All categories

2 years ago

Find the area of the figure below

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

5 0

It should be 153! Hopefully that helps

You might be interested in

In your own words, define Quadratic Equation. How many solutions does a Quadratic Equation have?

dmitriy555 [2]

Answer: an equation that has one term which is nameless and squared also no term which gets raised to higher power.

Step-by-step explanation:

5 0

3 years ago

the morning tempature was -7C. By noon the tempature was 8C. how many degrees has the tampature risen

ddd [48]

It has risen 15 degrees .....

7 0

3 years ago

Read 2 more answers

Help look at screenshot

ipn [44]

1/3 is the correct answer simplified

4 0

2 years ago

Read 2 more answers

What is the value of x in the solution to this system of equations?

Airida [17]

Answer:

5 is the answer

Step-by-step explanation:

7 0

3 years ago

Find the solution of the given initial value problem:<br><br> y''- y = 0, y(0) = 2, y'(0) = -1/2

igor_vitrenko [27]

Answer: The required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Step-by-step explanation: We are given to find the solution of the following initial value problem :

$y^{\prime\prime}-y=0,~~~y(0)=2,~~y^\prime(0)=-\dfrac{1}{2}.$

Let $y=e^{mx}$ be an auxiliary solution of the given differential equation.

Then, we have

$y^\prime=me^{mx},~~~~~y^{\prime\prime}=m^2e^{mx}.$

Substituting these values in the given differential equation, we have

$m^2e^{mx}-e^{mx}=0\\\\\Rightarrow (m^2-1)e^{mx}=0\\\\\Rightarrow m^2-1=0~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mx}\neq0]\\\\\Rightarrow m^2=1\\\\\Rightarrow m=\pm1.$

So, the general solution of the given equation is

$y(x)=Ae^x+Be^{-x},$ where A and B are constants.

This gives, after differentiating with respect to x that

$y^\prime(x)=Ae^x-Be^{-x}.$

The given conditions implies that

$y(0)=2\\\\\Rightarrow A+B=2~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

and

$y^\prime(0)=-\dfrac{1}{2}\\\\\\\Rightarrow A-B=-\dfrac{1}{2}~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Adding equations (i) and (ii), we get

$2A=2-\dfrac{1}{2}\\\\\\\Rightarrow 2A=\dfrac{3}{2}\\\\\\\Rightarrow A=\dfrac{3}{4}.$

From equation (i), we get

$\dfrac{3}{4}+B=2\\\\\\\Rightarrow B=2-\dfrac{3}{4}\\\\\\\Rightarrow B=\dfrac{5}{4}.$

Substituting the values of A and B in the general solution, we get

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

Thus, the required solution of the given IVP is

$y(x)=\dfrac{3}{4}e^x+\dfrac{5}{4}e^{-x}.$

4 0

3 years ago