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Ulleksa [173]
2 years ago
10

2/3+(2/3)^2+(2/3)^3 + ... = x, find x.

Mathematics
1 answer:
DedPeter [7]2 years ago
3 0

Let S be the sum of the first n terms of the left side:

S = \dfrac23 + \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \cdots + \left(\dfrac23\right)^n

Multiply both sides by 2/3 :

\dfrac23 S = \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \left(\dfrac23\right)^4 + \cdots + \left(\dfrac23\right)^{n+1}

Subtract this from S :

S - \dfrac23 S = \dfrac23 - \left(\dfrac23\right)^{n+1}

Solve for S :

\dfrac13 S = \dfrac23 - \left(\dfrac23\right)^{n+1}

S = 2 - 3 \left(\dfrac23\right)^{n+1}

As n gets larger and larger, S converges to the given sum, and the term (2/3)ⁿ⁺¹ converges to zero, which leaves us with

\displaystyle \lim_{n\to\infty} S = \boxed{x = 2}

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Find the first four terms and the tenth 12-n
Ad libitum [116K]

Answer:

Q1: 7, 10, 13, 16           10th Term = 34

Q2: -1, 3, 7, 11              10th Term = 35

Q3: 11, 10, 9, 8             10th Term = 2

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Step-by-step explanation:

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3 0
3 years ago
Given right triangle MNL, what is the value of cos(M)? Three-fifths Three-fourths Four-fifths Five-thirds
lidiya [134]

By applying the Pythagorean Theorem and the Trigonometry ratio, CAH, the value of cos(M) = 4/5

<em><u>Recall:</u></em>

Trigonometry ratios, <em>SOH CAH TOA</em> can be applied to solve a right triangle.

  • Pythagorean Theorem can also be applied which is: c² = b² + a², where c is the longest side (hypotenuse length).

<em><u>Given:</u></em>

ΔMNL is a right triangle

ML = 25

NL = 15

Find MN using the Pythagorean Theorem:

MN = √(ML² - NL²)

MN = √(25² - 15²)

MN = 20

To find the value of cos(M), apply the trigonometry ratio, CAH, which is:

cos ∅ = Adj/Hyp

  • Where,

∅ = M (reference angle)

Hypotenuse = ML = 25

Adjacent = MN = 20

  • Plug in the values:

cos(M) = 20/25

cos(M) = 4/5

Learn more about Trigonometry ratios on:

brainly.com/question/10417664

5 0
3 years ago
Hey anyone mind helping out
natulia [17]
Your answer is A :))))
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3 years ago
Read 2 more answers
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