The answer to 9!/3! Is 60480
a.
The polynomial w^2+18w+84 cannot be factored
The perfect square trinomial is w^2+18w + 81
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The reason the original can't be factored is that solving w^2+18w+84=0 leads to no real solutions. Use the quadratic formula to see this. The graph of y = x^2+18x+84 shows there are no x intercepts. A solution and an x intercept are basically the same. The x intercept visually represents the solution.
w^2+18w+81 factors to (w+9)^2 which is the same as (w+9)(w+9). We can note that w^2+18w+81 is in the form a^2+2ab+b^2 with a = w and b = 9
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b.
The polynomial y^2-10y+23 cannot be factored
The perfect square trinomial is y^2-10y + 25
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Using the quadratic formula, y^2-10y+23 = 0 has no rational solutions. The two irrational solutions mean that we can't factor over the rationals. Put another way, there are no two whole numbers such that they multiply to 23 and add to -10 at the same time.
If we want to complete the square for y^2-10y, we take half of the -10 to get -5, then square this to get 25. Therefore, y^2-10y+25 is a perfect square and it factors to (y-5)^2 or (y-5)(y-5)
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
Give me more points, and I will ;)
