Since AB is parallel to CD, then angle A and angle D are supplementary
And D = (1/2)A
So
A + (1/2)A = 180
(3/2)A =180
A = (2/3)(180) = 120°
*** edit ***
(23)x +10 = (25 )?
x2 -5x + 6 = 0 Factor
(x-3)(x-2) = 0 at least one of these factors must = 0
x-3 = 0
x = 3 or x-2 = 0
x = 2
You can also use the Quadratic Formula to solve this
a =1 b = -5 c = 6
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {-(-5) \pm \sqrt{(-5)^2-4(1)(6)} \over 2(1)}\)
Part of the question is missing.....
23 +10 = (25 )?
She is now 7+11 =18 m west
and 38 - 22 = 16 meters north
Pythag theorem
(distance from start)2 = 182 + 162
d2 = 324 +256
d = sqrt ( 580) = 24.08 m
guest, i think you're right, there is only 4 3 by 3 squares that include that black square. There are 9 squares in total, and the answerer might have been confused.
So the total number of squares that include that black square is 1 + 4 +4 +4 + 1 = 14 squares.
Neither digit is zero, 2 of each one
9C2*4!/(2!2!) = 9C2* 6 = 36*6 = 216
Zero plus one other number, 2 of each
9*3!/2! = 27
neither digit is zero, 3 of one and 1 of the other.
9C2*2*4 = 36*8 = 288
One digit is zero, 3 of one and 1 of the other.
9*3 = 27
216+27+288+27 = 558
Note: There is room here for error.
Q(2+i) = a(2+i) + b = 2a+b + ai
Since Q(2+i) = P(2+i) we must have 2a+b + bi = 4 - 3i
Equate real and imaginary parts Real: 2a + b = 4
Imag: b = -3
I'll leave you to finish this.
a^2 + bc = b^2 + ac
Rearrange as a^2 - b^2 = ac - bc
Factor as (a+b)(a-b) = (a-b)c
Simplify, a + b = c
Can you take it from here?
Use the calculator on the Home page here. Hover over the log button and you will see lg2 appear.
I have the same answer as Crypto, my method is just a bit different.
x=10mod 20 .... x=10, 30, 50, ............ , 230 ......
x=5mod45 .... x= 5, 50, 95, 140, 185, 230 ......
since these are both true and 230-50 = 180
x= 50+180k where k is an integer great or equal to 0
\(\frac{x}{6}=8+30k+\frac{2}{6}\\ so\\ x\mod6 = 2\)
10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 =
(5 * 2) * (3 * 3) * ( 4 * 2) * 7 * (3 * 2) * 5 * (2 * 2) * 3 * 2 =
(5 ) * (3^2) * 7 * 3 * 5 * 3 =
3^4 * 5^2 * 7^1
The number of odd divisors = add 1 to each exponent ...take the product of these sums = (4 + 1)(2 + 1) (1 + 1) = (5) (3) ( 2) = 30
(Just as the Guest said !!! )
Justin walks 1/10 of a lap in one minute
Keiko walks 1/6 of a lap in one minute
The LCM of 6 and 10 = 30 = the number of minutes until they meet again at the starting line
In that time Justin walks (1/10) * 30 = 3 laps
And Justin walks (1/6)*30 = 5 laps
The height of the triangle is the y coordinate of (6,4) = 4
80 = (1/2) base * height
80 =(1/2) base * 4
80 = 2 * base
80/2 = base = 40 = distance from (0,0) to the third vertex
The third vertex is (-40 , 0)
A =( 3,4) B =( 3 , -4) C = ( 3, 2 - - 4) = (3,6)
No triangle is formed....see here : https://www.desmos.com/calculator/kyasobika1
60
So,,,,, What
They are back at starting line:
J: 10 20 30 40 minutes
K: 6 12 18 24 30 36
So they meet again at 30 minutes at the starting line..... I think you can take it from here.......
THX, jugoslav !!!
angle YZX = 57.5 degrees
You get the minimum value when a = -2, b = 0, and c = 2. Then |a + bw + cw^2| = 2*sqrt(3).
(1, 3, 5, 7, 9, 15, 21, 25, 27, 35, 45, 63, 75, 81, 105, 135, 175, 189, 225, 315, 405, 525, 567, 675, 945, 1575, 2025, 2835, 4725, 14175)>>Total==30 such numbers
it's in all 9 of the 3 x 3 squares. How is that !? I see only 4 ......
Do you mean \(=\) or \(\equiv\)? If you mean \(\equiv\) then I can help you.
From the second equation, since \(x \equiv 9 \pmod{20}\), we know \(x\) is odd. Then \(r\) can be one of \(1\), \(3\), and \(5\). From the third equation, \(x \equiv 4 \pmod{45}\). We see that \(x \equiv 1 \pmod{3}\) (since \(x = 45y+4\), for some integer \(y\)). Then we see that \(r\) can only be \(\boxed1\) (since if \(x \equiv 3 \mod{6}\), it will have remainder of \(0\) when divided by \(3\) and if \(x \equiv 5 \mod{6}\), it will have a remainder of \(2\) when divided by \(3\).
Well, let's take a deeper look.
It's in the \(5\times5\) square.
it's in all \(4\) of the \(4\times4\) squares.
it's in all \(9\) of the \(3\times3\) squares.
It's in \(4\) of the \(2\times2\) squares.
It's in \(1\) of the \(1\times1\) squares.
That would leave us with a total of \(\boxed{19}\) squares.
