recall the order of operations or PEMDAS:
P: Operations within parentheses must be done first. On a side note, do parentheses before brackets.
E: Within the parentheses, if exponents are present, do them before all other operations.
M/D: Multiplication and division next, whichever comes first.
A/S: Addition and subtraction next, whichever comes first.
(Note: This is how the order of operations is traditionally taught and how it was to me. If this is different for you, I do apologize. However, the answer should be the same.)
Thus, we should do the operations inside the parentheses first. Therefore:
![12-[20-2(6^2\div3\times2^2)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D)
The parentheses is:

Square the 6 and the 4:

Do the operations from left to right. 36 divided by 3 is 12. 12 times 4 is 48:

Therefore, the original equation is now:
![12-[20-2(6^2\div3\times2^2)]\\=12- [20-2(48)]](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%5C%5C%3D12-%20%5B20-2%2848%29%5D)
Multiply with the brackets:
![=12-[20-96]](https://tex.z-dn.net/?f=%3D12-%5B20-96%5D)
Subtract with the brackets:
![=12-[-76]](https://tex.z-dn.net/?f=%3D12-%5B-76%5D)
Two negatives make a positive. Add:

Therefore:
![12-[20-2(6^2\div3\times2^2)]=88](https://tex.z-dn.net/?f=12-%5B20-2%286%5E2%5Cdiv3%5Ctimes2%5E2%29%5D%3D88)
Answer:
The law of large numbers
Step-by-step explanation:
The Law of large numbers states that if we observe more and more reptitiions of any chance process, the proportion of times that a specific outcome occurs approaches a single value.