Ln calculator How to calculate K value in LNK = 35.26 with calculator?

Ln calculator How to calculate K value in LNK = 35.26 with calculator?

Ln is not based on the natural logarithm e, the logarithm of K, solving the equation k = e to the power of 35.26 =

How to calculate LN in mathematics?
For example, 1 / 2ln (115 / 100) = 6.9881%?
Please be as detailed as possible. Thank you

With calculator: (1 / 2) ln (115 / 100) = (1 / 2) ln 1.15 = 0.1398 / 2 = 0.069881 = 6.9881%

The line y = x - 1 intersects two coordinate axes a and B respectively
If the line y = X-1 intersects two coordinate axes a and B respectively, and the point C is on the coordinate axis, if the triangle ABC is isosceles triangle, how many points c can satisfy the condition at most? 5, 6, 7 or 8?

When AB is the bottom edge, there is a triangle with point C at the origin of the coordinate. When AC is the bottom edge, you can draw a circle with point a and point C, and AC is the radius. If there are several points on the coordinate axis where the focus of the two circles are, how many isosceles triangles can be made. When BC is the bottom edge and AC is the bottom edge, the same result will come out,

Students participate in tree planting activities, 18 people complete this task, the average person to plant 35 trees, if reduce 3 people, the average person to plant how many trees?

18x35/(18-3)
=630/15
=42 (trees)
A: on average, 42 trees are planted per person

Given - 1 ≤ a ≤ 1, - 1 ≤ B ≤ 1, the probability that the equation x2 + ax + B2 = 0 of X has real roots?

If the equation x2 + ax + B2 = 0 has real roots, then △ = a2-4b2 ≥ 0, that is | B | ≤ 12 | a |. In the coordinate plane AOB, real numbers (a, b) form a square region with (1,1), (1, - 1), (- 1, - 1), (- 1,1) as vertices, and its area is 4. The region | B | ≤ 12 | a | is a square region with points (0,0), (1,12)

A store sold 190 bottles of large and small drinks a day
It is known that the number of large bottles of drinks sold is three times that of small bottles, less than 10 bottles, so how many bottles of large bottles and small bottles of drinks are sold?
What's the formula

Equation:
If x bottles of small drinks are sold, 190-x bottles of large drinks are sold
3X-10=190-X
3X+X=190+10
4X=200
X=50
190-50 = 140 (bottle)
A: 140 large bottles and 50 small bottles were sold
Formula:
4 drinks in large and small bottles: 3 + 1 = 4
Calculate a multiple of 1, that is, a small bottle of beverage: (190 + 10) △ 4 = 50 (bottle)
Large bottle beverage: 190-50 = 140 (bottle)
A: 140 large bottles and 50 small bottles were sold

Simplify (A's square + 3) / (A's Square-1) - (a + 1) / (A-1) - 1 to be equal to

(a²+3)/(a+1)(a-1)-(a+1)²/(a+1)(a-1) -1
=(a²+3-a²-2a-1)/(a+1)(a-1)-1
=(2-2a)/(a+1)(a-1)-1
=-2/(a+1)-1
=(-2-a-1)/(a+1)
=-(3+a)/(a+1)

The distance between city a and city B is 600 km. The freight cars drive from city a to city B at the speed of 40 km per hour. Five hours later, the passenger cars drive from city B to city A. after another four hours, the two cars meet. How many kilometers does the passenger car travel per hour?
NO x

(600-40×5)÷4-40
=400÷4-40
=100-40
=60 (km)

It is known that the solution of the equation AX + 2 = 2 (A-X) about X satisfies (the absolute value of X-1 / 2) = 1

By solving the equation AX + 2 = 2 (A-X), we get: ax + 2 = 2a-2xax + 2x = 2a-2 (a + 2) x = 2a-2  x = (2a-2) / (a + 2) ∵ the solution of the equation AX + 2 = 2 (A-X) satisfies (the absolute value of X-1 / 2) = 1 ∵ | X-1 / 2 | X-1 / 2 = ± 1, we get: X1 = 1.5, X2 = - 0.5 when x = 1.5, that is: (2a-2) / (a + 2) =

A and B start from ab at the same time and travel in opposite directions. Moreover, the speed of car a is 1 / 5 faster than that of car B. when they meet, car a is still 80 kilometers away from B. how many kilometers is the distance between AB and B?

The speed of car a is 1 / 5 faster than that of car B, that is to say, the speed ratio of car a and car B is 6 to 5, so their distance ratio in the same time is 6 to 5. Car a is 80 kilometers away from ground B, that is to say, car B has traveled 80 kilometers, so car a has traveled 96 kilometers. The distance between the two places is 176 kilometers
Let the time of meeting be t, the velocity of a be V, and the velocity of B be (1 + 1 / 5) vs total = VT + (1 + 1 / 5) VT = 11 / 5 vt. according to a series of equations which are 80km away from B at the time of meeting: s total_ SA = 80
11\5Vt-Vt=80
Vt=200\3
S=440\3
This is wrong. We should set the speed of B as V, and that of a as (1 + 1 / 5) V. It's a fast, not B fast