If the value of 1-m is greater than - 1 and not greater than 3, then the value range of M is () A. -1＜m≤3B. -3≤m＜1C. -2≤m＜2D. -2＜m≤2

If the value of 1-m is greater than - 1 and not greater than 3, then the value range of M is () A. -1＜m≤3B. -3≤m＜1C. -2≤m＜2D. -2＜m≤2

The value of ∵ algebraic formula 1-m is greater than - 1, but not greater than 3, ∵ 1 − m ＞ − 1, ① 1 − m ≤ 3, ②, from ①, m < 2, from ②, m ≥ - 2, so the value range of M is: - 2 ≤ m < 2
How to prove that the square of cosx plus the square of SiNx equals 1
It is proved by the unit circle that the unit circle is the function value of the angle between the radius of a circle made by unit 1 and the x-axis in the rectangular coordinate system
sinβ=y/1=y;cosβ=x/1=x
sin^2β+cos^2β=x^2+y^2=1
Similarly, for angles larger than 90 & amp; ordm;, sin α = y; Cos α = x, only x is negative, which is why the cosine value of obtuse angle is negative. Similarly, sin ^ 2 α + cos ^ 2 α = x ^ 2 + y ^ 2 = 1
So the square of cosx plus the square of SiNx equals 1, X ∈ R
Five and three eighths is a fraction
forty-three-eighths
Simplification: Tan (π / 4 + a) cos2a / 2cos ^ 2 (π / 4-A) online, etc
The solution [1] Tan [(π / 4) + a] = (1 + Tana) / (1-tana) = (COSA + Sina) / (COSA Sina) [2] cos2a = cos & # 178; a-SiN & # 178; a = (COSA + Sina) (COSA Sina) [3] 2cos & # 178; [(π / 4) - A] = 1 + cos {2 [(π / 4) - A]} = 1 + cos [(π / 2) - 2A] = 1 + sin2a = (COSA + Sina) & #
Why is the square of SiNx + the square of cosx equal to 1
I know that the square of SiNx + the square of cosx is 1,
Let the opposite side of X be a, the adjacent side B, and the hypotenuse C
sinx=a/c,sin＾2x=a＾2/c＾2
cosx=b/c,cos＾2x=b＾2/c＾2
∴sin＾2x＋cos＾2x＝a＾2/c＾2＋b＾2/c＾2＝（a＾2+b＾2)/c＾2
And a ＾ 2 + B ＾ 2 = C ＾ 2
∴sin＾2x＋cos＾2x＝1
Six centimeters is a fraction of a decimeter
Why, why
One decimeter equals 10 cm. Six tenths of a centimeter is three fifths of a meter
Six out of ten?
1 decimeter = 10 cm
So 6cm = 6 / 10 decimeters. = three fifths of a decimeter
6/10=3/5
three out of five
10 cm = 1 decimeter
6/10=3/5
3/5
Because 10 cm = 1 decimeter
6 cm = 6 / 10 decimeter = 3 / 5 decimeter
That's 10 / 5 decimeters,
Because 10 centimeters is one decimeter
3 / 5 decimeter
1 meter = 10 decimeters
1 decimeter = 10 cm
So 6 cm = 6 divided by 10 = 3-5 decimeters
6/10=3/5
Simplification: (sin α) ^ 2 + (sin β) ^ 2 - (sin α) ^ 2 (sin β) ^ 2 + Cos2 α Cos2 β
sinα=(1-cos2α)/2
(sinα)^2+(sinβ)^2-(sinα)^2(sinβ)^2+cos2αcos2β
=(1-cos2α)/2+(1-cos2β)/2-(1-cos2α)/2*(1-cos2α)/2+cos2αcos2β
=(1+cos2αcos2β)*0.75
Integral of 1 / (1-cosx) (SiNx) square
It's the square of SiNx at the denominator,
(sinx)2=1-(cosx)2=(1+cosx)*(1-cosx)
Expression = 1 + cosx
On the integral of 1 + cosx
The result is x + SiNx + C
(SiNx) 2 is the square of the sine
(cosx) 2
Results = Cotx + (Cotx) ~ 3 / 3 - (CSCX) ~ 3 / 3
The original formula = ︴ (1 + cosx) DX / (SiNx) ~ 4
=︴dx/(sinx)~4+︴cosxdx/(sinx)~4
=︴dx/(sinx~4)+︴d(sinx)/(sinx)~4
=︴dx/(sinx)~4-1/3(sinx~3)
Now find ︴ DX / (SiNx) ~ 4
... unfold
Results = Cotx + (Cotx) ~ 3 / 3 - (CSCX) ~ 3 / 3
The original formula = ︴ (1 + cosx) DX / (SiNx) ~ 4
=︴dx/(sinx)~4+︴cosxdx/(sinx)~4
=︴dx/(sinx~4)+︴d(sinx)/(sinx)~4
=︴dx/(sinx)~4-1/3(sinx~3)
Now find ︴ DX / (SiNx) ~ 4
=︴(cscx)~4dx
=︴(cscx)~2[1+(cotx)~2]dx
=︴[1+(cotx)~2]d(cotx)
=cotx+(cotx)~3/3
So the original formula is finally Cotx + (Cotx) ~ 3 / 3 - (CSCX) ~ 3 / 3
24 cm is a fraction of 5 decimeters?
5 decimeters = 50 cm
24/50=12/25
24 cm is 12 / 25 of 5 decimeters
Using Tan α / 2 = sin α / 1 + cos α = 1-cos α / sin α to simplify 1 + sin2 α - Cos2 α / 1 + sin2 α + Cos2 α
(1+sin2α-cos2α)/(1+sin2α+cos2α)=(1+sin2α-cos2α)/(1+sin2α+cos2α)=[(sinα)^2+(cosα)^2+2sinαcosα-(cosα)^2+(sinα)^2]/(1+sin2α+cos2α)=2sinα(sinα+cosα)/(1+sin2α+cos2α)=2sinα(sinα+cosα...