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olchik [2.2K]
2 years ago
13

What is the rule for the following geometric sequence? 64,128,256,512...

Mathematics
2 answers:
professor190 [17]2 years ago
8 0

Answer:

B

Step-by-step explanation:

The explicit rule for a geometric sequence is

a_{n} = a₁ (r)^{n-1}

where a₁ is the first term and r the common ratio

Here a₁ = 64 and r = \frac{a_{2} }{a_{1} } = \frac{128}{64} = 2 , then

a_{n} = 64 (2)^{n-1} → B

antiseptic1488 [7]2 years ago
6 0

Answer:

b).

{ \tt{a _{n}  = a( {r}^{n - 1} )}} \\ { \tt{a _{n} = 64( {2}^{n - 1}) }}

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nikklg [1K]

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no. sorry. :(

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3 years ago
Add​​ using a number line.<br><br>-2 + 4<br><br>5__ 5
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Answer: i.. do.... not... know..

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1 year ago
Dennis needs to fix a leaky roof on his house but does not own a ladder. He thinks that a 25-foot ladder will be
Mars2501 [29]

Answer:

Step-by-step explanation: 2/8 =h/2

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2 years ago
At a fast food​ restaurant, 10 cheeseburgers​ w/onion and 5 orders of french fries contain 6900 calories. 5 cheeseburgers​ w/oni
Leona [35]

Write the equations,

10 c + 5 f = 6900

5c + 3f = 3660

Double,

10c + 6f = 7320

Subtract,

f = 7320 - 6900 = 420

10 c = 6900 - 5f = 6900 - 2100 = 4800

c = 480

Answer: Cheeseburger 480 calories, fries 420 calories

Check:

10 (480) + 5 (420) = 4800 + 2100 = 6900 good

5(480) + 3(420) = 2400 + 1260 = 3660 good

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B5x%2F8y%7D" id="TexFormula1" title="\sqrt[4]{5x/8y}" alt="\sqrt[4]{5x/8y}" al
Furkat [3]

Answer:  \frac{\sqrt[4]{10xy^3}}{2y}

where y is positive.

The 2y in the denominator is not inside the fourth root

==================================================

Work Shown:

\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\

The idea is to get something of the form a^4 in the denominator. In this case, a = 2y

To be able to reach the 16y^4, your teacher gave the hint to multiply top and bottom by 2y^3

For more examples, search out "rationalizing the denominator".

Keep in mind that \sqrt[4]{(2y)^4} = 2y only works if y isn't negative.

If y could be negative, then we'd have to say \sqrt[4]{(2y)^4} = |2y|. The absolute value bars ensure the result is never negative.

Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.

3 0
2 years ago
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