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2 years ago

I need to know what the equation is for this graph..

Mathematics

1 answer:

Lostsunrise [7]2 years ago

3 0

Answer:

The choice second

$y = \frac{1}{3} \times ( \frac{1}{4} ) ^{x}$

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Alg1 i’ll give u points please help

3241004551 [841]

Answer= -9
It shows the y intercept on the graph

4 0

3 years ago

What is the circumference of a circle with a diameter of 6 feet ?Use 3.14

Illusion [34]

Answer:

18.84

Step-by-step explanation:

C=2πr

d=2r

Solving for C

C=πd=π·6=18.84ft

5 0

3 years ago

Read 2 more answers

What is the coordinate?

faltersainse [42]

Remember that the equation of a circle is:
$(x - h)^{2} + (y - k)^{2} = r^{2}$
Where (h, k) is the center and r is the radius.
We need to get the equation into that form, and find k.

$x^{2} - 6x + y^{2} - 10y = 56$

Complete the square. We must do this for x² - 6x and y² - 10y separately.

x² - 6x
Divide -6 by 2 to get -3.
Square -3 to get 9. Add 9,
x² - 6x + 9

Because we've added 9 on one side of the equation, we have to remember to do the same on the other side.

$x^{2} - 6x + 9 + y^{2} - 10y = 65$

Now factor x² - 6x + 9 to get (x - 3)² and do the same thing with y² - 10y.

y² - 10y
Divide -10 by 2 to get -5.
Square -5 to get 25.
Add 25 on both sides.

$(x - 3)^{2}+ y^{2} - 10y + 25= 90$

Factor y² - 10y + 25 to get (y - 5)²

$(x - 3)^{2}+ (y - 5)^{2} = 90$

Now our equation is in the correct form. We can easily see that h is 3 and k is 5. (not negative because the original equation has -h and -k so you must multiply -1 to it)

Since (h, k) represents the center, (3, 5) is the center and 5 is the y-coordinate of the center.

7 0

3 years ago

Find the midpoint of the line segment whose endpoints are (-2, 5) and (6, -9).

Katen [24]

We know that when $A=(x_A,y_A)$ and $B=(x_B,y_B)$ the midpoint of the line segment AB is given by:

$M=\left(\dfrac{x_A+x_B}{2},\dfrac{y_A+y_B}{2}\right)$

Here $A=(-2,5)$ and $B=(6,-9)$ so:

$M=\left(\dfrac{x_A+x_B}{2},\dfrac{y_A+y_B}{2}\right)=\left(\dfrac{-2+6}{2},\dfrac{5+(-9)}{2}\right)=\\\\\\=\left(\dfrac{4}{2},\dfrac{5-9}{2}\right)=\left(2,\dfrac{-4}{2}\right)=\boxed{\left(2,-2\right)}$

8 0

3 years ago

Determine the above sequence converges or diverges. If the sequence converges determine its limit

marshall27 [118]

Answer:

This series is convergent. The partial sums of this series converge to $\displaystyle \frac{2}{3}$ .

Step-by-step explanation:

The $n$ th partial sum of a series is the sum of its first $n\!\!$ terms. In symbols, if $a_n$ denote the $n\!$ th term of the original series, the $\! n$ th partial sum of this series would be:

$\begin{aligned} S_n &= \sum\limits_{k = 1}^{n} a_k \\ &= a_1 + a_2 + \cdots + a_{k}\end{aligned}$ .

A series is convergent if the limit of its partial sums, $\displaystyle \lim\limits_{n \to \infty} S_{n}$ , exists (should be a finite number.)

In this question, the $n$ th term of this original series is:

$\displaystyle a_{n} = \frac{{(-1)}^{n+1}}{{2}^{n}}$ .

The first thing to notice is the ${(-1)}^{n+1}$ in the expression for the $n$ th term of this series. Because of this expression, signs of consecutive terms of this series would alternate between positive and negative. This series is considered an alternating series.

One useful property of alternating series is that it would be relatively easy to find out if the series is convergent (in other words, whether $\displaystyle \lim\limits_{n \to \infty} S_{n}$ exists.)

If $\lbrace a_n \rbrace$ is an alternating series (signs of consecutive terms alternate,) it would be convergent (that is: the partial sum limit $\displaystyle \lim\limits_{n \to \infty} S_{n}$ exists) as long as $\lim\limits_{n \to \infty} |a_{n}| = 0$ .

For the alternating series in this question, indeed:

$\begin{aligned}\lim\limits_{n \to \infty} |a_n| &= \lim\limits_{n \to \infty} \left|\frac{{(-1)}^{n+1}}{{2}^{n}}\right| = \lim\limits_{n \to \infty} {\left(\frac{1}{2}\right)}^{n} =0\end{aligned}$ .

Therefore, this series is indeed convergent. However, this conclusion doesn't give the exact value of $\displaystyle \lim\limits_{n \to \infty} S_{n}$ . The exact value of that limit needs to be found in other ways.

Notice that $\lbrace a_n \rbrace$ is a geometric series with the first term is $a_0 = (-1)$ while the common ratio is $r = (- 1/ 2)$ . Apply the formula for the sum of geometric series to find an expression for $S_n$ :

$\begin{aligned}S_n &= \frac{a_0 \cdot \left(1 - r^{n}\right)}{1 - r} \\ &= \frac{\displaystyle (-1) \cdot \left(1 - {(-1 / 2)}^{n}\right)}{1 - (-1/2)} \\ &= \frac{-1 + {(-1 / 2)}^{n}}{3/2} = -\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\end{aligned}$ .

Evaluate the limit $\displaystyle \lim\limits_{n \to \infty} S_{n}$ :

$\begin{aligned} \lim\limits_{n \to \infty} S_{n} &= \lim\limits_{n \to \infty} \left(-\frac{2}{3} + \frac{2}{3} \cdot {\left(-\frac{1}{2}\right)}^{n}\right) \\ &= -\frac{2}{3} + \frac{2}{3} \cdot \underbrace{\lim\limits_{n \to \infty} \left[{\left(-\frac{1}{2}\right)}^{n} \right] }_{0}= -\frac{2}{3}\end{aligned}}_$ .

Therefore, the partial sum of this series converges to $\displaystyle \left(- \frac{2}{3}\right)$ .

8 0

3 years ago