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3 years ago

How much string would you need to go around the bottom of a hat shaped like a cone if the radius is 4 inches? (Use 3.14 for π)

Mathematics

1 answer:

Softa [21]3 years ago

7 0

Answer:

25.12 inches

Step-by-step explanation:

To find the circumference or the perimeter of a circle it is the diameter times pi. (pi = 3.14) Or you can do radius times 2 times pi.

Answer is 4 times 2 times 3.14 which is 25.12

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iogann1982 [59]

Every function is a relation because the numbers have relationships but not every relation is a function because for it to be a function there has to be one y fo every x if any x(input) has more than one y(output) it's not a function

3 0

3 years ago

13 5 11 True/False...the given triangle is a right triangle. O True O False

Gala2k [10]

Answer:

False

Step-by-step explanation:

11 times 11 is 121

5 times 5 is 25

13 times 13 is 169

121+25 doesnt equal 169

this means it is not a right angled triangle as it does not follow pythag

7 0

3 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

In the derivation of Newton’s method, to determine the formula for xi+1, the function f(x) is approximated using a first-order T

dimaraw [331]

Answer:

Part A.

Let f(x) = 0;

suppose x= a+h

such that f(x) =f(a+h) = 0

By second order Taylor approximation, we get

f(a) + hf'±(a) + $\frac{h^{2} }{2!}$ f''(a) = 0

$h = \frac{-f'(a) }{f''(a)}$ ± $\frac{\sqrt[]{(f'(a))^{2}-2f(a)f''(a) } }{f''(a)}$

So, we get the succeeding equation for Newton's method as

$x_{i+1} = x_{i} + \frac{1}{f''x_{i}} [-f'(x_{i})$ ± $\sqrt{f(x_{i})^{2}-2fx_{i}f''x_{i} } ]$

Part B.

It is evident that Newton's method fails in two cases, as:

1. if f''(x) = 0

2. if f'(x)² is less than 2f(x)f''(x)

Part C.

In case $x_{i+1}$ is close to $x_{i}$ , the choice that shouldbe made instead of ± in part A is:

f'(x) = $\sqrt{f'(x)^{2} - 2f(x)f''(x)}$ ⇔ $x_{i+1} = x_{i}$

Part D.

As given $x_{i+1}$ = $x_{i}$ = h

or h = $x_{i+1}$ - $x_{i}$

We get,

f(a) + hf'(a) +(h²/2)f''(a) = 0

or h² = -hf(a)/f'(a)

Also, ( $x_{i+1}$ - $x_{i}$ )² = -( $x_{i+1}$ - $x_{i}$ )(f( $x_{i}$ )/f'( $x_{i}$ ))

So, f(a) + hf'(a) - (f''(a)/2)(hf(a)/f'(a)) = 0

It becomes h = -f(a)/f'(a) + (h/2)[f''(a)f(a)/(f(a))²]

Also, $x_{i+1}$ = $x_{i}$ -f( $x_{i}$ )/f'( $x_{i}$ ) + [( $x_{i+1}$ - $x_{i}$ )f''( $x_{i}$ )f( $x_{i}$ )]/[2(f'( $x_{i}$ ))²]

6 0

3 years ago

Prove that the value of the expression (125^2+25^2)(5^2–1) is divisible by 3

Radda [10]

Step-by-step explanation:

$A =(125^{2} + 25^{2} ) (5^{2} - 1)\\A = [(5^{3})^{2} + (5^{2})^{2} ] . (5^{2} - 1)\\A = (5^{6} + 5^{4} ). (5^{2} - 1)\\A = [5^{4} . ( 5^{2} + 5)].(5^{2} - 1) \\A = 5^{4} . (25+ 5). (5^{2} - 1)\\A = 5^{4} . 30. (5^{2} - 1)\\$

Since 30 is divisible by 3

Thus, A is divisible by 3

Good luck!

5 0

3 years ago

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