Y=6, X=72
Y=?,X=8
12Y=X
12Y/9=X/9
4/3Y=8
Y=8*3/4
Y=6
1.(√3)+(√2)
2.(√2)+(√5)
3.(√5)-(√3)
4.(√9)+(√2)
5.(√5)+(√4)
6.(√10)-(√3)
7.(√4)-(√2)
Cuz √[(a+b)±2√(ab)]=(√a)±(√b)
Answer:
19+19+10+10= 58 is the answer
The difference between the sum of all eight positive integral divisors of 66 and the sum of all eight positive integral divisors of 70 is zero.
<h3>How to find the difference between the integral divisors?</h3>
First let's find the integral divisors. We can write 66 as a product of prime numbers as:
66 = 33*2 = 2*3*11
Then the integral divisors of 66 are:
2
3
11
2*3 = 6
2*11 = 22
3*11 = 33
1 (trivially)
66 (trivially)
The sum gives:
2 + 3 + 11 + 6 + 22 +33 + 1 + 66 = 144
For 70 we have:
70 = 7*10 = 2*5*7
Then the integral divisors are:
1
70
2
5
7
2*5 = 10
2*7 = 14
5*7 = 35
The sum gives:
1 + 70 + 2 +5 + 7 + 10 + 14 + 35 = 144
Then the difference between these two sums is:
144 - 144 = 0
If you want to learn more about integral divisors:
brainly.com/question/4785696
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<span>These are points where f ' = 0. Use the quiotent rule to find f '.
f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2
Set f ' (x) = 0 and solve for x.
f ' (x) = 0 = (2-2x^3) / (x^3+2)^2
Multiply both sides by (x^3+2)^2
(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2]
0 = 2 - 2x^3
Add 2x^3 to both sides
2x^3 + 0 = 2x^3 + 2 - 2x^3
2x^3 = 2
Divide both sides by 2
2x^3 / 2 = 2 / 2
x^3 = 1
Take cube roots of both sides
cube root (x^3) = cube root (1)
x = 1. This is our critical point
2) Points where f ' does not exist.
We know f ' (x) = (2-2x^3) / (x^3+2)^2
You cannot divide by 0 ever so f ' does not exist where the denominator equals 0
(x^3 + 2)^2 = 0. Take square roots of both sides
sqrt((x^3+2)^2) = sqrt(0)
x^3 + 2 = 0. Add -2 to both sides.
-2 + x^3 + 2 = -2 + 0
x^3 = -2. Take cube roots of both sides.
cube root (x^3) = cube root (-2)
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point.
3) End points of the domain.
The domain was clearly stated as [0, 2]. The end points are 0 and 2.
Therefore, our only options are: 0, 1, 2.
Check the intervals
[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign.
In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive.
In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative.
Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum.
f (0) = 0
f (1) = 1/3
f (2) = 1/5
Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
