Answer:
The average value of over the interval is .
Step-by-step explanation:
Let suppose that function is continuous and integrable in the given intervals, by integral definition of average we have that:
(1)
(2)
By Fundamental Theorems of Calculus we expand both expressions:
(1b)
(2b)
We obtain the average value of over the interval by algebraic handling:
The average value of over the interval is .
9 is in the hundred thousands place, 8 is in the ten thousands place, 7 is in the thousands place, 1 is in the hundreds place, 6 is in the tens place, 4 is in the ones place.
Answer: One POSSIBLE answer is (0,9)
Step-by-step explanation: This may look hard but it's actually super easy. All you have to do is substitute a value for x and see what you get as the y value. The first thing I would do is rearrange the equation so that the equation is in the format y = something. So the first thing you should do is subtract 6x from both sides of the equation and you should get -y=-6x+9. Right now, the y is being multiplied by -1, and to undo multiplication, you have to divide. So divide both sides of the equation by -1 (remember, what you do on one side of the equation, YOU HAVE TO DO THE SAME THING ON THE OTHER SIDE). So your new equation would be y=6x-9. Then, all you have to do is substitute a value for x and see what you get for y (x is basically the input and whatever you get for y is the output). S for example, let's say that x=0. So substitute a 0 in the equation wherever you see an x. So in this case, you would get y=(6x0)+9. Then you just simplify. So 6x0 is 0 and 0+9=9 So y=9
TO RECAP: Rearrange the equation so that the equation is in the format y = something. Then, input any number for x, and whatever you get for y is what you put in the brackets as your y value for your final answer.
Hope this helps!