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3 years ago

Solve for x: 3x − 24 = 81

2 answers:

4 0

If you would like to solve for x: 3x - 24 = 81, you can do this using the following steps:

3x - 24 = 81
3x = 81 + 24
3x = 105
x = 105 / 3
x = 35

The correct result would be 35.

Mekhanik [1.2K]3 years ago

4 0

Well you add 24 to both sides then divide by 3 answer 35

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Find the coordinates of the midpoint of the segment connecting points (-3,-6) and ( 2,5)

Pavlova-9 [17]

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3 years ago

Solve. 5x + 10 = −4x − 17 x =

Pani-rosa [81]

5x+ 10= -4x -17
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3 years ago

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3 years ago

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Alexandra [31]

Answer:

$x = 135(being \: opposite \: angles \: of \: a \: paralleogram)$

Step-by-step explanation:

x =135(being opposite angles of a parallelogram)

3 0

2 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 are $y = k \pm \dfrac{b}{a}(x - h)$

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.

7 0

3 years ago