I NEED HELPP?!?! SEQUENCE OF Transformations
All rotations are about the origin
Translate
(x – 2, y + 1) then
rotate 90°
counterclockwise
(-1, 1 )
Translate
(x -3, y - 3) then
translate
(x– 6, y – 6)
(5, 0)
Rotate 90°
counterclockwise
then reflect
across the x-axis
(-3, -3)
Rotate 180° then
reflect across
the y-axis
(5, -4)
Rotate 90°
clockwise then
translate 7 units
up
(0, 5)
Rotate 180° then
rotate 270°
counterclockwise
(2, -5)
Reflect across
the y-axis then
rotate 90°
clockwise
(-3, -5)
Translate
(x + 5, y – 3) then
rotate 90°
counterclockwise
(2, 5)
Translate
(x – 1, y + 3) then
reflect across
the x-axis
(3, -1 )
Reflect across
the y-axis then
translate 9 units
to the right
(-3, -1 )
Reflect across
the x-axis then
reflect across
the y-axis.
(0, 4)
Rotate 270°
counterclockwise
then translate
(x – 2, y + 9)
(-2, -4)
Rotate 180° then
translate
(x + 4, y)
(5, -2)
Reflect across
the x-axis then
rotate 90°
clockwise
(-4, 1 )
Translate 7 units
down then rotate
270°
counterclockwise
(5, 4)
Rotate 180° then
reflect across
the y-axis
(2, 6)
Translate
(x + 5, y – 1) then
translate
(x, y + 4)
(5, -2)
Reflect across
the y-axis then
rotate 180°
(3, -6)
Translate 4 units
to the left then
reflect across
the x-axis
(-1, 4)
(2, 3)
(-4, -9) (-3, 3)
(2, -1)
(-2, -3)
(2, -3) (5, -2) (3, -3) (3, 3) (5, -2)
(5, 2)
(5, -3)
(2, -7) (0, 0)
(-5, -2)
(3, 8) (0, -4) (-5, -3) (5, 2)
(-2, 7)
(2, -2) (3, -10) (-4, 0)
(-2, 6) (-2, -6) (0, 4)
(-1, 4)
(4, -2) (-2, 2) (12, -1)
(4, -1) (-1, -4) (6, 2)
(4, -4) (-5, -4) (2, -6) (1, 4) (-3,-6)
(1, -4)
(0, 1)
(10, 1) (3, 6) (5, -4)

v: l•w•h÷2.

v: 20•6•8 ÷2

v: 960÷2

v: 480

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 : )

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

where are the choices?