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2 years ago

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A triangle has vertices at B(-3,0), C(2, -1), D(-1, 2). Which transformation would produce an image with vertices B"(1, -2), C"(

0, 3), D"(3, 0) (1 point)
(X,Y) - (x + 1, Y + 1), (X,Y) - (-X, Y)
(x,y) - (x, y), (x, y) - (x + 2, Y + 2)
(x,y) - (-X, Y), (X,Y) - (x + 2, Y + 2)
(x,y) - (x + 1, Y + 1), (X,Y) - (y,x)
Will mark brainalist

2 answers:

telo118 [61]2 years ago

3 0

Answer:

(x,y) - (x + 1, Y + 1), (X,Y) - (y,x)

Step-by-step explanation:

Take B (-3, 0) as example

(-3,0) -> (-2, 1) -> (1, -2)

aniked [119]2 years ago

3 0

Answer:

(x,y) - (x + 1, Y + 1), (X,Y) - (y,x)

Step-by-step explanation:

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Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea

lord [1]

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a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

$\int\limits^y_0 {kx^{2} } \, =1$

where y=23

Clearing k=0.00025

a) $Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5$

b) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2} =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6$

c) The function is equal to:

f(x)=k(1+2x)

$\int\limits^y_0 {k(1+2x)} \, =1$

where y=20

k=0.0024

$Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3$

d) $Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2} =198.4-176.89=21.51$

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