The image of (triangle) abc is (triangle) a'b'c. what transformations would result in this image? a.(triangle)abc is reflected over the y-axis, then is rotated -90° around the origin. b.(triangle)abc is reflected over the line y = x. c.(triangle)abc is rotated -90° around the origin, then is reflected over the x-axis. d.(triangle)abc is rotated 90° around the origin, then is reflected over y-axis.

The triangle ABC is reflected over the line y = x ⇒ answer B

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

- If point (x , y) rotated about the origin by angle 90° anti-clock wise

∴ Its image is (-y , x)

- If point (x , y) rotated about the origin by angle 90° clock wise

∴ Its image is (y , -x)

- If point (x , y) rotated about the origin by angle 180°

∴ Its image is (-x , -y)

* There is no difference between rotating 180° clockwise or  

anti-clockwise around the origin

* Lets find the vertices of ABC and A'B'C' to solve the problem

- In Δ ABC

# A = (5 , -5) , B = (5 , -4) , C = (2 , -4)

- In Δ A'B'C'

# A' = (-5 , 5) , B = (-4 , 5) , C = (-4 , 2)

∵ The image of (5 , -5) is (-5 , 5)

∵ The image of (5 , -4) is (-4 , 5)

∵ The image of (2 , -4) is (-4 , 2)

∴ The point (x , y) is (y , x)

- From the rule above

∴ The triangle ABC is reflected over the line y = x

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

a. 32 degrees

step-by-step explanation:

here we have to use the rule.

m∠e = 1/2 (major arc - minor arc)

m∠e = 1/2 (212 - 148)

m∠e = 1/2 (64)

m∠e = 32 degrees

answer: a. 32 degrees

you