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3 years ago

15

The radius of a circle r in centimeters can be found by solving r to the second power = 70. What is the radius of the circle r,

in centimeters?

1 answer:

slavikrds [6]3 years ago

3 0

Answer:

bhb jkabdjkea 7- 70 bhakbhkd

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1 unit by 2 units by<br> units

Julli [10]

Answer:

2

Step-by-step explanation:

1 unit by 2 unit means area.

So, know that in area you multiply length times width, you multiply 2 and 1.

$1$ · $2$

Which is 2

8 0

3 years ago

The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 5, 8, and 9 is . The probabi

Tems11 [23]

Answer:

1/15

Step-by-step explanation:

When we form such three-digit numbers with distinct digits using the digits 1 , 2 , 3 , 5 , 8 and 9 (in all six digits all different), observe that the order of digits does matter. For example, if we make a number using digits 1 , 2 , and 3 , we can have 123 , 132 , 231 , 213 , 312 or 321 .

Hence we have to find number of 3 digit numbers that can be made from these six digits using permutation and answer is ⁶ P ₃ = 6 × 5 × 4 = 120 .

.How haw many of them will have first digit as even, we have two choices 2 and 8 . Once we have chosen 2

for hundreds place, we can have only 8 in units place and any one of remaining 4 can be used in tens place. Hence four choices, with 2 in hundreds place and another four choices when we have 8 in hundreds place (and 2 in units place) i.e. total 8 possibilities.

Hence, the probability, that both the first digit and the last digit of the three digit number are even numbers, is 8 /120 = 1 /15

.

4 0

3 years ago

Which of the following matches the graph of f (x) = 2(x – 1)2(x + 1)(x + 2)?

nadya68 [22]

Answer:

where is the rest of the question.......

6 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 $\neq$ 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

What is the area of the rectangle?

avanturin [10]

C
I believe is the right one

6 0

3 years ago