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lubasha [3.4K]
2 years ago
7

Please help with this I will mark you as brainliest

Mathematics
2 answers:
Liono4ka [1.6K]2 years ago
8 0

Answer:

t = 0.25

Step-by-step explanation:

LuckyWell [14K]2 years ago
3 0

Answer:

The answer is t = 0.25

Step-by-step explanation:

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I need the area of a rectangle, the middle is 5ft the right is 6ft and the bottom is 5ft. thank u
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length times width for area so its 5x6=30

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1-cos^2 A / sec^2 A - tan^2 A
tiny-mole [99]
I hope this helps you



✔cos^2A+sin^2A=1


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secA=1/cosA


tgA=sinA/cosA


sin^2A/1/cos^2A-sin^2A/cos^2A


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6 0
2 years ago
Find the length of the curve given by ~r(t) = 1 2 cos(t 2 )~i + 1 2 sin(t 2 ) ~j + 2 5 t 5/2 ~k between t = 0 and t = 1. Simplif
xxMikexx [17]

Answer:

The length of the curve is

L ≈ 0.59501

Step-by-step explanation:

The length of a curve on an interval a ≤ t ≤ b is given as

L = Integral from a to b of √[(x')² + (y' )² + (z')²]

Where x' = dx/dt

y' = dy/dt

z' = dz/dt

Given the function r(t) = (1/2)cos(t²)i + (1/2)sin(t²)j + (2/5)t^(5/2)

We can write

x = (1/2)cos(t²)

y = (1/2)sin(t²)

z = (2/5)t^(5/2)

x' = -tsin(t²)

y' = tcos(t²)

z' = t^(3/2)

(x')² + (y')² + (z')² = [-tsin(t²)]² + [tcos(t²)]² + [t^(3/2)]²

= t²(-sin²(t²) + cos²(t²) + 1 )

................................................

But cos²(t²) + sin²(t²) = 1

=> cos²(t²) = 1 - sin²(t²)

................................................

So, we have

(x')² + (y')² + (z')² = t²[2cos²(t²)]

√[(x')² + (y')² + (z')²] = √[2t²cos²(t²)]

= (√2)tcos(t²)

Now,

L = integral of (√2)tcos(t²) from 0 to 1

= (1/√2)sin(t²) from 0 to 1

= (1/√2)[sin(1) - sin(0)]

= (1/√2)sin(1)

≈ 0.59501

8 0
3 years ago
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