I've attached a plot of one such cross-section (orange) over the region in the x-y plane (blue), including the bounding curves (red). (I've set
for this example.)
The length of each cross section (the side lying in the base) has length determined by the horizontal distance
between the y-axis
and the curve
. In terms of
, this distance is
. The height of each cross section is twice the value of
, so the area of each rectangular cross section should be
.
This means the volume would be given by the integral
X = t/2. t = 2x
y = 3t^3 - 1
y = 3(2x)^3 -1
y = 24x^3 -1
<span>slope
m = (1 - 1)/(0 + 2) = 0
(x1, y1) as (-2, 1)
</span><span>point-slope form</span><span>
y - y1 = m(x - x1)
y - 1 = 0(x + 2)
y - 1 = 0
y = 1</span>
3x^2 + 11x - 4 = 0
x^2 + (11/3)x = 4/3
(x+11/6)^2 = 169/36
Answer is B.
Answer:
See below
Step-by-step explanation:
Here is how to start....you didn't include the equations in question, so I do not know what the answer may be
solve for x
y-10 = 5 x^2 divide thru by 5
(y-10)/5 = x^2 'sqrt' both sides
± sqrt( (y-10) / 5 = x change x's and y's
± sqrt ((x-10)/5) = y