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swat32
3 years ago
9

Plz help with mathmatics 1 hour till test

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0
1) 6x-5 = 2x + 7

4x=12 and x =3

2) since PQ//SR, Ange S is the supplement of angle P
 31° + P° =180°, and P° = 149°

3) The kite has a symmetric axis that is y-axis and
the coordinate of E(w,0)

4) The midpoint of WX:. Let M be the Midpoint
Formula to find the coordinates of Midpoint, knowing the coordinates
of W(0,4b) and X(2a,0) is:
the x of M = (x₁+x₂)/2 = (0+ 2a)/2 =a
and the y of M =(y₁+y₂)/2 = (4b+0)/2 =2b, so M(a,2b)

You might be interested in
The room can hold a maximum of 400 people. If there are already 134 people
Olegator [25]

Answer:

x+ 134 is less than or equal to 400 is the answer

6 0
3 years ago
Sam borrows from the bank $12,000 to buy a car paying in simple interest $ 2,844 per year. Sam paid the loan in 3 years, what is
12345 [234]

Answer:

3,468

Step-by-step explanation:

i multiplied 2,844 by 3 and got 8,532 i then subtracted that from 12,000 and got 3,468

5 0
3 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
A hot air balloon can hold up to 300 pounds of weight. The balloon's pilot weighs 140 pounds. How many 20-pound bags or sand can
VARVARA [1.3K]

Answer:

8 sand bags can be carried along.

Step-by-step explanation:

First you want to subtract 300-140. That subtracts the pilots weight giving you the maximum weight capacity left. (160) Then you want to divide 160 (the weight that can be carried) by 20 (the weight of the sand bags) after doing that you will get 8. And that is how i got 8 sand bags as my answer.

7 0
3 years ago
Read 2 more answers
3250 - M = 1500 + 2(M - 500)
bija089 [108]

Answer:

M=916 2/3

Step-by-step explanation:

3250-m=1500+2(m-500)

3250-M=1500+2m-1000

3250-M=2M+500

3M=2750

M=916 2/3

5 0
3 years ago
Read 2 more answers
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