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3 years ago

H(x)=x•2+1 k(x)=x-2 (H+k)(2)

Mathematics

1 answer:

slavikrds [6]3 years ago

7 0

Is it a true or false? If yes, the answer is false

If u have to solve the answer will be hx=x.2+kx=x-hk2

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Doug is 2 more than twice the age of Joey.<br> The sum of their ages is 74. How old is Doug?

Citrus2011 [14]

The answer would be 37

4 0

3 years ago

Solve |x + 7| < 6<br> its for homework

Semmy [17]

Ok so if you follow the process of pemdas is should be easy pemdas equals: parentheses, exponents, multiple and divide from left to right, lastly add and subtract from left to right

3 0

2 years ago

The difference of a number and 17 is more than 33

madreJ [45]

Answer:

In algebraic terms this means x-17>33

Step-by-step explanation:

to solve it x−17>33

1 Add 17 to both sides.

x>33+17

2 Simplify 33+17 to 50

x>50

5 0

4 years ago

Read 2 more answers

What is the median of this data set? {29, 22, 23, 31, 29, 29, 22, 23} Enter your answer in the box.

GaryK [48]

Answer:

The median is 26

Step-by-step explanation:

A "median" is equivalent to, when the terms are arranged in the proper order, whichever term is in the "middle".

So- first, arrange the terms in ascending order.

[22, 22, 23, 23, 29, 29, 29, 31]

There are an even number- eight- terms in total, so we will take the average of the two "middle" terms.

(29+23)/2; 29 and 23 are the "middle" terms, and there are two of them.

(52)/2 = 26

Therefore, the median is 26.

I hope this helped! :)

4 0

2 years ago

A torus is formed by rotating a circle of radius r about a line in the plane of the circle that is a distance R (> r) from th

jeyben [28]

Consider a circle with radius $r$ centered at some point $(R+r,0)$ on the $x$ -axis. This circle has equation

$(x-(R+r))^2+y^2=r^2$

Revolve the region bounded by this circle across the $y$ -axis to get a torus. Using the shell method, the volume of the resulting torus is

$\displaystyle2\pi\int_R^{R+2r}2xy\,\mathrm dx$

where $2y=\sqrt{r^2-(x-(R+r))^2}-(-\sqrt{r^2-(x-(R+r))^2})=2\sqrt{r^2-(x-(R+r))^2}$ .

So the volume is

$\displaystyle4\pi\int_R^{R+2r}x\sqrt{r^2-(x-(R+r))^2}\,\mathrm dx$

Substitute

$x-(R+r)=r\sin t\implies\mathrm dx=r\cos t\,\mathrm dt$

and the integral becomes

$\displaystyle4\pi r^2\int_{-\pi/2}^{\pi/2}(R+r+r\sin t)\cos^2t\,\mathrm dt$

Notice that $\sin t\cos^2t$ is an odd function, so the integral over $\left[-\frac\pi2,\frac\pi2\right]$ is 0. This leaves us with

$\displaystyle4\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}\cos^2t\,\mathrm dt$

Write

$\cos^2t=\dfrac{1+\cos(2t)}2$

so the volume is

$\displaystyle2\pi r^2(R+r)\int_{-\pi/2}^{\pi/2}(1+\cos(2t))\,\mathrm dt=\boxed{2\pi^2r^2(R+r)}$

6 0

3 years ago