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andreyandreev [35.5K]

2 years ago

9

The product of a rational and irrational number is rational. always sometimes never

1 answer:

V125BC [204]2 years ago

3 0

Answer:

"The product of a rational number and an irrational number is SOMETIMES irrational." If you multiply any irrational number by the rational number zero, the result will be zero, which is rational. Any other situation, however, of a rational times an irrational will be irrational.

Step-by-step explanation:

PLZ MARK BRAINLIEST

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A 8-foot ladder touches the side of a building at a point 6 feet above the ground. At what height would a 22-foot ladder touch t

charle [14.2K]

Answer:Katie is 100 cm tall and is standing so that her mother's shado covers her shadow. She is 90 cm from her mother and her mother's shadow is 225cm long. How tall is her mother?

Step-by-step explanation:

7 0

3 years ago

One's intelligence quotient, or IQ, varies directly as a person's mental age and inversely as that person's chronological age.

Olin [163]

Answer:

The chronological age is 30 years

Step-by-step explanation:

Q= (kM)/C

132= (22k)/ 25

3300= 22k

k= 150

Q= (150M)/C

100= (150 x 20)/C

100= 3000/C

100C= 3000

C= 30

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4 0

2 years ago

I need help with area homework

aksik [14]

Answer:

1. 44.8

2. 9.6

3. 61.11

4. 50.41

5. 169

12. 300

13. 1.44

14. id k

15. a. 21

b. 28

c. 35

d. 31.5

Step-by-step explanation:

sry abt the stuff i didn't answer, i couldn't find the answer

7 0

2 years ago

Can anyone help me with this one question??

AnnyKZ [126]

The correct answer is 14yds

7 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