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ICE Princess25 [194]
2 years ago
5

Find the 14 term of the following geometric sequence. 10, 20, 40, 80,

Mathematics
2 answers:
MA_775_DIABLO [31]2 years ago
8 0

Answer:

the 14th term in the sequence would be 81920.

Step-by-step explanation:

because you are multiplying by 2 each time to find the next multiply 80 by 2 giving you 160 again giving you 320 again giving you 640 than 1280 , 2560 ,5120 ,10240 ,20480 ,40960 ,finally the 14th number in the sequence multiplying 40960 by 2 would give you your answer of 81920.

vagabundo [1.1K]2 years ago
4 0

Answer:

<u>81920</u>

Step-by-step explanation:

Finding the common ratio (r) :

  • 20/10
  • 2

Finding the 14th term :

  • a₁₄ = ar¹⁴⁻¹
  • a₁₄ = ar¹³
  • a₁₄ = (10)(2)¹³
  • a₁₄ = (10)(8192)
  • a₁₄ = <u>81920</u>
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Rafeeq bought a field in the form of a quadrilateral (ABCD)whose sides taken in order are respectively equal to 192m, 576m,228m,
Valentin [98]

Answer:

a. 85974 m²

b. 17,194,800 AED

c. 18,450 AED

Step-by-step explanation:

The sides of the quadrilateral are given as follows;

AB = 192 m

BC = 576 m

CD = 228 m

DA = 480 m

Length of a diagonal AC = 672 m

a. We note that the area of the quadrilateral consists of the area of the two triangles (ΔABC and ΔACD) formed on opposite sides of the diagonal

The semi-perimeter, s₁,  of ΔABC is found as follows;

s₁ = (AB + BC + AC)/2 = (192 + 576 + 672)/2 = 1440/2 = 720

The area, A₁, of ΔABC is given as follows;

Area\, of \, \Delta ABC = \sqrt{s_1\cdot (s_1 - AB)\cdot (s_1-BC)\cdot (s_1 - AC)}

Area\, of \, \Delta ABC = \sqrt{720 \times (720 - 192)\times  (720-576)\times  (720 - 672)}

Area\, of \, \Delta ABC = \sqrt{720 \times 528 \times  144 \times  48} = 6912·√(55) m²

Similarly, area, A₂, of ΔACD is given as follows;

Area\, of \, \Delta ACD= \sqrt{s_2\cdot (s_2 - AC)\cdot (s_2-CD)\cdot (s_2 - DA)}

The semi-perimeter, s₂,  of ΔABC is found as follows;

s₂ = (AC + CD + D)/2 = (672 + 228 + 480)/2 = 690 m

We therefore have;

Area\, of \, \Delta ACD = \sqrt{690 \times (690 - 672)\times  (690 -228)\times  (690 - 480)}

Area\, of \, \Delta ACD = \sqrt{690 \times 18\times  462\times  210} = \sqrt{1204988400} = 1260\cdot \sqrt{759} \ m^2

Therefore, the area of the quadrilateral ABCD = A₁ + A₂ = 6912×√(55) + 1260·√(759) = 85973.71 m² ≈ 85974 m² to the nearest meter square

b. Whereby the cost of 1 meter square land = 200 AED, we have;

Total cost of the land = 200 × 85974 = 17,194,800 AED

c. Whereby the cost of fencing 1 m = 12.50 AED, we have;

Total perimeter of the land = 576 + 192 + 480 + 228 = 1,476 m

The total cost of the fencing the land = 12.5 × 1476 = 18,450 AED

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And then take the square root of both sides. Your final answer should be: 
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