Hyperbolic properties of Nike function In y = ax + B / x, what are the vertices of the imaginary axis of the line where the focus real axis is located
When a and B are positive, the shape in the first quadrant is a check sign shape. B / a increases monotonically from negative infinity to negative root sign, B / a decreases monotonically from negative root sign to 0, B / a decreases monotonically from 0 to root sign, and B / a increases monotonically from root sign to positive infinity
If the line y = - 2x intersects the hyperbola y = - 3 / X at the point (m, 3), the analytic expression of the function after translation is ()
Y = - 3 / x, y = 3 to get m = - 1
Y = - 2 (x + D) passing (- 1,3) point
The solution is d = - 1 / 2
So y = - 2x + 1
A variable is defined in one function. How to use this variable in another function
In C / C + +, a variable defined in a function body has a function scope and cannot be used externally. Because this definition allocates space in stack memory, once a variable is created, it will be destroyed automatically after the function ends. Therefore, it is also called auto variable
A simple calculation of dividing 2 / 119 by 1 / 120
A simple calculation of dividing 2 / 119 by 1 / 120
2 / 119 divided by 1 / 120
=(2/119)x120
=(2/119)x(119+1)
=2+2/119
=2 and 2 / 119
Simple calculation of 2 / 120 * 119
120 * 2 / 119
=(119 + 1) 2 / 119
=2 + 2 / 119
=2 / 119
Simple calculation of 120x25 / 119
120x25/119
=(119+1)x25/119
=119x25/119+25/119
=25+25/119
=25 and 25 / 119
Planting 101 trees beside a 100 meter long path proves that no matter what kind of planting, there are always two trees with a distance of less than 1 meter
Divide 100 meters into 100 sections, each section is 1 meter long
These 100 sections are regarded as 100 drawers and 101 trees as 101 elements
If 101 elements are put into 100 drawers, there must be two elements in the same drawer, that is, at least two trees are not more than 1 meter away
It is proved that 7 / 12 is less than 1 / 100 + 1 / 101 +. + 1 / 199 is less than 5 / 6
Let a = 1 / 100 + 1 / 101 +. + 1 / 199 = (1 / 100 + 1 / 199) + (1 / 101 + 1 / 198) + +(1/149+1/150)=299/(100*199)+299/(101*198)+…… +A > [299 / (145 * 150)] * 50 = 299 / 447 > 7 /
The integral part of (1 / 100 + 1 / 101 + 1 / 102 +... + 1 / 120) × 11
Solving the problem by the method of expansion and contraction
(1/120+1/120+…… +1 / 120) × 11 < formula < (1 / 100 + 1 / 100 +...) +1/100)×11
220 / 120 < formula < 220 / 100
1.83 < formula < 2.2
Please explain (1) 1-2 + 3-4 + 5 - ·· - 100 + 101 = (2) - 1 + 3-5 + 7-9 + · - 97 + 99=
(1)1-2+3-4+5-···-100+101
=(1-2)+(3-4)+…… +(99-100)+101
= -1×50+101
= -50+101
=51
(2)-1+3-5+7-9+···-97+99
=( -1+3)+(-5+7)+…… +(-97+99)
= 2×25
= 50