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3 years ago

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The coordinates of the vertices of triangle ABC are A (1,-1),B (1,4), and C (8,4). what is the length in units of the line segme

nt that connects vertex A and vertex B

1 answer:

Vinil7 [7]3 years ago

4 0

The length of the line segment that connects vertex A and vertex B is 5 units

Step-by-step explanation:

Let us revise some facts about the horizontal and vertical segments

The segment is horizontal if the y-coordinates of all points on the segment are equal
The length of the horizontal segment whose endpoints are $(x_{1},y)$ and $(x_{2},y)$ is $x_{2}-x_{1}$
The segment is vertical if the x-coordinates of all points on the segment are equal
The length of the vertical segment whose endpoints are $(x,y_{1})$ and $(x,y_{2})$ is $y_{2}-y_{1}$

In Δ ABC

∵ A = (1 , -1)

∵ B = (1 , 4)

∵ C = (8 , 4)

∵ The x-coordinate of point A = 1

∵ The x-coordinate of point B = 1

∴ The x-coordinates of points A and B are equal

- The x-coordinates of A and B are equal, then the line AB is a

vertical segment

∴ The length of AB is the difference between the y-coordinates

of points A and B

∵ The y-coordinate of point A = -1

∵ The y-coordinate ob point B = 4

∴ The length of AB = 4 - (-1)

∴ The length of AB = 4 + 1

∴ The length of AB = 5 units

The length of the line segment that connects vertex A and vertex B is 5 units

Learn more:

You can learn more about the length of the segments in brainly.com/question/6564657

#LearnwithBrainly

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Jaylen’s beta fish tank holds 4 1/2 gallons of water. It takes him 1 7/8 minutes to fill it. What is the flow rate of the water

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3 years ago

N is the in center of △ABC. Use the given information to find NS.<br><br><br> NQ=2x<br> NR=3x−2

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Given that N is the incenter of △ABC, NS = 4.

<h3>What is the Incenter of a Triangle?</h3>

The incenter of a triangle is the point of concurrency of the three angle bisectors of a triangle.
The perpendicular distances from each sides of the triangle to the incenter are equal.

Given the image where:

NQ = 2x

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NQ = NR = NS (equidistant from the incenter)

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2 years ago

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Read 2 more answers

If the integral of the product of x squared and e raised to the negative 4 times x power, dx equals the product of negative 1 ov

Nataly_w [17]

Answer:

$A + B + E = 32$

Step-by-step explanation:

Given

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Required

Find $A +B + E$

We have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

Using integration by parts

$\int {u} \, dv = uv - \int vdu$

Where

$u = x^2$ and $dv = e^{-4x}dx$

Solve for du (differentiate u)

$du = 2x\ dx$

Solve for v (integrate dv)

$v = -\frac{1}{4}e^{-4x}$

So, we have:

$\int {u} \, dv = uv - \int vdu$

$\int\limits {x^2\cdot e^{-4x}} \, dx = x^2 *-\frac{1}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x} 2xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} - \int -\frac{1}{2}e^{-4x} xdx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

-----------------------------------------------------------------------

Solving

$\int xe^{-4x} dx$

Integration by parts

$u = x$ ---- $du = dx$

$dv = e^{-4x}dx$ ---------- $v = -\frac{1}{4}e^{-4x}$

So:

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} - \int -\frac{1}{4}e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} + \int e^{-4x}\ dx$

$\int xe^{-4x} dx = -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}$

So, we have:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} \int xe^{-4x} dx$

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} +\frac{1}{2} [ -\frac{x}{4}e^{-4x} -\frac{1}{4}e^{-4x}]$

Open bracket

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{x^2}{4}e^{-4x} -\frac{x}{8}e^{-4x} -\frac{1}{8}e^{-4x}$

Factor out $e^{-4x}$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{x^2}{4} -\frac{x}{8} -\frac{1}{8}]e^{-4x}$

Rewrite as:

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{4}x^2 -\frac{1}{8}x -\frac{1}{8}]e^{-4x}$

Recall that:

$\int\limits {x^2\cdot e^{-4x}} \, dx = -\frac{1}{64}e^{-4x}[Ax^2 + Bx + E]C$

$\int\limits {x^2\cdot e^{-4x}} \, dx = [-\frac{1}{64}Ax^2 -\frac{1}{64} Bx -\frac{1}{64} E]Ce^{-4x}$

By comparison:

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

$-\frac{1}{8} = -\frac{1}{64}E$

Solve A, B and C

$-\frac{1}{4}x^2 = -\frac{1}{64}Ax^2$

Divide by $-x^2$

$\frac{1}{4} = \frac{1}{64}A$

Multiply by 64

$64 * \frac{1}{4} = A$

$A =16$

$-\frac{1}{8}x = -\frac{1}{64}Bx$

Divide by $-x$

$\frac{1}{8} = \frac{1}{64}B$

Multiply by 64

$64 * \frac{1}{8} = \frac{1}{64}B*64$

$B = 8$

$-\frac{1}{8} = -\frac{1}{64}E$

Multiply by -64

$-64 * -\frac{1}{8} = -\frac{1}{64}E * -64$

$E = 8$

So:

$A + B + E = 16 +8+8$

$A + B + E = 32$

4 0

3 years ago