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3 years ago

The steps for determining the center and radius of a circle using the completing the square method are shown below:

1 answer:

7 0

The answer is option c.
That is, the wrong step in step 6. It was written that the center of the circunference is the point (2.1). However, the general equation of a circumference is:
(X- (a)) ^ 2 + (Y- (b)) ^ 2 = r ^ 2
Where the point (a, b) is the center of the circle.
So for this case the point for the center is: (-2, -1)

You might be interested in

How the heck do I do this? And what’s the answer?

stira [4]

Answer:

Option A

Step-by-step explanation:

We need to find two expressions that, when simplified, give the same results.

1) First, simplify the expression stated in the question. Multiply each of the terms in the parentheses by the number that is next to them. This would mean you have to multiply both 9x and -6 by $\frac{2}{3}$ . You also have to multiply $\frac{1}{2}x$ and $-\frac{1}{2}$ by 4. Then, simplify.

$\frac{2}{3}(9x-6) + 4 (\frac{1}{2} x - \frac{1}{2})\\\frac{18}{3}x- \frac{12}{3} + \frac{4}{2}x -\frac{4}{2} \\6x - 4 + 2x - 2$

2) Now, combine the like terms.

$6x - 4 + 2x - 2$

$8x - 6$

So, we need to find which of the expressions listed equal 8x - 6.

3) Let's try option A. Do the same as before. Multiply each of the terms in the parentheses by the number that is next to them. So, multiply 4x and -12 by $\frac{3}{4}$ . Also, multiply 30x and 18 by $\frac{1}{6}$ . Then, combine like terms and simplify.

$\frac{3}{4} (4x-12) + \frac{1}{6}(30x + 18)\\\frac{12}{4}x-\frac{36}{4} + \frac{30}{6} x + \frac{18}{6} \\3x - 9 + 5x + 3 \\8x - 6$

This also equals 8x - 6. Therefore, option A is the answer.

5 0

2 years ago

Each bus for Philips Elementary School transports 40 students to school. There are 12 buses that transport students to school ea

8090 [49]

Answer:

480 students are transported to school each day.

Step-by-step explanation:

Since there are 12 buses and each one of them transports 40 students that means that the total amount of students will be equal to...

12 x 40 = 480

7 0

3 years ago

Find the slope of the line passing through the pairs of points and describe the line as rising, falling, horizontal or vertical.

Anna [14]

Answer:

$a.\ m=2,\ \text{the line is rising}\\\\b.\ m=-\dfrac{5}{4},\ \text{the line is falling}\\\\c.\ m=0,\ \text{the line is horizontal}\\\\d.\ m\ is\ unde fined,\ \text{the line is vertical}$

Step-by-step explanation:

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

m > 0, then a line is rising

m < 0, then a line is falling

m = 0, then a line is horizontal

m is undefined, then a line is vertical

(2, 1) and (4, 5)

$m=\dfrac{5-1}{4-2}=\dfrac{4}{2}=2>0\to\text{rising}$

(-1, 0) and (3, -5)

$m=\dfrac{-5-0}{3-(-1)}=\dfrac{-5}{4}=-\dfrac{5}{4}$

(2, 1) and (-3, 1)

$m=\dfrac{1-1}{-3-2}=\dfrac{0}{-5}=0\to\text{horizontal}$

(-1, 2) and (-1, -5)

$m=\dfrac{-5-2}{-1-(-1))}=\dfrac{-7}{0}\ \text{UNDEFINED}\to\text{vertical}$

7 0

3 years ago

Bob gets paid 64 hour heard that is 2/3 Melissa makes how much is Alicia paid per hour

Thepotemich [5.8K]

Answer:

Step-by-step explanation:

4 0

2 years ago

Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the

Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A= $\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]$ , to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B= $\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$

the system Bx=0 can be represented in matrix form as:

$\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have $B_{0}$ :

$B_{0}$ = $\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]$ . The rank of $B_{0}$ can be found by using the second column and third column pair as follows:

| $B_{0}$ |=(3*0)-(0*2)=0 i.e, $B_{0}$ is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is $\neq$ 0.

Comparing the rank of both B and $B_{0}$ , it is obvious that

Rank of B $\neq$ Rank of $B_{0}$ since (-2)<1.

Therefore, we can conclude that equation(1) is <em>inconsistent and thus has no solution. </em>

(2) If B= $\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ is the transpose of matrix A= $\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right]$ , then

Then the equation Bx=0 is represented as:

$\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.

$B_{0}$ = $\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right]$ ,

| $B_{0}$ |=(5*0)-(0*10)=0-0=0 i.e $B_{0}$ has a rank of order 1.

we can therefor conclude that since

rank B=rank $B_{0}$ =1, equation(2) is <em>consistent</em> and has 2 solutions for the 2 unknown ( $X_{1}$ and $X_{2}$ ).

<u>Summary:</u>

Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
Determine the rank of both the coefficients matrix B and $B_{0}$ which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.

5 0

3 years ago