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Gnom [1K]
3 years ago
8

What's the recursive formula of the geometric sequence below?

Mathematics
1 answer:
yKpoI14uk [10]3 years ago
5 0

Answer:

Hey there!

We have an=a1(r^n-1), which is the formula for a geometric sequence.

The common ratio is 3, and 2 is the 1st term.

Thus, we have a1=2(3)^(n-1)

Hope this helps :)

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Find the area of the circle<br> Pls help quick
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Answer:

153.86.

Step-by-step explanation:

Formula for area of a circle: A= Pi R (R is radius) squared. First fill in the numbers: 3.14x7^2 Then Solve for the squared: 7^2=49, then finally multiply 49x3.14 which gets you to 153.86.

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9(7/4)-2= what in fraction form ? Can someone please help ?
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55/4

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Simplify the expression using trigonometric identities: sec (–θ) – cos θ.
ololo11 [35]

Answer:

sin θ . tan θ

Step-by-step explanation:

Note : -

sec ( - θ ) = sec θ

Formula / Identity : -

sec θ = 1 / cos θ

sec ( - θ ) - cos θ

= [ 1 / cos θ ] - cos θ

{ LCM = cos θ }

= [ 1 / cos θ ] - [ cos²θ / cos θ ]

= [ 1 - cos²θ ] / cos θ

{ 1 - cos²θ = sin²θ }

= sin²θ / cos θ

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4 0
2 years ago
TIMED QUESTION NEED HELP FASTWhat values of b satisfy 3(2b + 3)2 = 36?
egoroff_w [7]

Answer:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

Step-by-step explanation:

3(2b+3)^{2}=36\\\frac{3(2b+3)^{2}}{3}=\frac{36}{3}\\(2b+3)^{2}=12

Taking square root both sides, we get

\sqrt{(2b+3)^{2}}=\pm \sqrt{12}\\2b+3=\pm \sqrt{4\times 3}\\2b+3=\pm 2\sqrt{3}\\2b+3-3=\pm 2\sqrt{3}-3\\2b=-3\pm 2\sqrt{3}\\b=\frac{-3\pm 2\sqrt{3}}{2}\\b=\frac{-3+ 2\sqrt{3}}{2}\textrm{ or }b=\frac{-3- 2\sqrt{3}}{2}

Therefore, the values of b are:

\frac{-3+ 2\sqrt{3}}{2}\textrm{ and }\frac{-3- 2\sqrt{3}}{2}

5 0
3 years ago
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