When constructing an inscribed square, what step comes after both diameters of the circle have been created?

What is the solution to 9e-7=7e-11<br> need it now

oee [108]

The answer is e = -2

I hope this helped! If so, please mark brainliest!

Step by step solution :

Step 1 :

Pulling out like terms :

1.1 Pull out like factors :

2e + 4 = 2 • (e + 2)

Equation at the end of step 1 :

Step 2 :

Equations which are never true :

2.1 Solve : 2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation :

2.2 Solve : e+2 = 0

Subtract 2 from both sides of the equation :

e = -2

8 0

3 years ago

Read 2 more answers

Write the slope-intercept form of the equation of each line given the slope and y-intercept

yuradex [85]

Answer:

y=4/3x-5.

y=-5x+1

4 0

2 years ago

Help please!!!!!!!!!

In-s [12.5K]

ANSWER

24

EXPLANATION

For a matrix A of order n×n, the cofactor $C_{ij}$ of element $a_{ij}$ is defined to be

$C_{ij} = (-1)^{i+j} M_{ij}$

$M_{ij}$ is the minor of element $a_{ij}$ equal to the determinant of the matrix we get by taking matrix A and deleting row i and column j.

Here, we have

$C_{11} = (-1)^{1+1} M_{11} = M_{11}$

M₁₁ is the determinant of the matrix that is matrix A with row 1 and column 1 removed. The bold entries are the row and the column we delete.

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \end{aligned}$

Since the determinant of a 2×2 matrix is

$\det\left( \begin{bmatrix} a & b \\ c& d \end{bmatrix} \right) = ad-bc$

it follows that

$\begin{aligned} A=\begin{bmatrix} \bf 1 & \bf -6 & \bf -4\\ \bf 7 & 0 & -3 \\ \bf -9 & 8 & -8 \end{bmatrix} \implies M_{11} &= \text{det}\left(\begin{bmatrix} 0&-3 \\ 8&-8 \end{bmatrix} \right) \\ &= (0)(-8) - (-3)(8) \\ &= -(-24) \\ &= 24 \end{aligned}$