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3 years ago

Center (7,-9); radius 3

1 answer:

6 0

The general form for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represent the centre and r represent the radius.

With centre (7, -9) and radius 3, we can simply substitute it into our form:

(x - 7)^2 + (y + 9)^2 = 3^2
(x - 7)^2 + (y + 9)^2 = 9

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GrogVix [38]

Answer:

12742π Km

Step-by-step explanation:

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3 years ago

Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.

Lilit [14]

For starters,

$\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k$

Consider the $n$ th partial sum, denoted by $S_n$ :

$S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n$

Multiply both sides by $\frac34$ :

$\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}$

Subtract $S_n$ from this:

$\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)$

Solve for $S_n$ :

$-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)$

$S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)$

Now as $n\to\infty$ , the exponential term will converge to 0, since $r^n\to0$ if $0$ . This leaves us with

$\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}$

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Neporo4naja [7]

Answer:

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3 years ago

Here is a situation: I have $5. I save $2 each week. How many weeks do I need to save until i have atleast $50? write the inequa

skad [1K]

Answer:

50 % 2 = 25 25 is the answer

Step-by-step explanation:

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2 years ago

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stiv31 [10]

Division is ur answer..

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