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MissTica
3 years ago
7

One endpoint of a line segment has coordinates represented by (x+4,1/2y). The midpoint of the line segment is (3,−2).

Mathematics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

(-x+2, -\frac{1}{2}y-4})

Step-by-step explanation:

We know that one of the endpoints of the line segment is (x+4, 1/2y)

The midpoint of the line segment is (3, -2).

And we want to find the other coordinates in terms of x and y.

To do so, we can use the midpoint formula:

M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Since we know that the midpoint is (3, -2), let's substitute that for M:

(3, -2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

Let's solve for each coordinate individually:

X-Coordinate:

We have:

3=\frac{x_1+x_2}{2}

We know that one of the endpoints is (x+4, 1/2y). So, let's let (x+4, 1/2y) be our (x₁, y₁). Substitute x+4 for x₁. This yields:

3=\frac{(x+4)+x_2}{2}

Solve for our second x-coordinate x₂. Multiply both sides by 2:

6=x+4+x_2

Subtract 4 from both sides:

2=x+x_2

Subtract x from both sides. Therefore, the x-coordinate of our second point is:

x_2=-x+2

Y-Coordinate:

We have:

-2=\frac{y_1+y_2}{2}

Substitute 1/2y for y₁. This yields:

-2=\frac{\frac{1}{2}y+y_2}{2}

Solve for y₂. Multiply both sides by 2:

-4=\frac{1}{2}y+y_2

Subtract 1/2y from both sides. So:

y_2=-\frac{1}{2}y-4

Therefore, the other coordinate expressed in terms of x and y is:

(-x+2, -\frac{1}{2}y-4})

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Step-by-step explanation:

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2 years ago
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Paul [167]

Answer:

-1000

Step-by-step explanation:

1. Do the Parenthesis problem first. (PEDMAS/PEMDAS)

You should then have -10^3

2. Solve with the exponent. (-10 x 10 x 10)

You should then get your final answer, which is -1000.

Any Questions/Concerns?

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3 years ago
Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

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\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

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Zigmanuir [339]

Answer:

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Step-by-step explanation:

From the picture attached,

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Since, AB is a diameter,

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m(arc EA) = 56°

Therefore, m(∠C) = \frac{1}{2}(56^{\circ}-20^{\circ})

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