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marissa [1.9K]
3 years ago
14

-6x+6y=6 -6x + 3y=-12 ???

Mathematics
2 answers:
Tcecarenko [31]3 years ago
7 0

Answer      

:x=5 y=6

Step-by-step explanation:

Darina [25.2K]3 years ago
5 0

Answer:

x=5 y=6

Step-by-step explanation:

Need to get rid of one variable to find one first. So I multiplied the second equation by -2 to get rid of y. So second equation becomes 12x-6y=24. add that to the first equation.

6x+6y=6

12x-6y=24

6x=30      then divide

x=5

then plug x into the original equation to get y.

-6(5) +6y=6

-30+6y=6    add 30 to both sides

6y=36       then divide

y=6

then put both into the original to make sure it makes sense

-6(5) +6(6)=6

-30+36=6

6=6

correct

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Last one thanks to whoever helps
marusya05 [52]

Answer:

see below

Step-by-step explanation:

take f(x)=5x-6, which is y=5x-6

switch the y for x and x for y and you get

x=5y-6

then add 6 on both sides and you get

x+6=5y

now divide both sides by 5 and you get

y=(x+6)/5

which is the inverse of the original function

5 0
3 years ago
Read 2 more answers
G(x) = 3x + 1; Find g(-8)
gregori [183]

Answer:

g(-8) = -23

Step-by-step explanation:

g(x) = 3x + 1    <em>Plug in g(-8)</em>

g(-8) = 3(-8) + 1  <em>Multiply 3 by -8</em>

g(-8) = -24 + 1   <em>Add 1 to -24</em>

g(-8) = -23

5 0
3 years ago
Read 2 more answers
Solve the following system of equations by using the inverse of a matrix.
Zepler [3.9K]

Answer:

(x, y, z) = (-8,4,-2)

Step-by-step explanation:

.......................................

6 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
"If a = − 9 and b = − 6, show that (a−b) ≠ (b−a)."
denis-greek [22]

Answer:

Step-by-step explanation:

LHS  a - b = -9 - (-6) = -9 +6 = -3

RHS  b-a = -6 - (- 9) =  -6 +9 = 3

as LHS not equal to RHS

a-b not equal to b-a

Thus proven

7 0
3 years ago
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