Answer:
Step-by-step explanation:
First we have to find the coordinates of the original ordered pairs then we will do the translation.
Now we will add 3 to every x value and add a -2 to every y value.
When you do that you will get the new prime points.
y = x^2 -4x
x = -1
y = (-1)^2 - 4×-1=1+4 = 5
x= 0
y = (0)^2 - 4×0 = 0
x = 1
y = 1^2 -4×1 = 1-4 = -3
x = 2
y = 2^2 -4×2 = 4-8 = -4
x=3
y = 3^2 - 4×3 = 9-12 = -3
x = 4
y = 4^2 - 4×4 = 16 - 16 = 0
now 2nd equation
y = 2x^2 + x
x = -2
y = 2 (-2)^2 + (-2)= 8-2 = 6
x = -1
y = 2 (-1)^2+(-1)= 2-1 = 1
x = 0
y = 2(0)^2 +0 = 0
x = 1
y = 2 (1)^2 + 1 = 3
x = 2
y = 2(2)^2+2= 8 + 2 = 10
Slope formula: y2-y1/x-x1
We can use the given points to solve.
3-(-3)/7-5
6/2
3
Best of Luck!
B a counterclockwise rotation about the origin of 90°
under a counterclockwise rotation about the origin
a point ( x , y ) → (- y, x)
figure Q to figure Q'
( 4,2 ) → (- 2, 4 )
(7, 5 ) → (- 5, 7 )
(3, 7 ) → (- 7 , 3 )
(2, 4 ) → (- 4, 2 )
(5, 4 ) → (- 4, 5 )
the coordinates of the original points of the vertices of Q map to the corresponding points on the image Q'
(a) First find the intersections of
and
:
So the area of
is given by
If you're not familiar with the error function
, then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.
(b) Find the intersections of the line
with
.
So the area of
is given by
which is approximately 1.546.
(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve
and the line
, or
. The area of any such circle is
times the square of its radius. Since the curve intersects the axis of revolution at
and
, the volume would be given by
