Answer:
2=k^1/x
3=k^1/y
12=k^1/z.......we get...
(atq)------k^1/x=k^1/y=k^1/z
(*a^m=b^m......b=a)
~1/x=1/y=1/z
1/z=1/y+1/x×2
(12=3+2×2)
1/z=1/y+2/x.....proved
Hi
2n³ + 4n² - 7 + (-n³ - 8n - 2) = 2n³ - n³ + 4n² - 8n - 7 - 2
n³ + 4n² - 8n - 9
Answer: n³ + 4n² - 8n - 9
36. you have to multiply the whole thing.
2/5, the formula is art over swimming, 16/40, then simplify 2/5
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1.) Will yield consecutive odd integers.
k+10, k+12, k+14
or
k+2, k+3, k+4
The answer is k + 10, k +12 , k + 14
Because consecutive integers have difference of 2.
If k is odd, then k+10, k+12, and k+14 are consecutive odd integers.
For example, assume k = 1, then,
k+10=11
k+12=13
k+14=15
And 11, 13 and 15 are consecutive odd integers.
2.) Will yield consecutive integers.
k+1, k+2, k+3
or
k+6, k+8, k+10
The answer is k+1, k+2, k+3
Let k be any integer number, k+1, k+2, k+3 are consecutive integers.
For example, let k = 23
k+1=24
k+2=25
k+3=26
24,25, and 26 are consecutive integers.
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