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Andreyy89
3 years ago
5

What two numbers does the square root of 161 lie between

Mathematics
2 answers:
aalyn [17]3 years ago
8 0

Answer:

12-13

Step-by-step explanation:

erastovalidia [21]3 years ago
4 0

Answer:

12 and 13

Step-by-step explanation:

use a calculator ツ

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1+(-2-5)^2+(14-17)×4 please answer
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When you simplify the expression you get 38. Hope this helps.

Answer = 38.
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What id the mean of the set? 10, 15, 14, 8, 18, 11, 12, 12, 10, 10, 17, 16
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The Mean is of the data is 12.75
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Goofy Golf ticket prices for kids under 12 years of age increased from $4.00 to $6.00 this year. What is the percent of increase
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There is a 50% increase in the two values.
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For this exercise assume that all matrices are ntimesn. Each part of this exercise is an implication of the form​ "If "statement
inna [77]

Answer:

C. True; by the Invertible Matrix Theorem if the equation Ax=0 has only the trivial solution, then the matrix is invertible. Thus, A must also be row equivalent to the n x n identity matrix.

Step-by-step explanation:

The Invertible matrix Theorem is a Theorem which gives a list of equivalent conditions for an n X n matrix to have an inverse. For the sake of this question, we would look at only the conditions needed to answer the question.

  • There is an n×n matrix C such that CA=I_n.
  • There is an n×n matrix D such that AD=I_n.
  • The equation Ax=0 has only the trivial solution x=0.
  • A is row-equivalent to the n×n identity matrix I_n.
  • For each column vector b in R^n, the equation Ax=b has a unique solution.
  • The columns of A span R^n.

Therefore the statement:

If there is an n X n matrix D such that AD=​I, then there is also an n X n matrix C such that CA = I is true by the conditions for invertibility of matrix:

  • The equation Ax=0 has only the trivial solution x=0.
  • A is row-equivalent to the n×n identity matrix I_n.

The correct option is C.

5 0
4 years ago
Find the inverse of the function y = x^2 + 12x
PilotLPTM [1.2K]

Answer:

y=x\\

x=y^{2} + 12y\\

y^{2} + 12y -x = 0\\

Delta = (12^{2}) - 4.1.(-x) = 144 +4X = 36.(4+x)

\sqrt{Delta} = 6 . \sqrt{(4+x)} \\

y' = \frac{-12 + 6.(\sqrt{(4+x)}}{2} = -6 + 3.\sqrt{(4+x)}\\

y" = -6 - 3.\sqrt{(4+x)}\\\\

y' = 3.\sqrt{(4+x)} - 6\\

y''= -3.\sqrt{(4+x)} - 6\\

7 0
3 years ago
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