All categories

3 years ago

The diameter of a circle is two times its radius. Which equation shows another way to represent the area of a circle?

1 answer:

4 0

If the area for a circle is $A= \pi r^{2}$ , you can write it in terms of its diameter like this: $A= \pi ( \frac{1}{2}d) ^{2}$

You might be interested in

What is the interquartile range of this data?

harkovskaia [24]

The interquartile range is 24...

7 0

3 years ago

Consider the differential equation y'' − y' − 20y = 0. Verify that the functions e−4x and e5x form a fundamental set of solution

KIM [24]

Answer:

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

Step-by-step explanation:

Given differential equation is

y''-y'-20y =0

Here P(x)= -1, Q(x)= -20 and R(x)=0

Let trial solution be $y=e^{mx}$

Then, $y'=me^{mx}$ and $y''=m^2e^{mx}$

$\therefore m^2e^{mx}-m e^{mx}-20e^{mx}=0$

$\Rightarrow m^2-m-20=0$

$\Rightarrow m^2-5m+4m-20=0$

$\Rightarrow m(m-5)+4(m-5)=0$

$\Rightarrow (m-5)(m+4)=0$

$\Rightarrow m=-4,5$

Therefore the complementary function is = $c_1e^{-4x}+c_2e^{5x}$

Therefore $e^{-4x}$ and $e^{5x}$ are fundamental solution of the given differential equation.

If $y_1$ and $y_2$ are the fundamental solution of differential equation, then

$W(y_1,y_2)=\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|\neq 0$

Then $y_1$ and $y_2$ are linearly independent.

$W(e^{-4x},e^{5x})=\left|\begin{array}{cc}e^{-4x}&e^{5x}\\-4e^{-4x}&5e^{5x}\end{array}\right|$

$=e^{-4x}.5e^{5x}-e^{5x}.(-4e^{-4x})$

$=5e^x+4e^x$

$=9e^x\neq 0$

Therefore $e^{-4x}$ and $e^{5x}$ are linearly independent, since $W(e^{-4x},e^{5x})=9e^x\neq 0$

Let the the particular solution of the differential equation is

$y_p=v_1e^{-4x}+v_2e^{5x}$

$\therefore v_1=\int \frac{-y_2R(x)}{W(y_1,y_2)} dx$

and

$\therefore v_2=\int \frac{y_1R(x)}{W(y_1,y_2)} dx$

Here $y_1= e^{-4x}$ , $y_2=e^{5x}$ , $W(e^{-4x},e^{5x})=9e^x$ ,and $R(x)=0$

$\therefore v_1=\int \frac{-e^{5x}.0}{9e^x}dx$

and

$\therefore v_2=\int \frac{e^{5x}.0}{9e^x}dx$

The the P.I = 0

The general solution of the differential equation is

$y=c_1e^{-4x}+c_2e^{5x}$

7 0

3 years ago

Y = -(x+2)* - 3<br> Plot on a graph

Orlov [11]

Answer:

y=-x-2*-3

Step-by-step explanation:

3 0

3 years ago

Name a real number that is an irrational number between 0 and 1

Veseljchak [2.6K]

Square roots that are not perfect squares are irrational.

An example could be √0.5 which equals 0.7071067811...

6 0

3 years ago

Given x^3 square root of 4y is equal to 5.

Musya8 [376]

The value of the square root function after differentiation is $\frac{dy}{dx}= \frac{3}{x}$ .

The given equation is :

x³√4y = 5

Now we will use implicit differentiation

$\implies3x^2\sqrt{4y} \ \ dx +x^3\times 2(-\frac{1}{\sqrt y}) =0\\\\\implies6x^2\sqrt{y}\ dx-2\frac{x^3}{\sqrt y}\ dy=0\\\\\implies \frac{dy}{dx}= \frac{3}{x}$

In mathematics, the derivative of a simple function that resembles a real variable measures how sensitive the function's value (or output value) is to changes in the underlying assumption (input value). The derivative is the fundamental tool in calculus.

The value of y from the equation is :

y = 25/4 x⁻³

now if we differentiate this function we get the same result.

y'=3/x

Hence we can see that the value of both the solution are equal.

A measure of how quickly an object's position changes over time is its velocity, which is the derivative of that object's point with regard to time.

Hence the required solution of the differentiation is $\frac{dy}{dx}= \frac{3}{x}$

to learn more about square root visit:

brainly.com/question/15208820

#SPJ9

6 0

1 year ago