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disa [49]
2 years ago
9

What is the value of the fourth term in a geometric sequence for which a1 = 30 and r = 1/2?.

Mathematics
1 answer:
KIM [24]2 years ago
8 0
For a Geometric sequence:
a n = a ( n-1 ) * r
In this case: a 1 = 30
a 2 = 30 * 1/2 = 15
a 3 = 15 * 1/2 = 7.5
a 4 = 7.5 * 1/2 = 3.75 = 3 3/4
Answer: The value of the fourth term is 3.75 or 3 3/4.
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Step-by-step explanation:

Rise over run

rise = 4 run=1

4/1=4

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3 0
2 years ago
In the previous part, we obtained dy dx = 3t2 − 27 −2t . Next, find the points where the tangent to the curve is horizontal. (En
mina [271]

Answer:

(27.55, 7.22), (-11.3, 3.21).

Step-by-step explanation:

When is the tangent to the curve horizontal?

The tangent curve is horizontal when the derivative is zero.

The derivative is:

\frac{dy}{dx} = 3t^{2} - 2t - 27

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

3t^{2} - 2t - 27 = 0

So

a = 3, b = -2, c = -27

Then

\bigtriangleup = b^{2} - 4ac = (-2)^{2} - 4*3*(-27) = 328

So

t_{1} = \frac{-(-2) + \sqrt{328}}{2*3} = 3.35

t_{2} = \frac{-(-2) - \sqrt{328}}{2*3} = -2.685

Enter your answers as a comma-separated list of ordered pairs.

We found values of t, now we have to replace in the equations for x and y.

t = 3.35

x = t^{3} - 3t = (3.35)^{3} - 3*3.35 = 27.55

y = t^{2} - 4 = (3.35)^2 - 4 = 7.22

The first point is (27.55, 7.22)

t = -2.685

x = t^{3} - 3t = (-2.685)^3 - 3*(-2.685) = -11.3

y = t^{2} - 4 = (-2.685)^2 - 4 = 3.21

The second point is (-11.3, 3.21).

8 0
3 years ago
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