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3 years ago

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Object A has a position as a function of time given by A(t) = (3.00 m/s)tî + (1.00 m/s2)t2ĵ. Object B has a position as a functi

on of time given by B(t) = (4.00 m/s)tî + (-1.00 m/s2)t2ĵ. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?

1 answer:

agasfer [191]3 years ago

3 0

Answer:

18.25m

Step-by-step explanation:

Subtracting the two Vectors we have

= 3ti - 4ti + t^2j - - t^2j

=- 1ti + 2 t^2j

= The distance between the lines is given by

= Squaroot of ( ( -3)^2 +( 2(3)^2 )^2

= Squaroot of 9 +324

= Squaroot of 333

= 18.25m

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In 1 year, Kevin will be 40 and Daniel will be 4. 40 is 10 times of 4.

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3 years ago

6x+4=4x-2 WILL GIVE BRAINLIEST 60 POINTS

vesna_86 [32]

Answer:

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Step-by-step explanation: Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

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STEP 1

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STEP 2

2.1 Solve : 2 = 0

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3 years ago

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1) If y = x3 + 2x and dx/ dt = 5, find dy/dt when x = 2. Give only the numerical answer. For example, if dy, dt = 3, type only 3

Semenov [28]

Answer:

a) 70

b) 10π ft²/ft

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Step-by-step explanation:

1) y = x³ + 2x

$\frac{dy}{dt}=\frac{d(x^3+2x)}{dt}$

or

$\frac{dy}{dt}$ = $3x^2\frac{dx}{dt}+2\frac{dx}{dt}$

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$\frac{dy}{dt}$ = $3(2)^2\times(5)+2\times(5)$

or

$\frac{dy}{dt}$ = 60 + 10 = 70

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or

$\frac{dA}{dr}$ = 2(πr)

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3) From Pythagoras theorem

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Thus,

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here, H = length of the ladder

P is the height of the wall

B is the distance from the wall at bottom

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20² + P² = 25²

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P² = 625 - 400

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or

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$2B\frac{dB}{dt}+2P\frac{dP}{dt}=0$

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or

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or

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If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

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The normal to the </span><span>paraboloid at any point is given by:

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Equating the two normal vectors, we have:
</span>
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