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Step2247 [10]
3 years ago
15

PLEASE HELP MEE (look at pic question 1 & 2)

Mathematics
2 answers:
S_A_V [24]3 years ago
5 0
The product of -3x^ (3) +2x-4 + 3x-2 is
-3x ^3-5x+6
and the product of 3x^(3)+2x-4 + 2x-3 is
-3x ^ (3)-4x+7
2 they are not equal
8090 [49]3 years ago
3 0

Answer

1 The product of -3x^(3)+2x-4 + 3x-2 is -3x^3-5x+6

and the product of 3x^(3)+2x-4 + 2x-3 is -3x^(3)-4x+7

2 they are not equal

Step-by-step explanation:

Combine like terms the compare the product -3x^3-5x+6   -3x^(3)-4x+7

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8(5) - 4y = 39
40 - 4y = 39
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So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit. (Brainliest? :3)

Step-by-step explanation:

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3 years ago
Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w
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Solution :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

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which is a family of ellipses.

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Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.

$r(t)=x(t)i+y(t)j$  is a parametrization of the most $\text{rapid increase curve}$ , then

$\frac{dx}{dt}=\frac{dx}{dt}i+\frac{dy}{dt}j$ is a tangent to the curve.

So then we have that $\frac{dr}{dt}=\lambda \triangledown C$

$\Rightarrow \frac{dx}{dt}=-\frac{2\lambda x}{10^4}e^{-(x^2+2y^2)/10^4}, \frac{dy}{dt}=-\frac{4\lambda y}{10^4}e^{-(x^2+2y^2)/10^4} $

∴ $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2y}{x}$

Using separation of variables,

$\frac{dy}{y}=2\frac{dx}{x}$

$\int\frac{dy}{y}=2\int \frac{dx}{x}$

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⇒ y = kx^2 for some constant k

but we know that $y(x_0)=y_0$

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∴ The path of the shark will follow is along the parabola

$y=\frac{y_0}{x_0^2}x^2$

$y=y_0\left(\frac{x}{x_0}\right)^2$

7 0
3 years ago
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6 0
3 years ago
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