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3 years ago

Find the coordinates of the midpoint of XY for X(-2,3) and T(-8,-9)

Mathematics

1 answer:

stellarik [79]3 years ago

8 0

Answer:

Midpoint of XY = ( - 5, - 3)

Step-by-step explanation:

$Midpoint \: of \: XY \\ = \bigg( \frac{ - 2 + ( - 8)}{2}, \: \frac{3 + ( - 9)}{2} \bigg) \\ = \bigg( \frac{ - 2 - 8}{2}, \: \frac{3 - 9}{2} \bigg)\\ = \bigg( \frac{ - 10}{2}, \: \frac{ - 6}{2} \bigg)\\ Midpoint \: of \: XY = ( - 5, \: - 3)$

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Solving the equation

creativ13 [48]

Answer:

9s+2=9s-5

Step-by-step explanation:

2 times 9s and 2 times 2 then subtract your 9sequals 3 times 3sand 3 times 6 so then you get 9s and 18 then you subtract 23 and you get 23. then you get your answer 9s-5=9s+2

5 0

3 years ago

Mexico City, Mexico, is the world's second largest metropolis and is also one of its fastest-growing cities with a projected gro

Cloud [144]

Answer:

Predicted population of Mexico City in year 2010 = 38386000

Step-by-step explanation:

Formula to be used for the population of Mexico,

P = 20.899 $e^{0.032t}$

Where t = Duration after year 1991

P = Final population

Number of years between 2010 and 1991 = 19 years

Predicted population of Mexico city after 2010 = 20.899 $e^{0.032\times 19}$

= 20.899 × (1.8368)

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= 38.386 million

Therefore, predicted population of Mexico City in year 2010 = 38386000

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3 years ago

How to solve for y.

NARA [144]

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so...
y^2 + 25^2 = 65^2
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3 years ago

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Alex777 [14]

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3 years ago

Read 2 more answers

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago