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3 years ago

Find (fof)(3) f(x) = x^2 - x A. 30 B. 6 C. 33 D. -6

Mathematics

2 answers:

EastWind [94]3 years ago

6 0

Sub x=3 into ((x^2)-x)^2 - (x^2 - x) = 30. So the answer is A.

posledela3 years ago

5 0

${3}^{2} - 3 \\ 9 - 3 \\ = 6$

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Evaluate the integral. W (x2 y2) dx dy dz; W is the pyramid with top vertex at (0, 0, 1) and base vertices at (0, 0, 0), (1, 0,

In-s [12.5K]

Answer:

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

Step-by-step explanation:

Given that:

$\iiint_W (x^2+y^2) \ dx \ dy \ dz$

where;

the top vertex = (0,0,1) and the base vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0), and (1, 1, 0)

As such , the region of the bounds of the pyramid is: (0 ≤ x ≤ 1-z, 0 ≤ y ≤ 1-z, 0 ≤ z ≤ 1)

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 \int ^{1-z}_0 (x^2+y^2) \ dx \ dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \int ^{1-z}_0 ( \dfrac{(1-z)^3}{3}+ (1-z)y^2) dy \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^3}{3} \ y + \dfrac {(1-z)y^3)}{3}] ^{1-x}_{0}$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz = \int ^1_0 \ dz \ ( \dfrac{(1-z)^4}{3}+ \dfrac{(1-z)^4}{3}) \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =\dfrac{2}{3} \int^1_0 (1-z)^4 \ dz$

$\iiint_W (x^2+y^2) \ dx \ dy \ dz =- \dfrac{2}{15}(1-z)^5|^1_0$

$\mathbf{\iiint_W (x^2+y^2) \ dx \ dy \ dz = \dfrac{2}{15}}$

7 0

3 years ago

<img src="https://tex.z-dn.net/?f=c-15%3D31%5Btex%5D%5Csqrt%5Bn%5D%7Bx%7D%20%5Csqrt%5Bn%5D%7Bx%7D" id="TexFormula1" title="c-15=

ExtremeBDS [4]

Answer:

C = 31

Step-by-step explanation:

Solve for c by simplifying both sides of the equation, then isolating the variable.

3 0

3 years ago

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Can someone help with #17

iVinArrow [24]

B, Because to be a parallelogram it has to have two sides that are both parallel and congruent

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zhuklara [117]

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garri49 [273]

Answer:

Step-by-step explanation:

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3 years ago

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