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3 years ago

Write an explicit formula for a^n the nth term of the sequence 1,3,9....

Mathematics

1 answer:

frutty [35]3 years ago

7 0

Answer:

a_n = 3^(n -1)

Step-by-step explanation:

The n-th term of a geometric sequence with first term a1 and common ratio r is given by ...

a_n = a1·r^(n-1)

Your sequence has first term 1 and ratio r=3, so the sequence is given by ...

a_n = 3^(n -1)

_____

Comment on sequences and series

The sequences we commonly study are "arithmetic" and "geometric." Each of these has an explicit formula for the n-th term, based on the first term and the common difference or ratio. Similarly, each series (sum of terms of a sequence) also has a formula. That's 4 formulas to keep track of; not difficult. One of them, the formula for the n-th term of a geometric sequence, is shown above.

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3 years ago

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$2/5lb find unit rate

shtirl [24]

Answer:

1$/2.5lb

Step-by-step explanation:

This is a fraction equal to 5 pounds / 2 dollars

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3 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

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3 years ago

Which of the following is a complex number? OA. 3V+ B. 5 - 207 oc. 9 - 1 OD. -88 - 4

Bogdan [553]

A complex number would be OA. 3V+

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3 years ago

Find the area of this circle. Use 3 for pi

inessss [21]

Answer:

243 in²

Step-by-step explanation:

Diameter = 18 in

Radius ( r ) = Diameter / 2 = 18 / 2

r = 9 in

π = 3

To find : Area of circle ( A )

Formula : -

A = πr²

A = 3 x 9²

= 3 x 9 x 9

A = 243 in²

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3 years ago