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3 years ago

6 hours 4 min 30 sec, 2 hours 43 min.

Mathematics

2 answers:

PtichkaEL [24]3 years ago

7 0

Answer: So whats the question?

Step-by-step explanation:

Luda [366]3 years ago

7 0

3 hours 2 mins 15 sec, 1 hour 21.5 mins

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Can someone help with number eleven? Please show work. 15 points by the way!!

attashe74 [19]

The irrigation arm is 300 ft. 300 ft is just the radius. The diameter is 600, because to get the diameter you must multiply the radius by two. So 600 is the land the irrigation arm covers.

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3 years ago

What is the distance between the following two points?<br> (-5, -1) and (-8, -4)

liberstina [14]

The answer is -3 , -3

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3 years ago

HELP NOW PLS EXTRA POINTS!!

aivan3 [116]

Answer:

X° + 44° = 2x° + 11° [ Sum of two adjacent interior angles of a triangle = opposite exterior angle]

=> - x° = 11°-44°

=> -x° = -33°

=> x° = 33°

Therefore the value of x° = 33°

Hope it helps

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3 years ago

The sum of the first n terms of a geometric series is 364? The sum of their reciprocals 364/243. If the first term is 1, find n

Afina-wow [57]

If the geometric series has first term $a$ and common ratio $r$ , then its $N$ -th partial sum is

$\displaystyle S_N = \sum_{n=1}^N ar^{n-1} = a + ar + ar^2 + \cdots + ar^{N-1}$

Multiply both sides by $r$ , then subtract $rS_N$ from $S_N$ to eliminate all the middle terms and solve for $S_N$ :

$rS_N = ar + ar^2 + ar^3 + \cdots + ar^N$

$\implies (1 - r) S_N = a - ar^N$

$\implies S_N = \dfrac{a(1-r^N)}{1-r}$

The $N$ -th partial sum for the series of reciprocal terms (denoted by $S'_N$ ) can be computed similarly:

$\displaystyle S'_N = \sum_{n=1}^N \frac1{ar^{N-1}} = \frac1a + \frac1{ar} + \frac1{ar^2} + \cdots + \frac1{ar^{N-1}}$

$\dfrac{S'_N}r = \dfrac1{ar} + \dfrac1{ar^2} + \dfrac1{ar^3} + \cdots + \dfrac1{ar^N}$

$\implies \left(1 - \dfrac1r\right) S'_N = \dfrac1a - \dfrac1{ar^N}$

$\implies S'_N = \dfrac{1 - \frac1{r^N}}{a\left(1 - \frac1r\right)} = \dfrac{r^N - 1}{a(r^N - r^{N-1})} = \dfrac{1 - r^N}{a r^{N-1} (1 - r)}$

We're given that $a=1$ , and the sum of the first $n$ terms of the series is

$S_n = \dfrac{1-r^n}{1-r} = 364$

and the sum of their reciprocals is

$S'_n = \dfrac{1 - r^n}{r^{n-1}(1 - r)} = \dfrac{364}{243}$

By substitution,

$\dfrac{1 - r^n}{r^{n-1}(1-r)} = \dfrac{364}{r^{n-1}} = \dfrac{364}{243} \implies r^{n-1} = 243$

Manipulating the $S_n$ equation gives

$\dfrac{1 - r^n}{1-r} = 364 \implies r (364 - r^{n-1}) = 363$

so that substituting again yields

$r (364 - 243) = 363 \implies 121r = 363 \implies \boxed{r=3}$

and it follows that

$r^{n-1} = 243 \implies 3^{n-1} = 3^5 \implies n-1 = 5 \implies \boxed{n=6}$

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2 years ago

Clara went to a music store and brought some CDs and DVDs. She paid $14 for each CD and $17 for each DVD. Let x represent the nu

Effectus [21]

14x+17y= total spent

7 0

3 years ago

6 hours 4 min 30 sec, 2 hours 43 min.​

6 hours 4 min 30 sec, 2 hours 43 min.