<h3>
<u>Explanation</u></h3>
- Convert the equation into slope-intercept form.

where m = slope and b = y-intercept.
What we have to do is to make the y-term as the subject of equation.


From y = mx+b, the slope is 3.
<h3>
<u>Answer</u></h3>

If you notice the picture below, the amount of fencing, or perimeter, that will be used will be 3w + 2l
now

solve for "w", to see what critical points you get, and then run a first-derivative test on them, for the minimum
notice the

so. you can pretty much skip that one, though is a valid critical point, the width can't clearly be 0
so.. check the critical points on the other
Answer:
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Step-by-step explanation:
Answer:
Step-by-step explanation:
Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2) P(G) = 5/8
3) P(G/E) = P(GE)/P(E)
GE = {G2, G4}
Hence P(G/E) = 2/5
4) GE = {G2, G4}
P(GE) = 2/8 = 1/4
5) P(G or E) = P(G)+P(E)-P(GE)
= 5/8 + 3/8-2/8 = 3/5
6) No there is common element as G2 and G4
Cannot be mutually exclusive