All categories

2 years ago

Sorry i didnt mean to post anything

Mathematics

1 answer:

Dmitry [639]2 years ago

4 0

Hi you seem nice I like hot cocoa and I’m a littleAnd 1+1 is two and 3+3 is six and 9+9 is 18

You might be interested in

YOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Molodets [167]

Hi yes yowww eeeeeeeee

4 0

3 years ago

How do i simplify it?

Lemur [1.5K]

Answer:

6 3/4x

Step-by-step explanation:

If you follow PEMDAS (Parenthesis, Exponent, Multiply, Divide, Add, Subtract) then you do what is inside the parenthesis first.

3 1/2 divided by 2= 1 3/4

1 3/4 plus 5= 6 3/5

Then, you add the x and since there is no other variable like x, then you just leave it with 6 3/4.

Hope that helped! :)

3 0

2 years ago

Can someone tell me the right answer pls and thank you

umka2103 [35]

B. -2 is the aswers

hope it help... :)

7 0

3 years ago

A man bought a car for Gh55,000.00 and sold it for Gh65,000.00. Find the percentage gain.

IgorLugansk [536]

Answer:

18.18%

Step-by-step explanation:

(65000-55000)/55000×100

5 0

2 years ago

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th

xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is $\bold{W=-3e^{7x}}$

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

$W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|$

Thus replacing the functions of the exercise we get:

$W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|$

Working with the determinant we get

$W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}$

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0

3 years ago