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bazaltina [42]
3 years ago
8

Find the area of the surface. the part of the paraboloid z=4-x^2-y^2 that lies above the xy-plane

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0
Parameterize the surface (call it \mathcal S) by

\mathbf s(u,v)=\langle u\cos v,u\sin v,4-u^2\rangle

with 0\le u\le2 and 0\le v\le2\pi. Then the surface element is

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

The area of \mathcal S is then given by the surface integral

\displaystyle\iint_{\mathcal S}\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv
=\displaystyle2\pi\int_{u=0}^{u=2}u\sqrt{1+4u^2}\,\mathrm du=\dfrac{(17^{3/2}-1)\pi}6\approx36.1769
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Let f(x) = 2x - 7 and g(x) = -6x - 3. Find f(x) + g(x) and state its domain.
Klio2033 [76]
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A baker wanted to enter the apple pie
dimaraw [331]

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3 years ago
Philip made a total of 9 bracelets and necklaces from 120 inches of cord. He used 8 inches of cord for each bracelet and 20 inch
Viktor [21]

Answer:

So Philip made 5 bracelets and 4 necklaces.

Step-by-step explanation:

Let x = number of bracelets and y = number of necklaces.

Since we have a total of 9 bracelets and necklaces,

x + y = 9 (1)

Also, we have 8 inches of cord for each bracelet and 20 inches of cord for each necklace, then the total length for the bracelet is 8x and that for the necklace is 20y.

So, the total length for both is 8x + 20y. Since the total length of cord used is 120 inches,

8x + 20y = 120 (2)

Simplifying it we have

2x + 5y = 30  (3).

Writing equations (1) and (3) in matrix form, we have

\left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}9\\30\end{array}\right]

Using Cramer's rule to solve for x and y,

x = det \left[\begin{array}{ccc}9&1\\30&5\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

x = (9 × 5 - 30 × 1) ÷ (1 × 5 - 1 × 2)

x = (45 - 30) ÷ (5 - 2)

x = 15 ÷ 3

x = 5

y = det \left[\begin{array}{ccc}1&9\\2&30\end{array}\right] /det \left[\begin{array}{ccc}1&1\\2&5\end{array}\right] \\

y = (30 × 1 - 9 × 2) ÷ (1 × 5 - 1 × 2)

y = (30 - 18) ÷ (5 - 2)

y = 12 ÷ 3

y = 4

So Philip made 5 bracelets and 4 necklaces.

3 0
3 years ago
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