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3 years ago

(40 Points + Brainliest for right answer only.)

Mathematics

1 answer:

Phantasy [73]3 years ago

4 0

The answer to this problem is D. y=-5x+7.
The slope intercept form is y=mx+b. Since you already know your m and b, you simply plug it in.

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Simplify 2⁰+5¹+4³/7

Mkey [24]

Answer:

Step-by-step explanation:

2^0 = 1

5^1 = 5

4^3 = 4 x 4 x 4

=> 16 x 4

=> 64

=> 1 + 5 + 64 / 7

=> 6 + 64 / 7

=> 70 / 7

=> 10

So, the answer is 10

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tangare [24]

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Answer:

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2 years ago

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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

3 years ago