If 9x^2 + kx + 36 = (3x+n)^2 = (3x)^2 + 2(3x)n + n^2 then equatiing coefficients of x^2, x^1 and x^0 (ie constant term) we have 9=9 (good), k=6n and 36=n^2. So n = +- 6 and given n > 0 we have n=6 and so k=36.
If n>0 and 9xsquared + kx + 36= (3x + n)squared, for all x, what is k-n?
another question i needed to check is this one:
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A family of five adults and two children are trying to cross a river. They have a lifeboat which can ONLY carry one of the following:
A) One Child
B) Two Children
C) One Adult
The boat may not carry two adults or a child and a adult
How many one-way trips for the entire family to cross the river.
How many one-way trips are required for a family of 1000 adults and two children
someone plz answer im a newbie so someone help me plz
9x^2 +kx +36 = (3x +n)^2
9x^2 +kx +36 = 9x^2 +6n +n^2
Since the lefthand side is equal to the righthand side, then we can say their corresponding terms are equal:
9x^2 = 9x^2 ----(1)
kx = 6n -----(2)
36 = n^2 -----(3)
From (1), x = 1 -------***
From (3), n = sqrt(36) = 6 ---***
So plug those into (2),
kx = 6n
k(1) = 6(6)
k = 36 ------***
Therefore, (k -n) = 36 -6 = 30 -------answer.
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River crossings.
i) 5 Adults and 2 Children, or, 5A and 2C
1st crossing, 2C, ---> across.
2nd crossing, 1C, <--- crossing back.
3rd..., 1A, ---> across. --------------1A already crossed ***
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4th..., 1C, <--- crossing back.
5th..., 2C, ---> across.
6th..., 1C, <--- crossing back.
7th..., 1A, ---> across. --------------2A already crossed ***
-------------------------------
8th..., 1C, <--- crossing back.
9th..., 2C, ---> across.
10th..., 1C, <--- crossing back.
11th..., 1A, ---> across. --------------3A already crossed ***
-------------------------------
12th..., 1C, <--- crossing back.
13th..., 2C, ---> across.
14th..., 1C, <--- crossing back.
15th..., 1A, ---> across. --------------4A already crossed ***
-------------------------------
16th..., 1C, <--- crossing back.
17th..., 2C, ---> across.
18th..., 1C, <--- crossing back.
19th..., 1A, ---> across. --------------5A already crossed. 1C is still left at the near side, so the 1C that has crossed will have to go back to fetch the other 1C.
20th..., 1C, <--- crossing back.
21st..., 2C, ---> across. -------------Now, all 5A and 2C are acrossed***
Therefore, for the entire family of 5 adults and 2 children, 21 one-way crossings are required for all of them to cross. -----answer.
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ii) For a family of 1000 adults and 2 children?
As seen above where there are 5 adults and 21 crossings were required,
(21 crossings) / (5 adults) = (4 crossings per adult) + (1 extra crossing)
So for 1000 adults,
(1000 adults)*(4 crossings/adult) = 4000 crossings
4000 + 1 extra = 4001 crossings ----------------------answer.