All categories

3 years ago

2(3)^n-1 what is the fifth term

Mathematics

1 answer:

Nitella [24]3 years ago

6 0

Answer:

The formula for an arithmetic series is shown below, where n = 1, 2, 3 ... f(n + 1) = f(n) + 8 If f(1) = 5, what are the fourth, fifth, and sixth terms

Step-by-step explanation:

You might be interested in

The diagram shows how cos θ, sin θ, and tan θ relate to the unit circle. Copy the diagram and show how sec θ, csc θ, and cot θ r

DIA [1.3K]

Copy the diagram and show how sec θ, csc θ, and cot θ relate to the unit circle.

The representation of the diagram is shown if Figure 1. There's a relationship between sec θ, csc θ, and cot θ related the unit circle. Lines green, blue and pink show the relationship.

a.1 First, find in the diagram a segment whose length is sec θ.

The segment whose length is sec θ is shown in Figure 2, this length is the segment $\overline{OF}$ , that is, the line in green.

a.2 Explain why its length is sec θ.

We know these relationships:

(1) $sin \theta=\frac{\overline{BD}}{\overline{OB}}=\frac{\overline{BD}}{r}=\frac{\overline{BD}}{1}=\overline{BD}$

(2) $cos \theta=\frac{\overline{OD}}{\overline{OB}}=\frac{\overline{OD}}{r}=\frac{\overline{OD}}{1}=\overline{OD}$

(3) $tan \theta=\frac{\overline{FD}}{\overline{OC}}=\frac{\overline{FC}}{r}=\frac{\overline{FC}}{1}=\overline{FC}$

Triangles ΔOFC and ΔOBD are similar, so it is true that:

$\frac{\overline{FC}}{\overline{OF}}= \frac{\overline{BD}}{\overline{OB}}$ 

∴ $\overline{OF}= \frac{\overline{FC}}{\overline{BD}}= \frac{tan \theta}{sin \theta}= \frac{1}{cos \theta} \rightarrow \boxed{sec \theta= \frac{1}{cos \theta}}$ 

b.1 Next, find cot θ

The segment whose length is cot θ is shown in Figure 3, this length is the segment $\overline{AR}$ , that is, the line in pink.

b.2 Use the representation of tangent as a clue for what to show for cotangent.

It's true that:

$\frac{\overline{OS}}{\overline{OC}}= \frac{\overline{SR}}{\overline{FC}}$

But:

$\overline{SR}=\overline{OA}$
$\overline{OS}=\overline{AR}$

Then:

$\overline{AR}= \frac{1}{\overline{FC}}= \frac{1}{tan\theta} \rightarrow \boxed{cot \theta= \frac{1}{tan \theta}}$

b.3 Justify your claim for cot θ.

As shown in Figure 3, θ is an internal angle and ∠A = 90°, therefore ΔOAR is a right angle, so it is true that:

$cot \theta= \frac{\overline{AR}}{\overline{OA}}=\frac{\overline{AR}}{r}=\frac{\overline{AR}}{1} \rightarrow \boxed{cot \theta=\overline{AR}}$

c. find csc θ in your diagram.

The segment whose length is csc θ is shown in Figure 4, this length is the segment $\overline{OR}$ , that is, the line in green.

3 0

4 years ago

All of the following expressions have the same value except

MrMuchimi

Answer:

1 - 4/5-(-1/5)

Step-by-step explanation:

1. - 4/5-(-1/5)

= -0.8+0.2 (if there is a minus sign inside and outside of the parenthesis, when we destroy that parenthesis, both the minus signs will be multiplied and becomes a (+).) (Making it to the percentile).

= -0.6

2. - 4/5-1/5

= -0.8-0.2

= -1

3. -(4/5+1/5)

= -(0.8+ 0.2) (Making those to the percentile).

= -(1) = -1 (When we destroy the parenthesis, if there is a minus sign outside the parenthesis, the result will become a minus.)

From the above expressions, we can see that after calculating, the second and third expressions have the same value, i. e. -1. However, the result of the first expression is different, and we get -0.6. So option one is the right choice.

6 0

3 years ago

Which of the following represents "eight times the difference of a number and seventeen is thirty"?

Simora [160]

B) 8(N-17)=30
8 times a the difference of a number and 17 is 30

3 0

4 years ago

Read 2 more answers

What is 3.098 written in words?

tatyana61 [14]

Three point zero nine eight

7 0

4 years ago

Read 2 more answers

Approximate the stationary matrix S for the transition matrix P by computing powers of the transition matrix P.

Scrat [10]

Answer:

S = [0.2069,0.7931]

Step-by-step explanation:

Transition Matrix:

$P=\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

Stationary matrix S for the transition matrix P is obtained by computing powers of the transition matrix P ( k powers ) until all the two rows of transition matrix p are equal or identical.

Transition matrix P raised to the power 2 (at k = 2)

$P^{2} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{2} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right]$

Transition matrix P raised to the power 3 (at k = 3)

$P^{3} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2203&0.7797\\0.2034&0.7966\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{3} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right]$

Transition matrix P raised to the power 4 (at k = 4)

$P^{4} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2086&0.7914\\0.2064&0.7936\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{4} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right]$

Transition matrix P raised to the power 5 (at k = 5)

$P^{5} =\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]X\left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2071&0.7929\\0.2068&0.7932\end{array}\right] X \left[\begin{array}{ccc}0.31&0.69\\0.18&0.82\end{array}\right]$

$P^{5} =\left[\begin{array}{ccc}0.2069&0.7931\\0.2069&0.7931\end{array}\right]$

P⁵ at k = 5 both the rows identical. Hence the stationary matrix S is:

S = [ 0.2069 , 0.7931 ]

6 0

4 years ago