X - 6< 3 please help
2 answers:
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Answer:
Step-by-step explanation:
We can solve this in either of two ways. I'll do both.
<u>Mathematically</u>
3x + 2y = 8
5x + 2y = 12
Rewrite either equation to isolate y. I'll use
3x + 2y = 8
2y = 8 - 3x
y = (8-3x)/2
Now use this value of y in the other equation:
5x + 2y = 12
5x + 2((8-3x)/2) = 12
5x + (8 - 3x) = 12
2x = 4
x = 2
Use x=2 to find y:
y = (8-3x)/2
y = (8-3*(2))/2
y = (8-6)/2
y = 1
The point is (2,1)
<u>Graphically</u>
Graph the two equations and find the point they intersect. See attachment.
We obtain the same answer: (2,1)
Answer:
Step-by-step explanation:
We are given the second derivative:
And we want to find its inflection points.
To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.
Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.
So first, finding the zeros, we see that:
So, we can draw the following number-line:
<----(-4)--------------(3)----(6)---->
Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.
Testing for x < -4, we can use -5. So:
Since we acquired a positive result, g(x) is concave up for x < -4.
For -4 < x < 3, we can use 0. So:
Since we acquired a negative result, g(x) is concave down for -4 < x < 3.
And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.
For 3 < x < 6, we can use 4. So:
Since we acquired a negative result, g(x) is concave down for 3 < x < 6.
Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.
And finally, for x > 6, we can use 7. So:
So, g(x) is concave up for x > 6.
And since we changed signs before and after the inflection point at x = 6, x = 6 is indeed an inflection point.