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2 years ago

How the hell did I get a notification that I answered a question i didn't create???

Mathematics

2 answers:

Elan Coil [88]2 years ago

7 0

Answer:

U (Followed) the question or U commented and somebody else commented and it notified u,its regular

Step-by-step explanation:

igomit [66]2 years ago

5 0

Answer:

i dont know same thing

Step-by-step explanation:

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I just need #21 and #23 answered please.

kykrilka [37]

#21: Even
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2 years ago

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It is known that the distance in metres through which a body falls freely varies directly as the square of the time taken. If a

erma4kov [3.2K]

Answer:

125 meters

Step-by-step explanation:

Let distance = d and time is t

Using the variation;

d = k t^2

So let’s us get k

80 = k * 4*2

16k = 80

k = 80/16 = 5

So for 5 seconds, the distance would be;

d = 5 * 5^2

d = 125 meters

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3 years ago

Tell whether the rates form a proportion.

pav-90 [236]

Answer: no

Step-by-step explanation: the answer is no

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2 years ago

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DaniilM [7]

Answer:

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Step-by-step explanation:

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5 0

2 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

2 years ago