There isn't a quotient of the division sign
Expand the following:
(5 a + b/5)^2
(5 a + b/5) (5 a + b/5) = (5 a) (5 a) + (5 a) (b/5) + (b/5) (5 a) + (b/5) (b/5):
5×5 a a + (5 a b)/5 + (5 b a)/5 + (b b)/(5×5)
(5 a b)/5 = 5/5×a b = a b:
5×5 a a + a b + (5 b a)/5 + (b b)/(5×5)
(b×5 a)/5 = 5/5×b a = b a:
5×5 a a + a b + b a + (b b)/(5×5)
Combine powers. (b b)/(5×5) = (b^(1 + 1))/(5×5):
5×5 a a + a b + b a + (b^(1 + 1))/(5×5)
1 + 1 = 2:
5×5 a a + a b + b a + (b^2/5)/5
5 a×5 a = 5×5 a^2:
5×5 a^2 + a b + b a + (b^2/5)/5
5×5 = 25:
Answer: 25 a^2 + a b + b a + (b^2/5)/5
Answer
4. - x ^ 2 + 13x - 42 <= 0524211r
5. x ^ 2 - 11x - 42 < 0
16 - x ^ 2 + 9x - 8 <= 0
17. x^ 2 -11x+80<0
18. - x ^ 2 + 3x + 4 <= 0
19. x ^ 2 - 12x + 36 <= 0
20. 2x ^ 2 - 32x + 56 < 0
21. x ^ 2 - 10x - 96 < 0
Answer:
A. (55 x 5) - (40 x 5)
Step-by-step explanation:
You are solving how much miles (further along) would the second car be after 5 hours.
The first car averages 40 miles per hour. 5 hours later, it will have averaged about 200 miles in 5 hours (40 x 5 = 200).
The second car averages 55 miles per hour. 5 hours later, it will have averaged about 275 miles in 5 hours (55 x 5 = 275)
Subtract: 275 - 200 = 75
The second car would have averaged 75 more miles than the first car.
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