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3 years ago

Plzzzzzzzzzzzzz hellllllllllllllllllllllllllpppppppppppppppppp

Mathematics

2 answers:

cestrela7 [59]3 years ago

3 0

Answer:

1.02 units3

Step-by-step explanation:

sukhopar [10]3 years ago

3 0

Answer is 1.02 ......

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What are the first three terms of the Arithmetic Sequence a_n = 15 – 2(n – 1)?

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3 years ago

Lim x → 0 sin3x)/5x^3 -4

JulsSmile [24]

$\bf \lim\limits_{x\to 0}\ \cfrac{sin^3(x)}{5x^3}-4
\\ \quad \\\\ \cfrac{sin^3(x)}{5x^3}-4\implies \cfrac{[sin(x)]^3}{5x^3}-4
\\ \quad \\
 

\cfrac{[sin(x)]^3}{x^3}\cdot \cfrac{1}{5}-4
\\ \quad \\
thus
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\lim\limits_{x\to 0}\cfrac{[sin(x)]^3}{x^3}\cdot \lim\limits_{x\to 0}\cfrac{1}{5}-4\qquad \boxed{recall \qquad \lim\limits_{x\to 0} \cfrac{sin(x)}{x}\implies 1}
\\ \quad \\
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3 years ago

Find all solutions to equations

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Step-by-step explanation:

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3 years ago

10. Write the numbers in expanded (or long) form.

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8 0

2 years ago

Find the volume of the given solid. Bounded by the cylinder y2 + z2 = 16 and the planes x = 2y, x = 0, z = 0 in the first octant

erastovalidia [21]

Answer:

$\mathbf{ \dfrac{128}{3}}$

Step-by-step explanation:

Given that:

$y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0$

$Replace \ z = 0 \ \text{in the given equation ;} y^2+z^2=16$

$Then;$

$y^2 = 16 \\ \\ y = \sqrt{16} \\ \\ y = \pm 4$

$\text{So; in 1st octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y$

∴

$\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy \\ \\ = \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\ V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy$

$Now, let:\\\\ 16 - y^2 = u \\ \\ -2ydy = du \\ \\ 2ydy = -du \\ \\ Hence, when \ y = 0; u = 16 \\ \\ and \\ \\ when \ y = 4 \ then \ u = 0$

∴

$V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy \\ \\ = \int ^0_{16} - \sqrt{u}\ du \\ \\ = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du \\ \\ = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg] \\ \\$

$= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )$

$= - \dfrac{2}{3} \Big(64\Big )$

∴

$\mathbf{V = \dfrac{128}{3}}$

5 0

2 years ago