Ask question

Ask question

All categories

4 years ago

8

Shelly and Terrence earned points in a game by completing various tasks. Shelly completed x tasks and scored 90 points on each o

ne. The expression below shows Terrence's total points in the game:
90x − 20

What does the constant term of the expression represent?

1 answer:

MaRussiya [10]4 years ago

7 0

I need points so this is not right i’m sry 70 s x 90

You might be interested in

The explicit formula for the geometric sequence Negative one-ninth, one-third, negative 1, 3, negative 9, ellipsis is f (x) = ne

storchak [24]

Answer:

Common ratio, r=-3

Recursive Formula

$f(n)=-3f(n-1),$ $n\geq 2\\f(1)=-\dfrac{1}{9},$

Step-by-step explanation:

The formula for the geometric sequence: $-\dfrac{1}{9} ,\dfrac{1}{3}, -1,3,-9,\cdots$ is given as:

$f(x)=-\dfrac{1}{9}(-3)^{x-1}$

<u>Common Ratio</u>

Dividing the next terms by the previous terms, we obtain:

$\dfrac{1}{3} \div -\dfrac{1}{9} = \dfrac{1}{3} \times -9 =-3\\3 \div -1 =-3\\-9 \div 3 =-3$

Therefore, the common ratio of the sequence, r=-3

<u>Recursive Formula</u>

We observe that the next term, $f(n)$ is obtained by the multiplication of the previous term. f(n-1) by -3.

Therefore, a recursive formula for the sequence is:

$f(n)=-3f(n-1),$ $n\geq 2\\f(1)=-\dfrac{1}{9},$

3 0

4 years ago

I found the interval of convergence, but I am not sure how to do the second part, finding the sum of the series as a function of

antiseptic1488 [7]

The given series is geometric with common ratio $6^x - 9$ , which converges if $|6^x - 9|$ (i.e. the interval of convergence). We have the well-known result

$\displaystyle |r| < 1 \implies \sum_{n=0}^\infty ar^n = \frac{a}{1-r}$

If you're not familiar with that result, it's easy to reproduce.

Let $S_N$ be the $N$ -th partial sum of the infinite series,

$\displaystyle S_N = \sum_{n=0}^N \left(6^x - 9\right)^n = 1 + \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \cdots + \left(6^x - 9\right)^N$

Multiply both sides by the ratio.

$\left(6^x - 9\right) S_N = \left(6^x - 9\right) + \left(6^x - 9\right)^2 + \left(6^x - 9\right)^3 + \cdots + \left(6^x - 9\right)^{N+1}$

Subtract this from $S_N$ to eliminate all the powers of the ratio between 0 and $N+1$ .

$\left(1 - \left(6^x - 9\right)\right) S_N = 1 - \left(6^x - 9\right)^{N+1}$

Solve for $S_N$ .

$S_N = \dfrac{1 - \left(6^x - 9\right)^{N+1}}{10-6^x}$

Now as $N\to\infty$ , the exponential term converges to 0 and we're left with

$\displaystyle \sum_{n=0}^\infty \left(6^x-9\right)^n = \lim_{N\to\infty} S_N = \boxed{\frac1{10-6^x}}$

7 0

2 years ago

VERY EASY WILL GIVE BRAINLEAST The data set represents the prices, in dollars, of the items students are selling for a fundraise

mrs_skeptik [129]

Answer:

The plot of the provided data is shown in the attached picture.

=>

A number line goes from 0 to 10.

The whiskers range from 1 to 10.

The box ranges from 2.25 to 5.

A line divides the box at 4.

7 0

3 years ago

Read 2 more answers

Which inequality represents this statement?

DanielleElmas [232]

Answer:

−6n≥−12

Step-by-step explanation:

6 0

3 years ago

What are the answers

MArishka [77]

I hope this helps you

8 0

3 years ago