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1 year ago

Find the first five terms. Please solve

Mathematics

1 answer:

KatRina [158]1 year ago

8 0

Answer:

a1=3, a2=6, a3=12, a4=24, a5=48

Step-by-step explanation:

a7=a*r^6=192

a10=a*r^9=1536, r^3=8, r=2 and a=3

a1=3, a2=6, a3=12, a4=24, a5=48

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Answer:

${(x + y)}^{2} - 1$

Step-by-step explanation:

${x}^{2} + y(y - 2x) - 1$

${x}^{2} + {y}^{2} - 2xy - 1$

Using the identity (a - b)^2 here ,

${(x + y)}^{2} - 1$

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2 years ago

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What is an equation of a vertical line that passes through (4,−6).

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Answer:

I think it would be x = 4

Step-by-step explanation:

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2 years ago

Two mechanics, Martin and Gordon, are working on your car. Martin can complete the work in 4 hours, while Gordon can complete th

ki77a [65]

Answer: 4/3 hours or 1 1/3 hours

Step-by-step explanation:

From the question, we are told that two mechanics, Martin and Gordon, are working on your car. Martin can complete the work in 4 hours, while Gordon can complete the work in 2 hours.

Since Martin can complete the work in 4 hours, that means in 1 hour, he will do 1/4 of the work.

Since Gordon can complete the work in 2 hours, that means in 1 hour, he'll do 1/2 of the job.

When they both work together, we will add their fraction of work done in one hour together. This will be:

= 1/4 + 1/2

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2 years ago

Here are yesterday's high temperatures (in Fahrenheit) in 12 U.S. cities. 48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80 Notice

irinina [24]

For the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

<h3>Measures of a Data </h3>

From the question, we are to determine the minimum, lower quartile, median, upper quartile, maximum, and interquartile range of the given data set

The given data set is

48, 50, 54, 56, 63, 63, 64, 68, 74, 74, 79, 80

Minimum = 48

Lower quartile = (54+56)/2

Lower quartile = 110/2

Lower quartile = 55

Median = (63+64)/2

Median = 127/2

Median = 63.5

Upper quartile = (74+74)/2

Upper quartile = 148/2

Upper quartile = 74

Maximum = 80

Interquartile range = Upper quartile - Lower quartile

Interquartile range = 74 - 55

Interquartile range = 19

Hence, for the given data set

Minimum = 48

Lower quartile = 55

Median = 63.5

Upper quartile = 74

Maximum = 80

Interquartile range = 19

Learn more on Measures of a Data here: brainly.com/question/15097997

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1 year ago

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Answer:

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Step-by-step explanation:

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2 years ago