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yuradex [85]
3 years ago
15

Infinite Algebra 2

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
The first problem, all you need to do is combine like terms then isolate the n:
4n-2n=4
~subtract 2n from 4n (2n)
2n=4
~then divide both sides of the equation by 2 to isolate the n
n=4

The second problem follows the same steps of combining like terms and isolating the variable. Here, you'll have to combine 2 like terms:
-12=2+5v+2v
~first combine the variables which is just 5v+2v which is 7v
-12=2+7v
~then subtract 2 from both sides to isolate the 7v
-14=7v
~then divide both sides by 7 to isolate the v and get your answer
-2=v

Hope that helped!
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Step-by-step explanation:

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iren2701 [21]
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3 years ago
Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
Radda [10]

Answer:

{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

Thus,  asymptote has the slope of

\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

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3 years ago
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