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3 years ago

Evaluate the geometric series. Help me!!

Mathematics

1 answer:

aleksklad [387]3 years ago

3 0

Answer:

255

Step-by-step explanation:

The n th term of a geometric series is

$a_{n}$ = a $(r)^{n-1}$

where a is the first term and r the common ratio

1 $(2)^{n-1}$ ← is the n th term

with a = 1 and r = 2

The sum to n terms of a geometric series is

$S_{n}$ = $\frac{a(r^n-1)}{r-1}$ , thus

$S_{8}$ = $\frac{1(2^8-1)}{2-1}$

= $2^{8}$ - 1 = 256 - 1 = 255

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Tomas bought 80 tickets for rides at an amusement park. Each ride costs 5 tickets, and Tomas has been on x rides so far. Which e

Alex Ar [27]

Answer:

Option A, C and D are correct choices.

Step-by-step explanation:

Let x be the number of rides that Tomas has been on so far. Each ride costs 5 tickets. This means that cost of x rides will be 5x.

We are told that Tomas bought 80 tickets for rides at an amusement park.

To find the number of of tickets that Tomas has left we will subtract cost of x rides from total number of tickets that Thomas bought.

We can represent this information in an expression as: $80-5x$

Now let us see which of our given choices in equivalent to our expression.

A. $80-5x$

Upon looking at option A we can see that it is same as our expression, therefore, option A is the correct choice.

B. $80+5x$

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option B is not a correct choice.

C. $5(16-x)$

Upon distributing 5 we will get,

$80-5x$

Now this expression is same same as our expression, therefore, option C is the correct choice.

D. $-5x+80$

This expression is also same as our expression as we can rearrange the terms in this expression as: $80-5x$ , therefore, option D is a correct choice as well.

E. $5(16+x)$

Upon distributing 5 we will get,

$80+5x$

We can see that in this expression cost of x rides in being added instead of subtraction, therefore, option E is not a correct choice.

6 0

3 years ago

I'm having a problem finding the points

Ivenika [448]

The points are (-3,3) and (2,2) then you use distance formula to find the distance between them

4 0

2 years ago

Read 2 more answers

How do you find y = x + 8. x + y = 2 in substitoun form

Arturiano [62]

You plug in x + 8 for y in the second equation.
answer: x + x + 8 = 2

4 0

2 years ago

Susie and Amelia solved the same equation using two separate methods. Their work is shown in the table below:

kvv77 [185]

Stephen and Aaron solved the same equation using two separate methods. Their work is shown in the table below:

Stephen Aaron:
3x - 2 = 5x - 6 3x - 2 = 5x - 6
3x - 2 + 2 = 5x - 6 + 2 3x - 3x - 2 = 5x - 3x - 6
3x = 5x - 4 -2 = 2x - 6
3x - 5x = 5x - 5x - 4 -2 - 6 = 2x
-2x = -4 -8 = 2x
x = 2 -4 = x

Identify who made the error and what he did wrong.
Aaron made the error when he subtracted 6.

Aaron made the error when he subtracted 3x.

Stephen made the error when he added 2.

Stephen made the error when he subtracted 5x.

answer:

In the Aaron`s work:
- 2 = - 2 x - 6
and after that:
- 2 - 6 = 2 x
It should be:
- 2 + 6 = 2 x
or: - 2 + 6 = 2 x - 6 + 6
Answer:
A ) Aaron made the error when he subtracted 6.

5 0

3 years ago

Write the given expression in terms of x and y only. sin(sin−1(x) + cos−1(y))

yan [13]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2308127

_______________

Write the expression below in terms of x and y only:

(I'm going to call it "E")

$\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}$

Let

$\begin{array}{lcl} \mathsf{\alpha=sin^{-1}(x)}&\qquad&\mathsf{then~-\,\dfrac{\pi}{2}\le \alpha\le \dfrac{\pi}{2}}\\\\\\ \mathsf{\beta=cos^{-1}(x)}&\qquad&\mathsf{then~0\le \beta\le \pi.} \end{array}$

so the expression becomes

$\mathsf{E=sin(\alpha+\beta)}\\\\ \mathsf{E=sin\,\alpha\,cos\,\beta+sin\,\beta\,cos\,\alpha\qquad\quad(ii)}$

• Finding $\mathsf{sin\,\alpha:}$

$\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}$

• Finding $\mathsf{cos\,\alpha:}$

$\mathsf{sin^2\,\alpha=x^2}\\\\ \mathsf{1-cos^2\,\alpha=x^2}\\\\ \mathsf{cos^2\,\alpha=1-x^2}\\\\ \mathsf{cos\,\alpha=\sqrt{1-x^2}\qquad\quad\checkmark}$

because $\mathsf{cos\,\alpha}$ is positive for $\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}$

• Finding $\mathsf{cos\,\beta:}$

$\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}$

• Finding $\mathsf{sin\,\beta:}$

$\mathsf{cos^2\,\alpha=y^2}\\\\
 \mathsf{1-sin^2\,\beta=y^2}\\\\ \mathsf{sin^2\,\beta=1-y^2}\\\\ 
\mathsf{sin\,\beta=\sqrt{1-y^2}\qquad\quad\checkmark}$

because $\mathsf{sin\,\beta}$ is positive for $\mathsf{\beta\in [0,\,\pi].}$

Finally, you get

$\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry

6 0

3 years ago