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3 years ago

what is the constant of variation for the quadratic variation of x |2|3|4|5|6| fx |28|63|112|175|252|

Mathematics

1 answer:

hjlf3 years ago

8 0

F(x) = k*x^2
You can find k using only the first point.
28 = k*2^2
7 = k

The constant of variation is 7.

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rjkz [21]

Answer:

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Step-by-step explanation:

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3 years ago

Which point is a solution to the linear equality y<-1/2x+2

Effectus [21]

Answer:

There are many points that are solutions to the equation, but one of which would be (1, 2)

Step-by-step explanation:

In order to tell if they are a solution, we use the values in the inequality and see if it results in a true statement.

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3 years ago

What is the answer to 55-3x-2x=10

patriot [66]

Answer:

x = 9

Step-by-step explanation:

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6 0

3 years ago

Please help me with the below question.

o-na [289]

Given that

$\vec r(t) = \left\langle 4t e^{-t}, 8 \arctan(t), 8 e^t \right\rangle$

we first differentiate with respect to t to get the tangent vector, $\vec T(t)$ :

$\vec T(t) = \dfrac{d\vec r}{dt} = \left\langle (4-4t) e^{-t}, \dfrac8{1+t^2}, 8e^t \right\rangle$

At t = 0, the tangent vector is

$\vec T(0) = \left\langle 4, 8, 8 \right\rangle$

To get the <em>unit</em> tangent vector, multiply this by 1/(norm of tangent vector) :

$\|\vec T(0)\| = \sqrt{4^2 + 8^2 + 8^2} = \sqrt{144} = 12$

Then the unit tangent vector is

$\dfrac{\vec T(0)}{\|\vec T(0)\|} = \left\langle \dfrac4{12}, \dfrac8{12}, \dfrac8{12}\right\rangle = \boxed{\left\langle\dfrac13, \dfrac23, \dfrac23\right\rangle}$

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2 years ago

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zepelin [54]

Answer: C

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P.S. just took the test

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3 years ago