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DedPeter [7]
3 years ago
5

Find the common ratio for the following sequence. 27, 9, 3, 1, ... User: Find the common ratio for the following sequence. 1/2,-

1/4, 1/8, -1/16 User: What is the seventh term of a sequence whose first term is 1 and whose common ratio is 3? 243 729 2,187
Mathematics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

My answer is 1/3 !!!!! Lmk if it's right or not

saveliy_v [14]3 years ago
4 0
R=common ratio= an / an-1
27,9,3,1.....;
 r=an/an-1= a₂/a₁=9/27=1/3

r=1/3.


1/2,-1/4,1/8,-1/16
r=an/an-1=(-1/4) / (1/2)=-2/4=-1/2

r=-1/2.

a₁=1
r=3
an=a₁*r^(n-1)
a₇=1.3⁷⁻¹=3⁶=729

a₇=729
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Answer:

Area = 13.15 square units

Step-by-step explanation:

Let the given vertices be represented as follows:

A(2, -1, 1) = 2i - j + k

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(i) Let's calculate the vectors of all the sides:

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AB = 5i + j + 4k - 2i + j - k                 [Collect like terms]

AB = 3i + 2j + 3k

BC = C - B =  (0i + j + k) - (5i + j + 4k)

BC = 0i + j + k - 5i - j - 4k                 [Collect like terms]

BC = -5i + 0j - 3k

CD = D - C =  (3i + 3j + 4k) - (0i + j + k)

CD = 3i + 3j + 4k - 0i - j - k                [Collect like terms]

CD = 3i + 2j + 3k

DA = A - D =  (2i - j + k) - (3i + 3j + 4k)

DA = 2i - j + k - 3i - 3j - 4k                [Collect like terms]

DA = -i - 4j - 3k

AC = C - A =  (0i + j + k) - (2i - j + k)

AC = 0i + j + k - 2i + j - k                [Collect like terms]

AC = -2i + 2j

BD = D - B = (3i + 3j + 4k) - (5i + j + 4k)

BD = 3i + 3j + 4k - 5i - j - 4k                [Collect like terms]

BD = -2i + 2j

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AB = CD This implies that AB || CD  [AB is parallel to CD]

AC = BD This implies that AC || BD  [AC is parallel to BD]

(iii) Therefore, ABDC is a parallelogram since opposite sides (AB and CD) are parallel. Hence, the points are vertices of a parallelogram

<u>Now let's calculate the area</u>

To find the area of the parallelogram, we find the magnitude of the cross product of any two adjacent sides.

In this case, we'll choose AB and AC

Area = |AB X AC|

Where;

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<u></u>

AB X AC = i(0 - 6) - j(0 + 6) + k(6 + 4)

AB X AC = - 6i - 6j + 10k

|AB X AC| = \sqrt{(-6)^2 + (-6)^2 + (10)^2}

|AB X AC| = \sqrt{172}

|AB X AC| = 13.15

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<u></u>

<u>PS: </u> ACBD is also a parallelogram. The diagram has also been attached to this response.

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