In triangle QRS, QR = 2x + 5, RS = 3x - 1,

Mathematics

1 answer:

Gemiola [76]3 years ago

7 0

Answer:

in triangle and the other is a good day to come over and play with the kids but I don't know if the voltage is reduced

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Determine whether the series is convergent or divergent. if it is convergent, find its sum. (if the quantity diverges, enter div

Lemur [1.5K]

The geometric series is convergent and the value is 16.

<h3>How to illustrate the information?</h3>

Recall that the sum of an infinite geometric series, S, given first term, t_1, and common ratio, r, is given by:

S = t_1/(1 - r)

Note that:

3 = (4)(3/4)

9/4 = (3)(3/4)

27/16 = (9/4)(3/4)

So this a geometric series with t_1 = 4 and r = 3/4. Therefore:

4 + 3 + 9/4 + 27/16 + ... = 4/(1 - 3/4) = 4/(1/4) = 16

The correct option is 16.

Learn more about geometric series on:

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Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum

4+3+ 9/4 +27/16 +???

choices are

1. 3/4

2. 12

3. 4

4. divergent

5. 16

7 0

1 year ago

The harbormaster wants to place buoys where the river bottom is 20 feet below the surface of the water. Complete the absolute va

Natalija [7]

Answer:

The answer is below

Step-by-step explanation:

The bottom of a river makes a V-shape that can be modeled with the absolute value function, d(h) = ⅕ ⎜h − 240⎟ − 48, where d is the depth of the river bottom (in feet) and h is the horizontal distance to the left-hand shore (in feet). A ship risks running aground if the bottom of its keel (its lowest point under the water) reaches down to the river bottom. Suppose you are the harbormaster and you want to place buoys where the river bottom is 20 feet below the surface. Complete the absolute value equation to find the horizontal distance from the left shore at which the buoys should be placed

Answer:

To solve the problem, the depth of the water would be equated to the position of the river bottom.

$d(h)=river \ bottom\\\\The \ river \ bottom=-20\ feet(below)\\\\d(h) = -20\\\\\frac{1}{5}|h-240|-48=-20\\ \\\frac{1}{5}|h-240|=-20+48\\\\\frac{1}{5}|h-240|=28\\\\|h-240|=28*5\\\\|h-240|=140\\\\h-240=140\ or\ h-240=-140\\\\h=140+240\ or\ h=-140+240\\\\h=380\ or\ h=100\\\\The\ buoys\ should\ be\ placed\ at\ 100\ feet\ and\ 380\ feet\ from\ left-hand\ shore$