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3 years ago

HELPPP PLSSS 18. A triangle has verticesA(4,3), B(6,9) and C(2, -3). The triangle is rotated 90°

Mathematics

1 answer:

Mariulka [41]3 years ago

8 0

Answer:

F and H are NOT true

Step-by-step explanation:

Rotation 90° CCW is represented by the transformation ...

(x, y) ⇒ (-y, x)

Then the rotated coordinates are ...

A(4, 3) ⇒ A'(-3, 4) . . . . . quadrant II

B(6, 9) ⇒ B'(-9, 6) . . . . . quadrant II

C(2, -3) ⇒ C'(3, 2) . . . . . quadrant I

The rotated figure will be congruent to the original, because rotation is a rigid transformation.

Consider the answer choices:

F: (-3, -4) is A' --- NOT true

G: triangles are congruent --- true

H: B is in quadrant III --- NOT true

J: (3, 2) is the coordinates of C' --- true

Both choices F and H are NOT true.

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2 years ago

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

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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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3 years ago