The figure shows two triangles on a coordinate grid: A coordinate plane with two triangles is shown. Triangle RST has vertices R at 3 comma 4, S at 1 comma 1, and T at 5 comma 1. Triangle R prime S prime T prime has vertices R prime at 2 comma negative 3, S prime at negative 1 comma negative 1 and T prime at negative 1 comma negative 5. What set of transformations is performed on triangle RST to form triangle R’S’T’? A 90-degree counterclockwise rotation about the origin followed by a translation 2 units left A 270-degree counterclockwise rotation about the origin followed by a translation 2 units to the right A translation 2 units down followed by a 90-degree counterclockwise rotation about the origin A translation 2 units down followed by a 270-degree counterclockwise rotation about the origin Question 4(Multiple Choice Worth 4 points) (02.03)Figure EFGH is transformed to E’F’G’H’, as shown: A coordinate plane is shown. Figure EFGH has vertices E at negative 5 comma 4, F at negative 5 comma 1, G at negative 3 comma 2, and H at negative 1 comma 6. Figure E prime F prime G prime H prime has vertices at E prime 7 comma 4, F prime 7 comma 1, G prime 5 comma 2 and H prime 3 comma 6. Which of the following sequences of transformations is used to obtain figure E’F’G’H’ from EFGH? Counterclockwise rotation by 90 degrees about the origin followed by a translation right by 2 units Counterclockwise rotation by 90 degrees about the origin followed by a translation left by 2 units Reflection about the y-axis followed by a translation right by 2 units Reflection about the x-axis followed by a translation left by 2 units Question 5(Multiple Choice Worth 4 points) (02.03)Which statement is true about the result of a rigid transformation? The image will be larger than the pre-image. The pre-image will be larger than the image. Each one is unique; no general statement can be made about size.

C,  "A 270-degree counterclockwise rotation about the origin followed by a translation 1 unit to the left" should be the right answer.

Step-by-step explanation:

Solution:

Vertices of Preimage that is Triangle ABC =A(4,-2), B(3,-2),C(3,-5)

Vertices of image , i.e Δ A'B'C'=A'(1,2),B'(2,2), C'(2,5)

Now, transalation by 5 units left of triangle ABC takes it to

A(4,-2)=(4-5,-2)=(-1,-2)

B(3,-2)=(3-5,-2)=(-2,-2)

C(3,-5)=(3-5,-5)=(-2,-5)

When preimage is rotated through an angle of 180 degree, counterclockwise vertices of triangle (-1,-2),(-2,-2) and (-2,-5) goes to image A'(1,2),B'(2,2), C'(2,5).

Option C: A translation 5 units to the left followed by a 180 degree counterclockwise rotation about the origin takes ΔABC to  ΔA'B'C'.

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 : )

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.