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2 years ago

What is an equation of the line that passes through the points (-2, 1) and (-8, 4)?

Mathematics

1 answer:

goldenfox [79]2 years ago

7 0

<h2>Answer: y = - ¹/₂ x OR y - 1 = - ¹/₂ (x + 2) </h2>

<h3>Step-by-step explanation: </h3>

For us to write the equation for this line, we need to (1) find the slope of the line, and (2) use one of the points to write an equation:

The question gives us two points, (-2, 1) and (-8, 4), from which we can find the slope and later the equation of the line.

<u>Finding the Slope </u>

The slope of the line (m) = (y₂ - y₁) ÷ (x₂ - x₁)

= (4 - 1) ÷ (-8 - (-2))

= - ¹/₂

<u>Finding the Equation</u>

We can now use the point-slope form (y - y₁) = m(x - x₁)) to write the equation for this line:

⇒ y - 1 = - ¹/₂ (x - (-2))

∴ y - 1 = - ¹/₂ (x + 2)

we could also transform this into the slope-intercept form ( y = mx + c)

since y - 1 = - ¹/₂ (x + 2)

⇒ y = - ¹/₂ x

<em>To test my answer, I have included a Desmos Graph that I graphed using the information provided in the question and my answer.</em>

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Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$