Answer:
Step-by-step explanation:
Given the functions
F(x) = 6-x²
G(x) = x²+4x-12
A) we are to find F(x)+g(x)
F(x)+G(x) = 6-x²+x²+4x-12
F(x)+G(x) = 6+0+4x-12
F(x)+G(x) = 4x+6-12
F(x)+G(x) = 4x-6
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
F(x)-g(x) = 6-x²-(x²+4x-12)
F(x)-g(x) = 6-x²-x²-4x+12
F(x)-g(x) = 6-2x²+4x+12
F(x)-g(x) = -2x²+4x+6
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
3) F(x)/G(x)
= 6-x²/x²+4x-12
The domain of the function is the values of x for which the expression exists. The expression exists on all real interval i.e xER
<u>Question</u>
In the diagram, angle OLM is twice as large as angle PON. What is the size of angle OLM?
Answer:
Step-by-step explanation:
Let Angle PON =x
Therefore: Angle OLM =2x
Therefore:
<span>Yes because it can be expressed as a fraction</span>
The area of the sector of a circle is expressed in the formula: A = 1/2 r^2 * theta where theta is expressed in radians. In this case, we are given with r equal to 6 inches and theta equal to pi/3. Hence the area of the sector is equal to 18.85 cm2 or 6 pi cm2.
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