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Travka [436]
3 years ago
14

Find the curvature of r(t) at the point (1, 0, 0). r(t) = et cos(t), et sin(t), 3t

Mathematics
1 answer:
Georgia [21]3 years ago
4 0
Plug in x&y use the equation given
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Help plz:)))I’ll mark u Brainliest
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If the bottom of the red triangle is equal to 4 then its

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2 years ago
5(3 - x) < -2x + 6 please solve for me <3
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3 0
3 years ago
Read 2 more answers
Question 2<br>0 Find 2/3 + 1/2​
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Answer:

1 1/6

Step-by-step explanation:

4 0
3 years ago
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Multiply: (3√6 - √2)(4√2 - √6)​
mixer [17]

Given multiplication:

(3√6 - √2)(4√2 - √6)

= 3√6(4√2-√6)-√2(4√2-√6)

= 3√6*4√2 - 3√6*√6 - √2*4√2 + √2*√6

= 12√12 -3√(6*6) -4√(2*2) + √(2*6)

= 12√(4*3) - 3√(36) - 4√(4) + √(12)

= 12*2√3 - 3*6 - 4*2 + 2√3

= 24√3 - 18-8 +2√3

= 24√3 + 2√3 - 18-8

= 26√3 - 18-8

= 26√3 - 26

= 26(√3-1).

<u>Answer</u><u>:</u><u>-</u> (3√6 - √2)(4√2 - √6) = 26(√3-1)

<u>also</u><u> read</u><u> similar</u><u> questions</u><u>:</u> Multiply. −3 2/3⋅(−2 1/4) ? −8 1/4 −6 1/6 6 1/6 8 1/4....

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3 0
2 years ago
After leaving the runway, a plane's angle of ascent is 15 degrees and its speed is 272 feet per second. How many minutes will it
Mice21 [21]
<h2>It takes 3.31 minutes the airplane to climb to a height of 14,000 feet</h2>

Step-by-step explanation:

Angle with horizontal = 15°

Velocity of plane = 272 feet per second angle of ascent is 15 degrees

Velocity of plane = 272 cos 15 i + 272 sin 15 j = 262.73 i + 70.40 j ft/s

Vertical velocity = 70.40 ft/s

Vertical displacement = 14000 ft

We have

      Vertical displacement = Vertical velocity x Time

      14000 = 70.40 x Time

      Time = 198.87 seconds = 3.31 minutes

It takes 3.31 minutes the airplane to climb to a height of 14,000 feet

4 0
3 years ago
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