If k is a positive two-digit integer, what is the tens digit of k ?
1 answer:
Answer:
The required positive two digit integer is 32.
Step-by-step explanation:
Given : If k is a positive two-digit integer.
1) The sum of the tens digit of k and the units digit of k is 5.
2) The tens digit of k + 6 is 3.
To find : What is the tens digit of k ?
Solution :
Let the two digit integer be
According to question,
By 1, The sum of the tens digit of k and the units digit of k is 5.
i.e.
The possible cases from counting are,
14, 41, 23, 32, 50 whose sum is 5.
By 2, The tens digit of k + 6 is 3.
Which means take number which is added by 6 and give tens digit as 3.
The only possible case is when k=32
As 32+6=38 where tens digit is 3.
Therefore, The required positive two digit integer is 32.
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Answer:
n = 59
Step-by-step explanation:
I find it easiest to work problems of this kind using a graphing calculator. That way, extraneous solutions can be avoided. It seems to work well to rewrite the problem, so you're looking for a value of n that makes the result zero. Here, that would mean you want ...
... f(n) = √(n+5) -√(n-10) -1
_____
The solution by hand involves eliminating the root symbols. You do that by squaring the equation:
... n +5 -2√((n+5)(n-10)) + n -10 = 1
Now, we isolate the remaining root and square again.
... 2n -6 = 2√((n+5)(n-10)) . . . collect terms, add 2√( ) -1
... n -3 = √(n²-5n-50) . . . . . . . divide by 2
... n² -6n +9 = n² -5n -50 . . . . square both sides
... 59 = n . . . . . . . . . . . . . . . . . add 50 +6n -n²
