Alexi transformed figure L such that its image is figure K after a 90degree counterclockwise rotation about the origin and a reflection over the y-axis. On a coordinate plane, triangle K has points (negative 3, 1), (negative 1, 2), (negative 2, 5). Triangle L has points (2, negative 1), (1, negative 3), (5, negative 2).

Jonathan used figure L to perform the same transformation, but he reflected figure L over the y-axis before performing the 90degree counterclockwise rotation. Which best compares Alexi’s and Jonathan’s transformed figures?
The change in order will not affect the final image. Their transformed figures will be identical.
The change in order will result in the transformed images having vertices in the same location, but in a different orientation.
The change in order will result in the transformed image of figure L being transformed back onto figure L.
The change in order will result in the transformed image of figure L having a different location.

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

step-by-step explanation: