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elena-14-01-66 [18.8K]

3 years ago

13

R^2p÷5 ; p = -5, r = - 5

1 answer:

marshall27 [118]3 years ago

5 0

Answer:

=-5^2×(-5)÷5

=25×(-5)÷5

=-125÷5

=-25

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Solve for n. (7^2)^4 = n^8

enyata [817]

7^2x4 = n^8
7^8 = n^8
n = 7

4 0

3 years ago

What percentage of videos on the streaming site are between 39 and 189 seconds?

PolarNik [594]

Answer:

B) 15.85%

Step-by-step explanation:

normalcdf(minimum,maximum,μ,σ)

normalcdf(39,189,264,75)

≈0.1573

≈15.73%

Therefore, the closest answer is B

4 0

2 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

What simple interest rate would allow $6000 to grow to an amount of $14550 in 10 years?

Harman [31]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{ \: 14.25 \: \% \: }}}}}$

Step-by-step explanation:

Given,

Principal ( P ) = $ 6000

Amount ( A ) = $ 14550

Time ( T ) = 10 years

Rate ( R ) = ?

<u>Finding </u><u>the </u><u>Interest</u>

The sum of principal and interest is called an amount.

From the definition,

$\boxed{ \sf{Amount = \: Principal + Interest}}$

plug the values

⇒ $\sf{14550 = 6000 + Interest}$

Swap the sides of the equation

⇒ $\sf{6000 + Interest = 14550}$

Move 6000 to right hand side and change its sign

⇒ $\sf{Interest = 14550 - 6000}$

Subtract 6000 from 14550

⇒ $\sf{Interest = \: 8550 \: }$

Interest = $ 8550

<u>Finding </u><u>the </u><u>rate </u>

${ \boxed{ \sf{Rate = \frac{Interest \times 100}{Principal \times Time}}}}$

plug the values

⇒ $\sf{ Rate = \frac{8550 \times 100}{6000 \times 10} }$

Calculate

⇒ $\sf{Rate = \frac{855000}{60000} }$

⇒ $\sf{Rate = 14.25 \: \% \: }$

Hope I helped!

Best regards!!

8 0

3 years ago

The graph shows the relationship between the price of frozen yogurt and the number of ounces of frozen yogurt sold at different

snow_lady [41]

Answer:

13

Step-by-step explanation:

If you look at where $5 crosses the line of best fit, you see it between 12 and 14 ounces. The answer is 13.

3 0

3 years ago

Read 2 more answers