To get the equivalent of 3/5, multiply both the numerator and denominator by the same number. So, multiply 3/5 by 2 and you will get 6/10. Multiply 3/5 by 3 and you will get 9/15. <em><u>The equivalent fractions of 3/5 are 6/10 and 9/15.</u></em>
Hope it helps :)
If Simone has 20 bills, then:
1/5 of 20 = 4 [hundreds]
1/4 of 20 = 5 [fifites]
1/10 of 20 = 2 [$20s]
4 * 100 + 5 * 50 + 2 * 20 = 400 + 250 + 40 = 690 [out of $780]
4 + 5 + 2 = 11 [out of 20 bills]
So now we can easily find out she had 9 bills of tens - because 20 - 11 = 9 or because (780 - 690) ÷ 10 = 90 ÷ 10 = 9.
Assuming R and H:
So volume is pir^2 * H = 1500 and H = 1500/(pir^2) while surface area is A= 2pir*H + 2pir^2
A = 2pir(r+h)= 2piR^2 + 2pir*1500/(pir^2)= 2piR^2 + 3000/r
For A to take minimum, get the derivative 4pir - 3000/R^2 and let it be 0
4pir^3 - 3000 = 0
r = cbrt(3000/(4pi)) ≈ 6.20
h = 1500/(pi(6.20)^2) ≈ 12.42
Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4
or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.
If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is
f(x) approx. equal to [f '(c)]x + f(c)].
Here a = c = 81.
Thus, the linearization of the given function at a = c = 81 is
f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4
Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.
What is the linearization of f(x) = (x^3)/4, if c = a = 81?
It will be f(x) (approx. equal to)