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3 years ago

Select the function that represents a geometric sequence.

Mathematics

2 answers:

jeka57 [31]3 years ago

6 0

Answer:

A(n) = P(1 + i)^n-1, where n is a positive integer

Flauer [41]3 years ago

6 0

Answer: A. A(n) = P(1 + i)n - 1, where n is a positive intege

Step-by-step explanation: I did the work and the test it is right.

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In a fruit cocktail for every 100ml Orange juice you need 12ml Apple juice and 8ml of coconut milk what is the ratio of Apple ju

gayaneshka [121]

12:8:100= 3:2:25 is the simplest form

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2 years ago

What is the solution of the equation<br> √3x+2=√2x-2

MakcuM [25]

Step-by-step explanation:

$\sqrt{3 x - \sqrt{2x = 4} }$

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3 years ago

Which of the sets of three dimensions could make a right triangle

hodyreva [135]

Answer:

B) 6, 8, 10

Step-by-step explanation:

All right triangles will follow the following theorem (Pythagorean theorem):

$a^2+b^2=c^2$ , where $a$ and $b$ are two legs of the triangle and $c$ is the hypotenuse. Out of all answer choices, only B) 6, 8, 10 follow this theorem.

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3 years ago

Read 2 more answers

Find a common multiple between 30 and 42 that is bigger than them.

Wittaler [7]

Answer:

210

Step-by-step explanation:

GCF(30,42) = 6

LCM(30,42) = ( 30 × 42) / 6

LCM(30,42) = 1260 / 6

LCM(30,42) = 210

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3 years ago

Read 2 more answers

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

4 0

3 years ago