Part a.
On the same xy axis coordinate system, graph the four inequalities. Each inequality divides the xy plane in two parts (shaded region vs unshaded region). The four shaded regions overlap to form what you see in the attached image below (figure 1).
To graph any of the inequalities, you graph the boundary line first which is simply a line through 2 points. Once you have the boundary formed, you'll shade either above or below depending on the inequality sign. Shade above if you have a "greater than" sign; shade below if you have a "less than" sign.
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Part b.
From part (a) earlier, we see that the four vertex corner points are:
P = (0,3)
Q = (2,2)
R = (3,0)
S = (0,0)
Now plug each of those points into z = 3x+y to see what the largest z value we can get. Check out figure 2 (also attached below) to see the work shown for this part. In that image, we see that z = 9 is the largest z value possible. so this is the final answer for part b since the other values were z = 3, z = 8, z = 0. This value results from plugging in (x,y) = (2,2).
Six hundred fifty eight and one hundred twenty nine thousandths
Assuming the spaces mean plus (+):
2(3y+5) = 3(5y+13)
6y+10 = 15y+39
10 = 9y+39
9y = -29
Therefore, y = -29/9 or -3.22
Answer:
she flew 1h 30m and ran 30m
Step-by-step explanation:
First we have to make 2 equations, one that represents the distance that it traveled and another that represents the time that it ran
x = time she ran
y =
time she flew
x + y = 2h
x * 12 + y * 76 = 120
we clear x in the first equation
x + y = 2
x = 2 - y
we replace x with (2 - y) in the second equation
x * 12 + y * 76 = 120
(2 - y) * 12 + y * 76 = 120
24 - 12y + 76y = 120
-12y + 76y = 120 - 24
64y = 96
y = 96/64
y = 3/2
this means that 1 hour and a half was flying
x = 2 - y
x = 2 - 3/2
x = 1/2
this means that half hour was runing