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attashe74 [19]
3 years ago
12

F(x)=x^2-3 domain and range

Mathematics
1 answer:
Bad White [126]3 years ago
6 0
Domain is all real numbers
Range is [-3,∞)
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Step-by-step explanation:

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If AM=10, then AC=10 because both are radii. TC is a diameter so that’ll just be the radius doubled. TC=20!
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Find the coordinates of the image of G(−3,9) after the translation (x,y)→(x+10,y−7) .
nataly862011 [7]

Answer:

D. (7,2)

Step-by-step explanation:

-3 +10 is 7

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2 years ago
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
3 years ago
Hey Mahfia, please help! Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at
Dafna1 [17]

Answer:

\huge\boxed{  \red{ \boxed{ \tt{ \frac{ {y}^{2} }{ {10}^{2} }  -  \frac{ {x}^{2} }{ {12}^{2} }  = 1}}}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • conic sections
  • PEMDAS
<h3>tips and formulas:</h3>
  • \sf hyperbola \:equation :  \\  \sf  \frac{ {x}^{2} }{ {a}^{2} }   -  \frac{ {y}^{2} }{ {b}^{2} }  = 1
  • vertices of hyperbola:(±a,0) and (0,±b) if reversed
  • \sf \: asymptotes :  \\ y =   \pm\frac{b}{a} x
<h3>given:</h3>
  • vertices: (0,±10)
  • the hyperbola equation is inversed since the vertices is (0,±10)
  • asymptotes:\pm \frac{5}{6}x
<h3>let's solve:</h3>
  • the asymptotes are in simplest and we know b is ±10

according to the question

  1. y =  \sf  \frac{5 \times 2}{6 \times 2} x \\ y =  \frac{10}{12} x

therefore we got

  • a=12
  • b=10

note: the equation will be inversed

let's create the equation:

  1. \sf substitute \: the \: value \: of \: a \: and \: b :  \\  \sf  \frac{ {y}^{2} }{ {10}^{2} }  -  \frac{ {x}^{2} }{ {12}^{2} }  = 1

3 0
3 years ago
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