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velikii [3]
3 years ago
5

What is the standard form for circles?

Mathematics
1 answer:
MrRissso [65]3 years ago
5 0

Answer:circles

Step-by-step explanation:

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A thin metal plate, located in the xy-plane, has temperature T(x, y) at the point (x, y). Sketch some level curves (isothermals)
Sophie [7]

Answer:

Step-by-step explanation:

Given that:

T(x,y) = \dfrac{100}{1+x^2+y^2}

This implies that the level curves of a function(f) of two variables relates with the curves with equation f(x,y) = c

here c is the constant.

c = \dfrac{100}{1+x^2+2y^2} \ \ \--- (1)

By cross multiply

c({1+x^2+2y^2}) = 100

1+x^2+2y^2 = \dfrac{100}{c}

x^2+2y^2 = \dfrac{100}{c} - 1 \ \  -- (2)

From (2); let assume that the values of c > 0 likewise c < 100, then the interval can be expressed as 0 < c <100.

Now,

\dfrac{(x)^2}{\dfrac{100}{c}-1 } + \dfrac{(y)^2}{\dfrac{50}{c}-\dfrac{1}{2} }=1

This is the equation for the  family of the eclipses centred at (0,0) is :

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

a^2 = \dfrac{100}{c} -1  \ \ and \ \ b^2 = \dfrac{50}{c}- \dfrac{1}{2}

Therefore; the level of the curves are all the eclipses with the major axis:

a =  \sqrt{\dfrac{100 }{c}-1}  and a minor axis b =  \sqrt{\dfrac{50 }{c}-\dfrac{1}{2}}  which satisfies the values for which 0< c < 100.

The sketch of the level curves can be see in the attached image below.

7 0
4 years ago
What is the answer???
frez [133]
B will b the answer that's the only answer yuh can get is b
4 0
3 years ago
Read 2 more answers
5a+4c-b<br><img src="https://tex.z-dn.net/?f=5a%20%2B%204c%20-%20b%20" id="TexFormula1" title="5a + 4c - b " alt="5a + 4c - b "
Tatiana [17]
Hi,

5a + 4c - b = 5a + ( - 4bc) \\ 5a - 4bc = 1abc

Hope this helps.
r3t40
8 0
3 years ago
Round 100.9158 to the nearest whole number
Natalka [10]
The answer would be 101
6 0
3 years ago
What is the simplest form of the equation ?<br><br> csc x<br> ——-<br> cos x
larisa86 [58]

Answer:

CBC

Step-by-step explanation:

ez

8 0
3 years ago
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