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3 years ago

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Given the function h(x)=x2+6x+6h(x)=x^2+6x+6h(x)=x2+6x+6, determine the average rate of change of the function over the interval

−7≤x≤2-7\le x \le 2−7≤x≤2.

1 answer:

Zanzabum3 years ago

6 0

Answer:

<h2>The answer is -11.</h2>

Step-by-step explanation:

The function is given by $h(x) = x^{2} - 6x + 6$ .

Rate of change refers to the ratio between the change of the dependent variable with the change of independent variable.

The minimum value of x is -7 and the maximum value is 2.

h(-7) = 49 + 42 + 6 = 97.

h(2) = 4 - 12 + 6 = -2.

The average rate of change is $\frac{h(-7) - h(2)}{-7 - 2} = \frac{97 + 2}{-9} = -11$ .

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Solve for y.<br> - 4+v=6

Leviafan [203]

Answer: v=10

Step-by-step explanation:

since you didn’t put a y on there, I solved for v.

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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

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

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

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3 years ago

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Anna11 [10]

Answer:

Step-by-step explanation:

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2 years ago

Put together, Albert and Jason have

Anon25 [30]

Answer:

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3 years ago

Christine is putting money into a savings account. She starts with $650 in the savings account, and each week she adds $60. Let

TEA [102]

Total amount of money in the savings account after 11 weeks is $1310.

Total money in the account =

Initial Money + Money added per week x Number of weeks

Given:

Initial Money = $650

Money added per week = $60

Total money in the account = S

Number of weeks = W

Substituting it in the above equation we get,

S = $650 + $60xW (General Equation)

Total amount of money in the savings account after 11 weeks

S = $650 + $60x11

S = $650 + $660

S = $1310

Thus total amount of money in the savings account after 11 weeks is $1310.

Learn more about Linear Equations here :

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2 years ago