I'm going to assume you mean F(x) = -2x2 + 10x + 6 as -2x plus 10x = 8x...
First differentiate -2x2 + 10x + 6:
dy/dx = -4x + 10
Set dy/dx = 0:
-4x + 10 = 0
Find x:
10 = 4x
2.5 = x
Find y at x =2.5:
y = -2(2.5)2 + 10(2.5) + 6 = -12.5 + 25 +6 = 18.5
So max or min is at (2.5, 18.5)
To find out whether it is max or min differentiate twice:
d2y/dx2 = -4
If its negative, point will be a maximum
If its positive, point will be a minimum.
In this case, a maximum occurs at (2.5, 18.5).
Wiki User
∙ 15y ago3.55p.m
This appears to be a number trick or joke. The answer is that 1 hour, plus 1 minute, plus 59 minutes, will equal 2 hours.
The min and max is when the first derivative , or slope at any point, is zero. For f of x = 2x first derivative is 2, so this is constant slope with no min or max as this is not zero; min is thus negative infinity and max is infinity
10 hours and 15 minutes
y = x2 + 12x + 21At the max or min point, the first derivative of the function = 0.2x + 12 = 02x = -12x = -6
add them all up underneath eachother so 311 526 446 820 524 _____ 2627 <-------- answer
3.55p.m
This appears to be a number trick or joke. The answer is that 1 hour, plus 1 minute, plus 59 minutes, will equal 2 hours.
The min and max is when the first derivative , or slope at any point, is zero. For f of x = 2x first derivative is 2, so this is constant slope with no min or max as this is not zero; min is thus negative infinity and max is infinity
take the max plus the min and divide the anser by to
10 hours and 15 minutes
y = x2 + 12x + 21At the max or min point, the first derivative of the function = 0.2x + 12 = 02x = -12x = -6
#include <stdio.h> #include <stdlib.h> #include <time.h> int GetRand( int min, int max) { // Swap min and max if wrong way around. if(max<min)min^=max^=min^=max; return(rand()%(max-min+1)+min); } int main() { // Seed the random number generator with current time to ensure // random numbers are different on each run. // Note: use srand(1) to generate the same sequence on every run. srand((unsigned)time(NULL)); printf("Number from 0 to 10 : %d\n", GetRand(0,10)); printf("Number from 1 to 100: %d\n", GetRand(1,100)); return(0); }
=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)=MAX(Z1:Z10)/MIN(Z1:Z10)
Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. Steps to perform MaxMin on a data set (2,4,6,3,8,1,9,7) are:(2,4,6,3) (8,1,9,7)((2,4)(6,3)) ((8,1)(9,7))In sublist (4,6), max is 6 and min is 4. In sublist (8,9), max is 9 and min is 8.Comparing max and min values of sublist (2,4) and sublist (6,3), value of max is 6 and min is 2.Therefore, for sublist (2,4,6,3) max is 6 and min is 2.Similarly, comparing max and min values of sublist (8,1) and sublist (9,7), value of max is 9 and min is 1.Therefore, for sublist (8,1,9,7) max is 9 and min is 1.Finally, comparing max and min values of sublist (2,4,6,3) and sublist (8,1,9,7), value of max is 9 and min is 1. sonika aggarwal GNIIT
32 min 9 sec 40 min 10 sec adding gives... 71 min 19 sec 1 hr 11 min 19 sec. ■
Use the median-of-three algorithm: int min (int a, int b) { return a<b?a:b; } int max (int a, int b) { return a<b?b:a; } int median_of_three (int a, int b, int c) { return max (min (a, b), min (max (a, b), c)); } Note that the algorithm does not cater for equal values which creates a problem when any two values are equal, because there are only two values to play with, neither of which can be regarded as being the middle value. If the equal value is the lower of the two values, the largest value is returned if and only if it is the last of the three values, otherwise the lowest value is returned. But when the equal value is the larger of the two values, the largest value is always returned. Lowest value is equal: Input: 0, 0, 1 = max (min (0, 0), min (max (0, 0), 1)) = max (0, min (0, 1)) = max (0, 1) = 1 Input: 0, 1, 0 = max (min (0, 1), min (max (0, 1), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Input: 1, 0, 0 = max (min (1, 0), min (max (1, 0), 0)) = max (0, min (1, 0)) = max (0, 0) = 0 Highest value is equal: Input: 0, 1, 1 = max (min (0, 1), min (max (0, 1), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 0, 1 = max (min (1, 0), min (max (1, 0), 1)) = max (0, min (1, 1)) = max (0, 1) = 1 Input: 1, 1, 0 = max (min (1, 1), min (max (1, 1), 0)) = max (1, min (1, 0)) = max (1, 0) = 1 The only way to resolve this problem and produce a consistent result is to sum all three inputs then subtract the minimum and maximum values: int median_of_three (int a, int b, int c) { return a + b + c - min (min (a, b), c) - max (max (a, b), c)); } Lowest value is equal: Input: 0, 0, 1 = 0 + 0 + 1 - min (min (0, 0), 1) - max (max (0, 0), 1) = 1 - 0 - 1 = 0 Input: 0, 1, 0 = 0 + 1 + 0 - min (min (0, 1), 0) - max (max (0, 1), 0) = 1 - 0 - 1 = 0 Input: 1, 0, 0 = 1 + 0 + 0 - min (min (1, 0), 0) - max (max (1, 0), 0) = 1 - 0 - 1 = 0 Highest value is equal: Input: 0, 1, 1 = 0 + 1 + 1 - min (min (0, 1), 1) - max (max (0, 1), 1) = 2 - 0 - 1 = 1 Input: 1, 0, 1 = 1 + 0 + 1 - min (min (1, 0), 1) - max (max (1, 0), 1) = 2 - 0 - 1 = 1 Input: 1, 1, 0 = 1 + 1 + 0 - min (min (1, 1), 0) - max (max (1, 1), 0) = 2 - 0 - 1 = 1 This makes sense because when we sort 0, 0, 1 in ascending order, 0 is in the middle, while 0, 1, 1 puts 1 in the middle.