33 and 242 least common multiple and greatest common factor 4, 15 and 9 least common multiple and greatest common factor

33 and 242 least common multiple and greatest common factor 4, 15 and 9 least common multiple and greatest common factor

33=11*3
242=11*22
Least common multiple = 11 * 22 * 3 = 726
Greatest common factor = 11
15=3*5
9=3*3
Least common multiple = 3 * 5 * 3 = 45
Greatest common factor = 3
The least common multiple of 1 and 7 is 7, and the greatest common factor is 1
The least common multiple of 9 and 7 is 63, and the greatest common factor is 1
The least common multiple of 4 and 8 is 8, and the greatest common factor is 4
The least common multiple of 6 and 8 is 24, and the greatest common factor is 2
The least common multiple of 5 and 10 is 10, and the greatest common factor is 5
The least common multiple of 9 and 10 is 90 and the greatest common factor is 1
33 and 242 LCM = 726
Greatest common factor = 11
4. 15 and 9 LCM = 180
Greatest common factor = 1
thank you
726 11
180 1
How to find the least common multiple and the greatest common factor of 4, 5 and 6
The least common multiple of 4,5,6 is 60 and the greatest common divisor is 1
If the complex Z satisfies 1-z / 1 + Z = I, then the value of | Z + 1 | is
Let z = a + bi (a, B are real numbers, and B ≠ 0) (1-z) / (1 + Z) = i1-z = (1 + Z) i1-a-bi = (1 + A + bi) I collate, we get (a-b-1) + (a + B + 1) I = 0a-b-1 = 0A + B + 1 = 0, we get a = 0, B = - 1z = - I, Z + 1 = 1-I | Z + 1 | = √ [1 | 178; + (- 1) | 178;] = √ 2 | Z + 1 |
Given that z is a complex number, Z + 2I and Z2 − I are both real numbers, where I is an imaginary number unit. (I) find the complex number Z; (II) if the point corresponding to the complex number (Z + AI) 2 in the complex plane is in the first quadrant, find the value range of the real number a
In this paper, Z + 2I = Z + 2I = a + bi + 2I = a + bi + 2I = a + bi + 2I = a + (B + 2) I = a + (B + 2) I = a + (B + 2) I = a + bi (a, B ∈ R) from the title, Z + 2I = Z + 2I = a + bi + bi (a, B ∈ R), Z + 2I = a + bi + bi + 2I = a + (B + 2) I = a + bi + bi + bi + bi + 2I = a + (B + bi + 2I + 2I = a + 2I = 4-4-2i. (Ⅱ) from (I) from (I) we can know that z = 4-4-4-2i, we can know z = 4-4-2i-4-2i, we know that z = 4-4-4-2i, we know that z = 4-4-2i, from (Ⅱ) from (Ⅱ) by (in the first quadrant of the complex plane, 16 -- (a − 2) 2 > 08 (a − 2) > 0, the range of a is 2 < a < 6
What is the exponential form of complex number Z = 1 / 2 + 1 / 2I and how to find the argument?
Let Z be a complex plane
If z = a + bi, then:
The form of Z = re ^ (I θ) is called the exponential form of complex number, where:
If R is Z, θ is the principal value of the radiation angle, and - π
z=1/2+1/2i
=√2/2(cosπ/4+isinπ/4)
=√2/2e^(iπ/4)
a+bi=pe^iθ
p= √(a^2+b^2)
tanθ=b/a
Here Tan θ = (1 / 2) / (1 / 2) = 1, θ = π / 4
p=√(0.5^2+0.5^2)=√2/2
z. If u ∈ complex number, Z ≠ u, | Z | = 1, then the value of | (z-u) / {1 - (conjugate complex number of Z) * u} is?
Use Z 'to denote the conjugate complex number of Z
|(z-u) / (1-z'u) | (numerator and denominator multiplied by Z at the same time)
=|(z-u)z/[z(1-z'u)]|
=|(z-u) Z / (z-zz'u) | (note that | Z | = 1, ZZ '= | Z | ^ 2 = 1)
=|(z-u)z/(z-u)|
=|z|
=1
That is | (z-u) / (1-z'u) | = 1
Argz is the argument of Z, where a = Arg (2 + I), B = Arg (- 1 + 2I), then sin (a + b) =?
The correct answer is 3 / 5!
tgA=1/2,tgB=-2
sinA=1/5^(1/2),cosA=2/5^(1/2)
sinB=2/5^(1/2),cosB=-1/5^(1/2)
sin(A+B)=sinAcosB+sinBcosA
=(1/5^(1/2))*(-1/5^(1/2))+(2/5^(1/2))*(2/5^(1/2))
=-1/5+4/5=3/5
An urgent plural question~
The corresponding points of complex z = (k2-3k + 2) + (4-k2) I in the complex plane are at the following positions:
1, on the imaginary axis excluding the origin of coordinates
2. In the second quadrant
Virtual axis excluding coordinate origin
k^2-3k+2=0
k=1,k=2
4-K ^ 2 is not equal to 0
So k = 1
In the second quadrant
k^2-3k+2
If | Z | = 1 is known, what is the value range of the module of the complex u = Z ^ 2-1 + I
Given | Z | = 1, what is the value range of the module of the complex u = Z ^ 2-1 + I? Ask for detailed explanation
A plural question in Senior High School
It is known that the equation 2x ^ 2 + BX + C = 0 (B, C ∈ R) about X has an imaginary root, which is root 2 - root 3I. Find another root of the equation and the value of B, C
The real coefficient equation is the radical conjugate imaginary number
So the other one is √ 2 + √ 3I
By Weida theorem
x1+x2=-b/2
x1x2=c/2
So B = - (x1 + x2) = - 2 √ 2
c=2x1x2=2(2+3)=10
The other root is root 2 + root 3I
From the Weida theorem
Two sum = 2 root sign 2 = - B / 2, so B = - 4 root sign 2
Two product = 5 = C / 2, so C = 10
It can be seen from the problem that this is a real coefficient equation. So there are two conjugate imaginary roots, so the other root is √ 2 + √ 3I. According to Weida's theorem, if the sum of the two is (- B / 2), B = - 4 √ 2 can be obtained. If the product of the two is (C / 2), C = 10 can be obtained
.... It's said that B / C belongs to R, and my answer should be the right one!!!! I have confidence