Just do #6, giving all I have!

The formula for any arithmetic sequence is an = a1 + d ( n - 1), where an represents the value of the n th term, a1 represents t

hoa [83]

10, 8, 6, 4,…
-2 -2 -2

a(n) = a₁ + d(n - 1)
a(n) = 10 - 2(n - 1)
a(n) = 10 - 2(n) + 2(1)
a(n) = 10 - 2n + 2
a(n) = -2n + 10 + 2
a(n) = -2n + 12

6 0

3 years ago

Represent the geometric series using the explicit formula.

user100 [1]

Answer: $f(n)=12(-3)^{(n-1)}$

Step-by-step explanation:

The Explicit formula in function notation for a geometric series is:

$f(n)=f(1)r^{(n-1)}$

Where $f(n)$ is the nth term of the sequence, $f(1)$ is the first term in the sequence and $r$ is the common ratio.

Given the following geometric serie:

$12, -36, 108, -324...$

The common ratio is:

$r=\frac{-36}{12}=-3$

And the first term is:

$f(1)=12$

Therefore, substituting values, we get that the Explicit formula that represents this geometric series is:

$f(n)=12(-3)^{(n-1)}$

3 0

4 years ago

A music store orders wooden drumsticks that weigh 96 grams per pair. The total weight of the box of drumsticks is 782 grams. How

Nikolay [14]

Hi there! The answer is 6.

The box including the drumsticks has a weight of 782 grams.
The box itself has a weight of 206 grams.
Therefore the drumsticks in the box have a weight of 782 - 206 = 576 grams.

Each drumstick has a weight of 96 grams.
The total amount of drumsticks have a weight of 576 grams.
Therefore there are 576 / 96 = 6 drumsticks in the box.

6 0

3 years ago

Read 2 more answers

Which expression is equivalent to (2to the third power) times -5?

Airida [17]

2 to the third power is 2³

so 2 to the third power times -5 is 2³ * (-5)

which on simplification will be 8* (-5) =-40

3 0

3 years ago

t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago