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ASHA 777 [7]
3 years ago
9

√12 + √10 − √2) is

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0

Hello!

\bf  \sqrt{12}  + \sqrt{10}  -  \sqrt{2} =

\bf  \sqrt{ {2}^{2}  \times 3}  +  \sqrt{10}  -  \sqrt{2}  =

\bf  \sqrt{ {2}^{2} }  \sqrt{ 3}  +  \sqrt{10}  -  \sqrt{2}  =

\bf \boxed{ 2  \sqrt{3}  +  \sqrt{10}  -  \sqrt{2}}

Answer: (c) An irrational number

Good luck! :)

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nlexa [21]

Answer:

70 cm^2

Step-by-step explanation:

The area of square is 5×10=50 plus the square of triangle, high of triangle is 18-10=8 and base of triangle is 5 because it common base with square, so the square of triangle is 1/2(8)(5)=20

The total area is 50+20=70 cm^2

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3 years ago
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For every 24meals Momoka prepared for customers last night, she prepared 2 meals for employees. Momoka prepared 156 meals.
Olenka [21]

Answer:

144 meals for customers and 12 meals for employees.

Step-by-step explanation:

156/24=6.5

6 x 24 =144

156-144=12

It adds up to be simple numbers, if it didn't it would make sense. For example, she wouldn't serve 6 and a <u>half</u> people. She would serve either 6 people or 7 people. She wouldn't serve 11.3973043 meals. She would serve either 11 meals or 12 meals.

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Compute the probability of randomly selecting a three or club.
Pavlova-9 [17]
Find the total number of cards in a deck. Count all the # 3’s total cards/#3 cards
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5 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 \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
Can u Simplify this problem???!? 3t-7t+14t
sleet_krkn [62]
Come on now, Tayala !

What if I say to you:
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Don't let the little 't' scare you.  It doesn't matter one bit what 't' really is.  It can be
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Hint:  If there are not ten of them, then go back and look at the cows again.
7 0
3 years ago
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