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3 years ago

What is the radius of a hemisphere with a volume of 165\text{ cm}^3,165 cm 3 , to the nearest tenth of a centimeter?

Mathematics

1 answer:

DanielleElmas [232]3 years ago

4 0

Answer:

4.3cm

Step-by-step explanation:

the radius of the hemisphere can be determined its volume

volume of a hemisphere = (2/3) x (n) x (r^3)

n = 22/7

r = radius

(2/3) x (22/7) x( r^3) = 165 cm^3

44/21 x r^3 = 165

divide both sides of the equation by 21/44

165 x 21/44

r^3 = 78.75

find the cube root

r = 4.2863 cm

the tenth is the first number after the decimal place. To convert to the nearest tenth, look at the number after the tenth (the hundredth). If the number is greater or equal to 5, add 1 to the tenth figure. If this is not the case, add zero

the 100th terms, 8 is greater than 6, so add 1 to 2. this give 4.3cm

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Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

5 0

3 years ago

PLEASE HELP! I'll give 5 out of 5 stars, give thanks, and give as many points as I can.

Liula [17]

Answer:

$\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}$

Step-by-step explanation:

$\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)$

$\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)$