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AleksAgata [21]
3 years ago
12

Find the sum of the geometric series in which a1=160, a6= 5 and r= ½

Mathematics
1 answer:
Luden [163]3 years ago
4 0

Answer:

Sn = 315

The sum of the first six terms of the series is 315

Completed question:

Find the sum of the first six terms of the geometric series in which a1=160, a6= 5 and r= ½

Step-by-step explanation:

The sum of a geometric series in with common ratio

r < 1 is;

Sn = a(1 - r^n)/(1-r)

Where;

r = common ratio

a = first term

n = nth term

Given;

r = 1/2

a = a1 = 160

n = 6

Substituting the values, we have;

Sn = 160(1 - (1/2)^6)/(1 - 1/2)

Sn = 315

The sum of the first six terms of the series is 315

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In a company 85% of the workers are men if 510 women work for the company how many workers are there in all
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3400 workers

Step-by-step explanation:

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3 years ago
F divides HJ in the ratio 3:1. If H = (-11,7) and J = (5,3), what are the<br> coordinates of F?
Hatshy [7]

Answer:

The coordinates of HF are (1, 4)

Step-by-step explanation:

The parameters of the line are;

The coordinate of the end points are H = (-11, 7), and J = (5, 3)

The ratio by which the point F divides the line = 3:1

The segments in the line are HF, and FJ

Therefore;

The fraction of the length of HJ that is represented by HF = 3/(3 + 1) × HJ = 3/4 × HJ

HF = 3/4 × HJ

Which gives the coordinates of the point F as follows;

Coordinate of F = (-11 +(5 - (-11))×3/4, 7 + (3 - 7)×3/4) =  (1, 4)

The coordinates of F are (1, 4)

We check the length of HF, from the equation for the length to of a line to get;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

l_{HF} = \sqrt{\left (4-7  \right )^{2}+\left (1-(-11)  \right )^{2}} = \sqrt{\left (-3  \right )^{2}+\left (12  \right )^{2}} = 3\cdot \sqrt{17}

Similarly, we check the length of HJ, to get;

l_{HF} = \sqrt{\left (3-7  \right )^{2}+\left (5-(-11)  \right )^{2}} = \sqrt{\left (-4  \right )^{2}+\left (16  \right )^{2}} = 4\cdot \sqrt{17}

The length of HF = 3·√(17)

The length of HJ = 4·√(17)

Therefore, from HF = 3/4× HJ, we have;

HF = 3/4 × 4·√(17) = 3·√(17)

Therefore, the coordinates of HF are (1, 4)

7 0
3 years ago
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