Which of the following explains how ΔBEI could be proven similar to ΔCEH using the AA similarity postulate? Quadrilateral ABDC, in which point F is between points A and C, point G is between points B and D, point I is between points A and B, and point H is between points C and D. A segment connects points A and D, a segment connects points B and C, a segment connects points I and H, and a segment connects points F and G. Segments AD, BC, FG, and IH all intersect at point E.

∠BEI ≅ ∠CEH because vertical angles are congruent; rotate ΔCEH 180° around point E, then translate point C to point B to confirm∠IBE ≅ HCE.
∠BEI ≅ ∠CEH because vertical angles are congruent; reflect ΔCEH across segment FG, then translate point C to point B to confirm∠IBE ≅ HCE.
∠BEI ≅ ∠CEH because vertical angles are congruent; rotate ΔCEH 180° around point E, then dilate ΔCEH to confirm segment EB ≅ segment EC.
∠BEI ≅ ∠CEH because vertical angles are congruent; reflect ΔCEH across segment FG, then dilate ΔCEH to confirm segment EH ≅ segment EI.

The correct option is;

∠BEI ≅ ∠CEH because vertical angles are congruent; rotate ΔCEH 180° around point E, then translate point C to point B to confirm ∠IBE ≅ ∠HCE

Step-by-step explanation:

The Angle-Angle (AA) similarity postulate states that two triangles are similar if two interior angles of one triangle are equal to two interior angles of the other triangle

The given triangles ΔCEH and ΔBEI have ∠CEH and ∠BEI as vertically opposite angles which are congruent

Rotating ΔCEH by 180° around point E will have segment EH coincide with segment EI

Translating point C to point B after rotation will have segment CH coincide with BI and segment CE coincide with segment BE, therefore, ∠ECH ≅ ∠EBI and ΔCEH and ΔBEI are then similia by having two congruent interior angles, ∠BEI ≅ ∠CEH and  ∠ECH ≅ ∠EBI by the AA similarity postulate.

A) ∠BEI ≅ ∠CEH because vertical angles are congruent; rotate ΔCEH 180° around point E, then translate point C to point B to confirm ∠IBE ≅ ∠HCE

Step-by-step explanation:

i have taken the test and it was correct.

all about the same size in the city the game was

step-by-step explanation:

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.

-h=2 ⇒h=-2

k=4

so, vertex= (-2, 4)

as, a=1 which is greater than 0 so parabola is opened upward and the graph has minimum.

hope this : )