Solve the following equation: x + 5 = 42 a -42, 42 b -5, 5 c -37, 47 d -47, 37
1 answer:
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Step-by-step explanation:
We have to find the point in the feasible region which maximizes the objective function. To find that point first we need to graph the given inequalities to find the feasible region.
Steps to graph 3x + y ≤ 12:
First we graph 3x + y = 12 then shade the graph for ≤.
plug any value of x say x=0 and x=2 into 3x + y = 12 to find points.
plug x=0
3x + y = 12
3(0) + y = 12
0 + y = 12
y = 12
Hence first point is (0,12)
Similarly plugging x=2 will give y=6
Hence second point is (2,6)
Now graph both points and joint them by a straight line.
test for shading.
plug any test point which is not on the graph of line like (0,0) into original inequality 3x + y ≤ 12:
3(0) + (0) ≤ 12
0 + 0 ≤ 12
0 ≤ 12
Which is true so shading will be in the direction of test point (0,0)
We can repeat same procedure to graph other inequalities.
From graph we see that ABCD is feasible region whose corner points will result into maximum or minimu for objective function.
Since objective function is not given in the question so i will explain the process.
To find the maximum value of objective function we plug each corner point of feasible region into objective function. Whichever point gives maximum value will be the answer
Answer:
Step-by-step explanation:
You need to use synthetic division to do all of these. The thing to remember with these is that when you start off with a certain degree polyomial, what you get on the bottom line after the division is called the depressed polynomial (NOT because it has to math all summer!) because it is a degree lesser than what you started.
a. 3I 1 3 -34 48
I'm going to do this one in its entirety so you get the idea of how to do it, then you'll be able to do it on your own.
First step is to bring down the first number after the bold line, 1.
3I 1 3 -34 48
_____________
1
then multiply it by the 3 and put it up under the 3. Add those together:
3I 1 3 -34 48
3
----------------------------
1 6
Now I'm going to multiply the 6 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18
_________________
1 6 -16
Same process, I'm going to multiply the -16 by the 3 after the bold line and add:
3I 1 3 -34 48
3 18 -48
___________________
1 6 -16 0
That last zero tells me that x-3 is a factor of that polynomial, AND that the depressed polynomial is one degree lesser and those numbers there under that line represent the leading coefficients of the depressed polynomial:
Factoring that depressed polynomial will give you the remaining zeros. Because this was originally a third degree polynomial, there are 3 zeros as solutions. Factoring that depressed polynomial gives you the remaining zeros of x = -8 and x = 2
I am assuming that since you are doing synthetic division that you have already learned the quadratic formula. You could use that or just "regular" factoring would do the trick on all of them.
Do the remaining problems like that one; all of them come out to a 0 as the last "number" under the line.
You got this!