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2 years ago

-8=-2(z+7) [solving linear equations ]

Mathematics

2 answers:

Sveta_85 [38]2 years ago

8 0

Z=-3 is the correct answer. Use math papa if you need help with the steps on how we got z=-3

hoa [83]2 years ago

7 0

-8=-2(z+7)

-8=-2z-14

-2z-8=-14

-2z=-6

Z=3

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kobusy [5.1K]

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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

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

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let's create the equation:

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

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