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e-lub [12.9K]
3 years ago
15

Math, homework, need help

Mathematics
1 answer:
Yanka [14]3 years ago
5 0
Essentially, what you are doing is taking each individual point and moving that point as specified.

So, for example, problem 1 has coordinates as follows: (be aware: I have terrible eyesight, so the points are being named based on what I think the letters are).

X: located at (2,0), move 1 unit left, new location: (1,0)
G: located at (4,0), move 1 unit left, new location: (3,0)
Q: located at (2,-2), move 1 unit left, new location: (1,-2)
C: located at (3,-4), move 1 unit left, new location: (2,-4)

Now, just plot the new shape.

Let me know if you need more help!

You might be interested in
Can someone PLEASE help me with this
Mekhanik [1.2K]

Answer:

4.5 u

Step-by-step explanation:

There are many ways to do it, the most easier is this:

Area=(b*h)/2

Area of ABD = (3*3)/2 = 4.5 u

3 0
4 years ago
Read 2 more answers
Mason, a hardwood flooring installer, needs to calculate the amount of hardwood flooring needed to cover a floor that's 33⁄4 yar
Aleksandr-060686 [28]

Answer:

20 5/8 yd^2

Step-by-step explanation:

width = 3 3/4 yd

length = 5 1/2 yd

First change both measurements to fractions.

3 3/4 = 3 + 3/4 = 12/4 + 3/4 = 15/4

5 1/2 = 5 + 1/2 = 10/2 + 1/2 = 11/2

area = length * width

area = 15/4 yd * 11/2 yd

area = (15 * 11)/(4 * 2) yd^2

area = 165/8 yd^2

165/8 = 20 remainder 5

area = 20 5/8 yd^2

3 0
4 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 \neq0.

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

Rank of B\neqRank 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
Solve algebra 2 bju press
Nitella [24]

Remark

It looks like all you want is question 6. If that is the case, there are two ways to do it.

Algebra

<u>First answer</u>

abs(b - 22) = 5      Equate to + 5

b - 22 = 5              Add 22 to both sides.

b = 5 + 22

b = 27

<u>Second Answer</u>

Equate to - 5

b - 22 = -5

b = 22 - 5

b = 17

Method Two

<u>Graph the question</u>

The graph y = abs(b - 22) is shown below in red.

The values of y when y =5 are shown in blue.

5 0
3 years ago
Jason bought 10 of the 30 raffle tickets for a drawing. assuming that he chooses the winning ticket each time, what is the proba
ikadub [295]

Answer: 2.96% rounded

Step-by-step explanation: 10/30) x (9/29) x (8/28)  =  720 / 24,360 = 6/203 = 2.96%

8 0
2 years ago
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