Answer:
Number of marbles with Williams at the starting = 192
Step-by-step explanation:
Let total number of marbles with Williams = m
He lost marbles = 25% of the total number
= 
= 
Remaining marbles with Williams = m -
= 
He gave marbles to his brother = 
th of the remaining marbles
= 
= 
Remaining marbles after giving marbles to his brother = 120
Therefore, 
m = 
m = 192
Number of marbles with Williams at the starting = 192
B- Not sure if it's a line
C- y=(-x/2)-6
D- y=4x-12
E- y=(2/5)x-10
The answer is C. 250.80
change the percent to a decimal (0.20) and multiply 1,254x0.20=250.80
B agree
A,c, d don’t make sense
Answer:
√3 is irrational
Step-by-step explanation:
The location of the third point of a triangle can be found using a rotation matrix to transform the coordinates of the given points.
<h3 /><h3>Location of point C</h3>
With reference to the attached figure, the slope of line AC is √3, an irrational number. This means the line AC <em>never passes through a point with integer coordinates</em>. (Any point with integer coordinates would be on a line with rational slope.)
<h3>Equilateral triangle</h3>
The line segments making up an equilateral triangle are separated by an angle of 60°. If two vertices are on grid squares, the third must be a rotation of one of them about the other through an angle of 60°. The rotation matrix is irrational, so the rotated point must have irrational coordinates.
The math of it is this. For rotation of (x, y) counterclockwise 60° about the origin, the transformation matrix is ...
![\left[\begin{array}{cc}\cos(60^\circ)&\sin(60^\circ)\\-\sin(60^\circ)&\cos(60^\circ)\end{array}\right] \left[\begin{array}{c}x\\y\end{array}\right]=\left[\begin{array}{c}x'\\y'\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ccos%2860%5E%5Ccirc%29%26%5Csin%2860%5E%5Ccirc%29%5C%5C-%5Csin%2860%5E%5Ccirc%29%26%5Ccos%2860%5E%5Ccirc%29%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dx%27%5C%5Cy%27%5Cend%7Barray%7D%5Cright%5D)
Cos(60°) is rational, but sin(60°) is not. For any non-zero rational values of x and y, the sum ...
cos(60°)·x + sin(60°)·y
will be irrational.
As in the attached diagram, if one of the coordinates of the rotated point (B) is zero, then one of the coordinates of its image (C) will be rational. The other image point coordinate cannot be rational.
