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)
3a= 3(5) = 15
15 - - 3 = 15+3 = 18
18/6 = 3
b + a = 5 + -3 = -3 + 5 = 2
3 x 2 = 6
Every square is a rectangle because it is a quadrilateral with all four right angle (a rectangle is not a square)
We are given that there
will be (1/2) a litre after the first pouring, so considering two successive
pourings (n and (n+1)) with 1/2 litre in each before the nth pouring occurs:
1/2 × (1/n) = 1/(2n)
1/2 - 1/(2n) = (n-1)/2n
1/2 + 1/(2n) = (n+1)/2n
(n-1)/2n and (n+1)/2n in
each urn after the nth pouring
Then now consider the
(n+1)th pouring
(n+1)/2n × 1/(n+1) =
1/(2n)
(n+1)/(2n) - 1/(2n) =
n/(2n) = 1/2
Therefore this means that after
an odd number of pouring, there will be 1/2 a litre in each urn
Since 1997 is an odd
number, then there will be 1/2 a litre of water in each urn.
Answer:
<span>1/2</span>
Answer:
850 but im not sure lol
Step-by-step explanation:
645 + 225 = 870
1,320 - 870 = 450
450 + 350 = 800
800 + 50 = 850
