If 7 / 2 minus 3x and 4 are opposite, then x = []

If 7 / 2 minus 3x and 4 are opposite, then x = []

(7-3x)/2+4=0
The equal sign is multiplied by two on both sides
7-3x+8=0
15=3x
x=5

7:4 is 3x out of 4

Solve the equation and get x = 7 / 3

3 / 4 x + 1 / 2 x is equal to 25, to the whole!
We need to work it out with the liberation process! X

3 out of 4 x + 1 out of 2 x is 25
5/4x=25
x=20

The monotone decreasing interval of the function y = sin (x + π 4) is___ ．

Let 2K π + π 2 ≤ x + π 4 ≤ 2K π + 3 π 2, K ∈ Z, obtain 2K π + π 4 ≤ x ≤ 5 π 4 + 2K π, so the monotone decreasing interval of function y = sin (x + π 4) is [2K π + π 4, 5 π 4 + 2K π], K ∈ Z, so the answer is: [2K π + π 4, 5 π 4 + 2K π], K ∈ Z

How to find the derivation of SiNx by definition?

According to the question: Δ y = sin (x + Δ x)
Then Δ Y / Δ x = [sin (x + Δ x) - SiNx] / Δ x = (sinxcos Δ x + sin Δ xcosx SiNx) / Δ x = cosxsin Δ X / Δ X
When Δ x → 0, sin Δ x = Δ x, so sin Δ X / Δ x = 1
So y '= cosx

The derivation process of SiNx

Use the definition
sinx'=lim△x->0 (sin(x+△x)-sinx)/△x
=lim△x->0 2cos[(2x+△x)/2]sin(△x/2)/△x
=lim△x->0 2cos[(2x+△x)/2]△x/2/△x
=cosx

When x belongs to (0, π / 2), we can prove x > SiNx by derivation

Let f (x) = x-sinx
f'(x)=1-cosx
If x belongs to (0, π / 2), then f '(x) > 0 is constant and f (x) increases in the domain of definition;
f(x)>f(0)=0
So x-sinx > 0 is x > SiNx

How to find the derivative of √ (SiNx & # 178;)?

=1/[2√(sinx²)]·(sinx²)'
=1/[2√(sinx²)]·cosx²·(x²)'
=1/[2√(sinx²)]·cosx²·2x
=xcosx²/√(sinx²)

Derivation: y = (SiNx) ^ n * SiNx ^ n

y'=((sinx)^n*sinx^n)'
=n(sinx)^(n-1)*cosx*sinx^n+(sinx)^n*(cosx^n)*n*x^(n-1)
=n(sinx)^(n-1)[cosxsinx^n+sinx*(cosx^n)*x^(n-1)]

Derivation y = SiNx ^ n

(sinx)'=cosx
(cosx)'=-sinx
x'=1
Constant '= 0
y'=cosx+sinx+1