Answer:
b. 4x^2 + 3x - 6
Step-by-step explanation:
The values of f(x) for the extremes of x are more positive than the value of f(x) for the middle x, so we know the parabola opens upward. That eliminates choice D.
It is probably easiest to evaluate the other expressions to see which one matches the given f(x) values. For the purpose, it is usually easier to use the Horner form of the equation.
a. f(-2) = (3(-2) +4)(-2) -6 = -2(-2) -6 = -2 ≠ 4
b. f(-2) = (4(-2) +3)(-2) -6 = -5(-2) -6 = 4 . . . . matches the given data point
c. Because (b) matches, we know this one will not.
The appropriate choice is B.
There's nothing preventing us from computing one integral at a time:
![\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B2-x%7D%20xyz%20%5C%2C%5Cmathrm%20dz%20%3D%20%5Cfrac12xyz%5E2%5Cbigg%7C_%7Bz%3D0%7D%5E%7Bz%3D2-x%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac12xy%282-x%29%5E2)
![\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B1-x%7D%5Cint_0%5E%7B2-x%7Dxyz%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%20%3D%20%5Cfrac12%5Cint_0%5E%7B1-x%7Dxy%282-x%29%5E2%5C%2C%5Cmathrm%20dy%20%5C%5C%5C%5C%20%3D%20%5Cfrac14xy%5E2%282-x%29%5E2%5Cbigg%7C_%7By%3D0%7D%5E%7By%3D1-x%7D%20%5C%5C%5C%5C%3D%20%5Cfrac14x%281-x%29%5E2%282-x%29%5E2)
![\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E%7B1-x%7D%5Cint_0%5E%7B2-x%7Dxyz%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%20%3D%20%5Cfrac14%5Cint_0%5E1x%281-x%29%5E2%282-x%29%5E2%5C%2C%5Cmathrm%20dx)
Expand the integrand completely:
![x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x](https://tex.z-dn.net/?f=x%281-x%29%5E2%282-x%29%5E2%20%3D%20x%5E5-6x%5E4%2B13x%5E3-12x%5E2%2B4x)
Then
![\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac14%5Cint_0%5E1x%281-x%29%5E2%282-x%29%5E2%5C%2C%5Cmathrm%20dx%20%3D%20%5Cleft%28%5Cfrac16x%5E6-%5Cfrac65x%5E5%2B%5Cfrac%7B13%7D4x%5E4-4x%5E3%2B2x%5E2%5Cright%29%5Cbigg%7C_%7Bx%3D0%7D%5E%7Bx%3D1%7D%20%5C%5C%5C%5C%20%3D%20%5Cboxed%7B%5Cfrac%7B13%7D%7B240%7D%7D)
It's suppose to be -18 so I guess no real number....
First add 2y to both sides
so 4 + 3x + 2y = 0
then subtract 4 from both sides
so 3x + 2y = -4
that is standard form
now subtract 3x from both sides
2y = 3x - 4
now divide by 2 on both sides
y = 1.5x - 2
This is solved for y and is slope intercept form
starting from original subtract 4 from both sides
so 3x = -2x -4
now divide by 3 on both sides
so x = -2/3x -4/3
this is solved for x
hope any of that is what you needed.
Answer:
A) 5.12mm³
Step-by-step explanation:
Step 1
Find the volume of the bigger pyramid
This is a triangular based pyramid.
The formula to use is given as Volume of Pyramid =
1/3 × Area of the triangle × Height
Area of the triangle = 1/2 × 3 mm × 4mm = 6mm²
Volume of the large pyramid = 1/3 × 6mm² × 5mm
= 10mm³
We are given the length of the small pyramid as 4mm
We would be using was is known as scale factor to find the volume of the small pyramid
The scale factor k = (Height of the small pyramid)³/ (Height of the Large pyramid)³
k = 4³/5³
k = Volume of small pyramid/ Volume of large pyramid
Volume of small pyramid = X
Volume of large pyramid = 10mm³
Hence,
4³/5³ = X/10
Cross Multiply
= 4³ × 10 = 5³ × X
X = (4³ × 10)/ 5³
X = 5.12mm³