Y2 = 900 what is the solution for the equation

Mathematics

1 answer:

charle [14.2K]3 years ago

5 0

Answer: -30 and 30

Step-by-step explanation:

You might be interested in

(BRAINLIEST TO THE BEST ANSWER + PLEASE ANSWER PROPERLY, OR YOUR ANSWER WILL BE REPORTED. DON'T FORGET TO THOROUGHLY EXPLAIN.) J

KatRina [158]

Hey there!

In order to answer this question, it's easiest to start with one of the money amounts and work your way to the other ones. Keep in mind that, no matter what, the amount of collective money won't exceed 1.50, meaning you can't have two 1 dollar bills, etc. When I wrote it out, I did it in this order:

1 = 1 dollar
0.5 = Half dollar
0.25 = Quarter

1 + 0.5 = 1.50
1 + 0.25 + 0.25 = 1.50
0.5 + 0.5 + 0.5 = 1.50
0.5 + 0.5 + 0.25 + 0.25 = 1.50
0.5 + (0.25*4) = 1.50
0.25*6 = 1.50

Since no other combinations add up to 1.50, your answer is 6.

Hope this helped you out! :-)

4 0

3 years ago

Maggie jogged 7/8 mile on monday and half of that on tuesday. How far did she run on tuesday?

mote1985 [20]

She ran .4375 on Tuesday!!

8 0

3 years ago

A = 4 0 0 1 3 0 −2 3 −1 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) Find the real ei

Illusion [34]

Answer:

Step-by-step explanation:

We are given the matrix

$A = \left[\begin{matrix}4&0&0 \\ 1&3&0 \\-2&3&-1 \end{matrix}\right]$

a) To find the characteristic polynomial we calculate $\text{det}(A-\lambda I)=0$ where I is the identity matrix of appropiate size. in this case the characteristic polynomial is

$\left|\begin{matrix}4-\lambda&0&0 \\ 1&3-\lambda&0 \\-2&3&-1-\lambda \end{matrix}\right|=0$

Since this matrix is upper triangular, its determinant is the multiplication of the diagonal entries, that is

$(4-\lambda)(3-\lambda)(-1-\lambda)=(\lambda-4)(\lambda-3)(\lambda+1)=0$

which is the characteristic polynomial of A.

b) To find the eigenvalues of A, we find the roots of the characteristic polynomials. In this case they are $\lambda=4,3,-1$

c) To find the base associated to the eigenvalue lambda, we replace the value of lambda in the expression $A-\lambda I$ and solve the system $(A-\lambda I)x =0$ by finding a base for its solution space. We will show this process for one value of lambda and give the solution for the other cases.

Consider $\lambda = 4$ . We get the matrix

$\left[\begin{matrix}0&0&0 \\ 1&-1&0 \\-2&3&-5 \end{matrix}\right]$

The second line gives us the equation x-y =0. Which implies that x=y. The third line gives us the equation -2x+3y-5z=0. Since x=y, it becomes y-5z =0. This implies that y = 5z. So, combining this equations, the solution of the homogeneus system is given by

$(x,y,z) = (5z,5z,z) = z(5,5,1)$