Excuse me for not understanding. If the sequence consists of all positive multiples of 3 that contain at least one digit that is a 1, shouldn't numbers like 111, 114, 141, 117, 171, etc. belong to the sequence as well?
The probability is 4/51.
The length of arc YZ is 6*pi.
The intersection point is (-5,-1).
By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4). Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i. Then the other roots work out as
4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,
4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and
4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.
(a) See the graph below.
(b) g(x) = 2/3*f(3x - 1).
(c) We stretch the graph horiztonally by a factor of 3, then stretch the graph vertically by a factor of 1/2, then shift down 3 units.
The value of x is 115 (degrees).
There are five different prime numbers.
c = 4 + 4*sqrt(5)/5.
The area of triangle BCD is 20/3.
5-12-13 right triangle
A
10
5 D
3
C 12 B
sin ABC = 5/13
Area of BCD = (1/2) ( BC * DB sin (ABC) =
(1/2) (12) (3) ( 5/13) =
180 / 26 =
90 / 13 units^2
See here : https://web2.0calc.com/questions/help_39113
Interior angles sum to 360°
128 + ( 128 - d ) + (128 - 2d ) +(128 - 3d ) = 360
512 - 6d = 360
512 - 360 = 6d
152 = 6d
152 / 6 = d = 76/3
Smallest angle = 128 - 3 ( 76/3) = 128 -76 = 52°
z = a + bi z (complement) = a - bi i^2 = -1
3 (a + bi) + 4i (a - bi) = 3 - 3i
3a + 3bi + 4ai - 4bi^2 = 3 - 3i
3a + (4a + 3b)i + 4b = 3 - 3i
(3a + 4b) + (4a + 3b) i = 3 - 3i implies that
3a + 4b = 3 ⇒ 12a + 16b = 12 (1)
4a + 3b = -3 ⇒ -12a -9b = 9 (2)
Add ( 1) and (2)
7b = 21
b = 3
3a + 4 (3) = 3
3a + 12 = 3
3a = 3 - 12
3a = -9
3a = -9 / 3
a = -3
z = -3 + 3 i
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With replacement = (1/13)^4 = 1 / 28561
( 23 + 16 + 30 + 33 + x + x + 2) / 6 = 27
(104 + 2x) = 27*6
104 + 2x = 162
2x = 162 - 104
2x = 58
x = 58 / 2 = 29 x + 2 = 31
So.......16 is discarded
(23 + 29 + 30 + 31 + 33) / 5 = 146 / 5 = 29.2 = new mean
See here : https://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.1047022.html
Note that when f(x) = y = 1
We have the following
f(-3) = 1 f(-2) = 1 f (-1) = 1 f(1) = 1 f(2) = 1 f(3) = 1
Sum of all c^s = 6
Using Binet's formula, c = 2 + sqrt(5)/5.
The number of possible ways is 484.
The interior angle that is missing is 58 degrees.
Using coordinates, the area of triangle BCD is 120/7.
slope = (11 - (-17))/(5 - (-2)) = 28/7 = 4
y + 17 = 4(x + 2)
y + 17 = 4x + 8
y = 4x - 9
+4 
