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a recipe to make 4 pancakes calls for 6 tablespoons of flour. Tracy wants to make 10 pancakes using this recipe. What equation w

bekas [8.4K]

The multiplication sign because "6 teaspoons of flour per pancake (6x) 10 pancakes she want to make (6x10=60) 60 tablespoons of flour!

5 0

4 years ago

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The data shown in the table below can be

kvv77 [185]

Answer:

6: 12

7: 14

8: 16

Step-by-step explanation:

You do the last (y) term given [18] minus the first (y) term [10] given divided by how many terms it takes to get from the last (y) term given minus the first (y) term given [4]. So the equation looks like this:

(18 - 10)/4= 8/4 = 2

4 0

3 years ago

help????? A sixth grade basketball team has 3 centers, 5 forwards, and 6 guards. Write a ratio using a colon for centers to tota

Alexus [3.1K]

Answer: 3:14

Step-by-step explanation: there are 3 centers and if you add all the players you get 14

4 0

3 years ago

In 1995 a gallon of gas used to cost $1.00 today a gallon of gas cost $4.00 what is the percent increase?

Wittaler [7]

Answer:

300%

Step-by-step

300% of 1.00$ is 4.00 because you are adding 1.00 + 300%

5 0

3 years ago

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(1) (10 points) Find the characteristic polynomial of A (2) (5 points) Find all eigenvalues of A. You are allowed to use your ca

Yuri [45]

Answer:

Step-by-step explanation:

Since this question is lacking the matrix A, we will solve the question with the matrix

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

so we can illustrate how to solve the problem step by step.

a) The characteristic polynomial is defined by the equation $det(A-\lambdaI)=0$ where I is the identity matrix of appropiate size and lambda is a variable to be solved. In our case,

$\left|\left[\begin{matrix}4-\lamda & -2 \\ 1 & 1-\lambda \end{matrix}\right]\right|= 0 = (4-\lambda)(1-\lambda)+2 = \lambda^2-5\lambda+4+2 = \lambda^2-5\lambda+6$

So the characteristic polynomial is $\lambda^2-5\lambda+6=0$ .

b) The eigenvalues of the matrix are the roots of the characteristic polynomial. Note that

$\lambda^2-5\lambda+6=(\lambda-3)(\lambda-2) =0$

So $\lambda=3, \lambda=2$

c) To find the bases of each eigenspace, we replace the value of lambda and solve the homogeneus system(equalized to zero) of the resultant matrix. We will illustrate the process with one eigen value and the other one is left as an exercise.

If $\lambda=3$ we get the following matrix

$\left[\begin{matrix}1 & -2 \\ 1 & -2 \end{matrix}\right]$ .

Since both rows are equal, we have the equation

$x-2y=0$ . Thus x=2y. In this case, we get to choose y freely, so let's take y=1. Then x=2. So, the eigenvector that is a base for the eigenspace associated to the eigenvalue 3 is the vector (2,1)

For the case $\lambda=2$ , using the same process, we get the vector (1,1).

d) By definition, to diagonalize a matrix A is to find a diagonal matrix D and a matrix P such that $A=PDP^{-1}$ . We can construct matrix D and P by choosing the eigenvalues as the diagonal of matrix D. So, if we pick the eigen value 3 in the first column of D, we must put the correspondent eigenvector (2,1) in the first column of P. In this case, the matrices that we get are

$P=\left[\begin{matrix}2&1 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}3&0 \\ 0 & 2 \end{matrix}\right]$

This matrices are not unique, since they depend on the order in which we arrange the eigenvalues in the matrix D. Another pair or matrices that diagonalize A is

$P=\left[\begin{matrix}1&2 \\ 1 & 1 \end{matrix}\right], D=\left[\begin{matrix}2&0 \\ 0 & 3 \end{matrix}\right]$

which is obtained by interchanging the eigenvalues on the diagonal and their respective eigenvectors

4 0

3 years ago