Answer:
Step-by-step explanation:
We cannot factor this function out which tells us that there are no zeros. The graph backs this up because we can see that there are none. The easiest way to graph this would start by plugging in points and making a table for yourself.
Lets start by plugging in -1.
(-1)^2-2(-1)+3 =6, this means we have a point at (-1,6).
Now lets plug in 0.
(0)^2-2(0)+3= 3 (0,3)
For plugging in 1
(1)^2-2(1)+3=2 (1,2) (this happens to be the vertex)
And lastly lets plug in 2
(2)^2-2(2)+3=3 (2,3)
Depending on how many points are needed, keep plugging in numbers like we did above.
3(2a+1)=-5(a+6)
First expand the brackets.
6a+3=-5a-30
Now add 5a to both sides.
11a+3=-30
Then subtract 3 from both sides.
11a=-33
Finally divide both sides by 11.
a=- 33/11
=-3
You would circle the 3 in 31, the 9 in 94, and the 1 in 17
Answer: B. (2,1)
Step-by-step explanation:
its not c because if you look at R you see that it is going 4,3 not 3,4 because when you doing ratios on a number line it goes x-axis then y-axis next not y-axis then the x-axis.
It's not D because if you look at Q you see that it is going 3,5 not 5,3 because when you doing ratios on a number line it goes x-axis then y-axis next not y-axis then x-axis.
That's the same thing for A so your answer is b.
We define the probability of a particular event occurring as:

What are the total number of possible outcomes for the rolling of two dice? The rolls - though performed at the same time - are <em>independent</em>, which means one roll has no effect on the other. There are six possible outcomes for the first die, and for <em>each </em>of those, there are six possible outcomes for the second, for a total of 6 x 6 = 36 possible rolls.
Now that we've found the number of possible outcomes, we need to find the number of <em>desired</em> outcomes. What are our desired outcomes in this problem? They are asking for all outcomes where there is <em>at least one 5 rolled</em>. It turns out, there are only 3:
(1) D1 - 5, D2 - Anything else, (2), D1 - Anything else, D2 - 5, and (3) D1 - 5, D2 - 5
So, we have

probability of rolling at least one 5.