If f (x) = 5x + 1 / (a)

If f (x) = 5x + 1 / (a)

If the function f (x) = 5x + 1 / (A-1) x ^ 2 + 2x-3 is always meaningful for any x ∈ R, then the value range of a is constant
Given by, denominator cannot be 0
Only (A-1) x ^ 2 ≠ 0
∴a≠1

The function f (x) = sin ^ 5x + 1 is known

Reference: given f (x) = SiNx ^ 5 + 1, calculate the integral from - π / 2 to π / 2
The answer is pi
SiNx ^ 5 is an odd function from - π / 2 to π / 2, the integral is 0
The integral of 1 from - π / 2 to π / 2 is 0 π
The answer is pi

Find the range of function f (x) = X-1 / 2x ^ 2-5x + 5

Let f (x) = t = (x-1) / (2x & # 178; - 5x + 5), then
→2tx²-(5t+1)x+(5t+1)＝0.
The discriminant is not less than 0
(5t+1)²-8t(5t+1)≥0
→(5t+1)(3t-1)≤0
∴-1/5≤t≤1/3.
So the definition field of function f (x) is: [- 1 / 5,1 / 3]

What is the monotone increasing interval of the function f (x) = 1 / 5x + 3?

Function y = 1 / 5x + 3 = f (x)
The definition fields of functions are (- ∞, 0) and (0, + ∞)
1) Let x1, X2 ∈ (- ∞, 0), and X1 > x2
Then f (x1) - f (x2)
=1/5x1 +3 -1/5x2 -3
=(x2-x1)/5x1x2
Because X1 > X2, so x2-x10
So (x2-x1) / 5x1x2, so x2-x10
So (x2-x1) / 5x1x2

Right triangle, three sides are 3.5, 1.35, 3.28, find two angles except right angle

68 degrees: 1.35,3.5/22 degrees: 3.5,3.28, calculated by drawing

The length of the right side of a right triangle is one 20 and one 30 to find the angles of the two angles except the right angle

Angle A: 33.69 degrees
The angle B is 56.31 degrees
C side length 36.06
The above figures are rounded off
Please set my answer to adopt,

How to calculate the degree of angle a when the length of three sides of a right triangle is known

It can be found by its sin or cos value, because it is unique

In △ ABC, if cosaa = cosbb = sincc, then △ ABC is ()
A. A right triangle with an internal angle of 30 ° B. an isosceles right triangle C. an isosceles triangle with an internal angle of 30 ° D. an equilateral triangle

∵ cosaa = sincc, ∵ combined with sine theorem sinaa = sincc, we can get Sina = cosa, so Tana = 1, we can get a = π 4. Similarly, we can get b = π 4 ∵ ABC is an isosceles right triangle with C as right angle, so we choose: B = π 4

Diagonal triangle
In triangle ABC, a = 60 °, B = 1, and the area of triangle is root 3, then the value of (a + B + C) / (Sina + SINB + sinc) is ()
A (8 root 3) / 3 B (2 root 39) / 3 C (26 root 3) / 3 D2 root 7

B (2 root 39) / 3
S=1/2sinAbc=√3
So C = 4
According to the cosine theorem (b ^ 2 + C ^ 2-A ^ 2) / 2BC = cosa = 1 / 2
a^2=13
So a = √ 13
（a+b+c)/(sinA+sinB+sinC)=a/sinA=2√39/3

High school solution of oblique triangle
Triangle ABC, known C = 60 degrees, AB is root 3, the height of AB side is 4 / 3, find BC + AC

BC = x, AC = y (xysin60) / 2 = (4 / 3AB) / 2 = > xy = 8 / 3 (sine theorem) AB ^ 2 = x ^ 2 + y ^ 2-2xycos60 = > x ^ 2 + y ^ 2-2xy / 2 = 3 x ^ 2 + y ^ 2 = 3 + XY (BC + AC) ^ 2 = (x + y) ^ 2 = x ^ 2 + y ^ 2 + 2XY = 3 + XY + 2XY = 3 + 3xy = 11bc + AC = radical 11