Help please! :) -Substitution Method-

What is 6r+70=-2(35+3r)

lesya [120]

6r+70=-2(35+3r)

multiply the bracket by -2

(-2)(35)=-70

(-2)(3r)=-6r

6r+70=-70-6r

move -70 to the left side

sign changes from -70 to +70

6r+70-70=-70-70-6r

6r=-70-70-6r

move-6r to the left side

6r=-140-6r

move -6r to the other side

6r+6r=-140-6r+6r

12r=-140

divide both sides by 12

12r/12=-140/12

r=-140/12

reduce- by dividing by 4

-140/4=-35

12/4=3

Answer:

r= -35/3 and r=-11 2/3

7 0

4 years ago

Solve for b<br> A=h(a+b)2

Elanso [62]

B=A/2h-a
Isolate the variable by dividing each side by factors that don’t contain the variable.

4 0

3 years ago

Read 2 more answers

You invest $1,300 in an account with an annual interest rate of 2.5%, compounded monthly. How much money is in the account after

galben [10]

A=1,300×(1+0.025÷12)^(12×15)
A=1,890.75

5 0

4 years ago

Read 2 more answers

Marco is reconstructing his expenses for the past two weeks. Here are the records of his expenses: Transaction Cost ($) T-shirt

Stella [2.4K]

The total expenses of Marco's account are $174. The ending balance is $81. So, the starting balance will be $255.

Option A is correct for starting balance.

Further Explanation on Starting and Ending Balance

Given that Marco’s bank statement says that he has an ending balance of $81. The expenses $ of Marco is given as,

T-shirt 20
Gas 22
Movie 13
Video game 39
Jeans 34
Hat 15
Books 31

The total expenses can be calculated as given below.

Total Expenses = Sum of all Expenses

Total Expenses = 20+22+13+39+34+15+31

Total Expenses = $174

The starting balance can be calculated as given below.

Starting Balance = $\$174 +\$81$

Starting Balance = $ 255.

Hence we can conclude that the starting balance of Marco's account is $255. Option A is correct.

To know more about expenses, follow the link given below.

brainly.com/question/24803457.

8 0

2 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

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

So, the base for this eigenspace is the vector (5,5,1).

If $\lambda = 3$ then the base is (0,4,3) and if $\lambda = -1$ then the base is (0,0,1)

3 0

3 years ago