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3 years ago

12

M-n÷4

smiddle" class="latex-formula">

1 answer:

igor_vitrenko [27]3 years ago

4 0

The expression you have is already in its simplest form. However, there is another way to write it.

m - n/4

This expression is the same as the original, apart from the fact that we wrote the division as a fraction.

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How do cows count<br> [riddle]

crimeas [40]

Answer:

Step-by-step explanation:

I think the answer is utterly perfect

Lol or its a 'cow'nter

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3 years ago

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What is the vertex of the graph y = x^2 + 6x - 5 ?

Dafna11 [192]

Given: <span>y = x^2 + 6x - 5. Then a = 1, b = 6 and c = -5.

The x-coord. of the vertex is given by x = -b / (2a), which here is x = -6 / (2*1) = -3.

Use the given formula </span><span>y = x^2 + 6x - 5 to find the value of y when x = -3:

y = (-3)^2 + 6(-3) - 5 = 9 - 18 - 5 = -14

Then the vertex is (-3, -14).</span>

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3 years ago

(3x2)(−2x4) = 3(−2)x2∙4 = 6x8 can you find whats wrong with this problem

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3 years ago

Given the function f(x) =-2x^2 + 4x -7 find f(-4)

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Correct the answer is -55

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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

siniylev [52]

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

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

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

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,

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

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

$\ln y=2 \ln x$

⇒ y = kx^2 for some constant k

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

$\Rightarrow kx_0^2=y_0$

$\Rightarrow k =\frac{y_0}{x_0^2}$

∴ 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