Find the quotient. 48 ÷ -6 =

Mathematics

1 answer:

Jlenok [28]3 years ago

7 0

Answer:

Step-by-step explanation:

-48/6 = -8

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Write an<br> explicit formula for an, the nth term of the sequence 39, 43, 47, ....

Inga [223]

9514 1404 393

Answer:

an = 39 +4(n -1)

Step-by-step explanation:

The formula for the general term of an arithmetic sequence is ...

an = a1 +d(n -1)

Here, the sequence has first term a1 = 39 and common difference d = 43-39 = 4. So, the general term is ...

an = 39 +4(n -1)

3 0

3 years ago

Find an equation of the tangent line to the hyperbola x2 a2 − y2 b2 = 1 at the point (x0, y0).

nalin [4]

The equation of the tangent line is $y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$

To find the tangent to the hyperbola $\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }$ at (x₀, y₀), we differentiate the equation implicitly to find the equation of the tangent at (x₀, y₀).

So, $\frac{d}{dx} (\frac{x^{2} }{a^{2} } - \frac{y^{2} }{b^{2} }) = \frac{d0}{dx}\\\frac{d}{dx} \frac{x^{2} }{a^{2} } - \frac{d}{dx}\frac{y^{2} }{b^{2} }= 0\\\frac{2x }{a^{2} } - \frac{dy}{dx}\frac{2y }{b^{2} } = 0\\\frac{2x }{a^{2} } = \frac{dy}{dx}\frac{2y }{b^{2} } \\\frac{dy}{dx} = \frac{b^{2}x }{a^{2}y }$

So, at (x₀, y₀)

$\frac{dy}{dx} = \frac{b^{2} x_{0} }{a^{2} y_{0} }$

So, the equation of the tangent line is gotten from the standard equation of a line in point-slope form

So, $\frac{y - y_{0} }{x - x_{0} } = \frac{b^{2} x_{0} }{a^{2} y_{0} } \\y - y_{0} = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) \\y = \frac{b^{2} x_{0} }{a^{2} y_{0} }(x - x_{0}) + y_{0} \\y = \frac{b^{2} x_{0}x }{a^{2} y_{0} } - \frac{b^{2} x_{0}^{2} }{a^{2} y_{0} } + y_{0}$