If a+2b=60 that is just like the problem of finding the maximum area enclosed on three sides by 60 ft of fencing. That is when a=30 and 2b=30. So, maximum ab = 30*15 = 450
f(x) = x^2 - 3x + 4
f(√3-√2) = (√3-√2)^2 - 3(√3-√2) + 4
The mean is (90+x)/5 = 18 + 0.2x
arranged in order (15,20,x,25,30)
x = 18+0.2x
x = 22.5
But x can only be one of 20,21,22,23,24,25
Only x=23 satisfies median=mean-0.4
the mean of (15,20,23,25,30) = 22.6
Given that a and b are nonnegative real numbers such that a+2b=60, what is the largest possible value of ab?
And
If f(x) = x^2 - 3x + 4, find the value of
f(sqrt3 - sqrt2).
And
The median of {20, x, 15, 30, 25} is 0.4 less than the mean. If x is a whole
number, what is the sum of all possible values of x?
4 answers
I still don't understand the second one, can you explain it more please?
I've gotten to 1-sqrt 3+sqrt 2
but I don't know what to do next, or if it's wrong.
I've gotten to 1-sqrt 3+sqrt 2
but I don't know what to do next, or if it's wrong.
(√3-√2)^2 = 3-2√6+2 = 5-2√6
so that means you have
5-2√6 - 3√3 + 3√2) + 4
= 9 + 3√2 - 3√3 - 2√6
so that means you have
5-2√6 - 3√3 + 3√2) + 4
= 9 + 3√2 - 3√3 - 2√6
Oh, that makes a lot more sense thank you