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3 years ago

X+1=3 x-4=5 solve by equation method

Mathematics

2 answers:

jolli1 [7]3 years ago

8 0

Answer:

X+1=3

x = 3 - 1

x = 2

Check:

2 + 1 = 3

x-4=5

x = 5 + 4

x = 9

Check:

9 - 4 = 5

Fiesta28 [93]3 years ago

8 0

Answer:

1)x=2

2)x=9

Step-by-step explanation:

x+1=3

=>x=3-1

therefore

x=2

second question

x-4=5

=>x=5+4

therefore

x=9

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The height of a cone is twice the radius of its base.

lara [203]

Answer:

$A. \frac{2}{3} \pi x^3$

Step-by-step explanation:

From the way the answers are presented, it can be seen that x refers to the radius of the base of the cone

radius: $x$

and we are told that the height is twice the radius, so:

height: $2x$

and now we use the formula to calculate the volume of a cone:

$V=\frac{\pi r^2h}{3}$

where $V$ is volume, $r$ is radius, and $h$ is the height. and $\pi$ is a constant

in this case

$r=x$

$h=2x$

so we substitute thisvalues in the formula for the volume:

$V=\frac{\pi x^2(2x)}{3}$

Rearranging the terms

$V=\frac{2\pi x^3}{3} \\V=\frac{2}{3} \pi x^3$

which is option A.

6 0

3 years ago

How to find the average squared distance between the points of the unit disk and the point (1,1)

ddd [48]

The unit disk can be parameterized by the function

$\mathbf p(r,\theta)=(r\cos\theta,r\sin\theta)$

where $0\le r\le 1$ and $0\le\theta\le2\pi$ . The squared distance between any point in this region $(x,y)=(r\cos\theta,r\sin\theta)$ and the point (1, 1) is

$(x-1)^2+(y-1)^2=(r\cos\theta-1)^2+(r\sin\theta-1)^2$
$=(r^2\cos^2\theta-2r\cos\theta+1)+(r^2\sin^2\theta-2r\sin\theta+1)$
$=r^2(\cos^2\theta+\sin^2\theta)-2r(\cos\theta-\sin\theta)+2$
$=r^2-2r(\cos\theta-\sin\theta)+2$

The average squared distance is then going to be the ratio of [the sum of all squared distances between every point in the disk and the point (1, 1)] to [the area of the disk], i.e.

$\dfrac{\displaystyle\iint_{x^2+y^2$
$=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}$
$=\dfrac{\frac{5\pi}2}\pi=\dfrac52$