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3 years ago

What is minutes spent on homework

Mathematics

1 answer:

Eduardwww [97]3 years ago

3 0

It appears that you have here a list of the # of minutes spent each day on HW, and that y ou need to sum up this list to answer the question. Unfortunately I cannot read the last numeral.

The answer (assuming that my guess is correct) would be found as follows:

Total hours spent on HW in one week = (18+20+22+11+19+18+ ?? )

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ANSWER FOR BRAINLIEST

dezoksy [38]

Answer:A

Step-by-step explanation: A is 8d+d, and since d=1, 8d,+1d is equal to 9d

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3 years ago

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What is the perimeter of a triangle with the side lengths 4 cm, 6 cm, and (3y-2) cm?

jek_recluse [69]

Answer:

Step-by-step explanation:

P = 4 + 6 + (3y - 2) Remove the brackets

P = 10 + 3y - 2 Combine

P = 8 + 3y

Note 3y must be less than 8. That's because 2 sides of any triangle must be larger than the 3rd side or else you don't have a triangle.

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3 years ago

Zenin earns $142 per shift at his new job. During a pay period, he works 12 shifts. What would his pay be for that period?

Degger [83]

Answer:

total pay for his work is $1704

Step-by-step explanation:

given data

Zenin earns = $142 per shift

total shifty = 12

to find out

What would his pay for 12 shift

solution

we have given per shift charge and total shift

total pay for his work = total shift × per shift charge ................1

put here value

total pay for his work = 12 × $142

total pay for his work = $1704

3 0

3 years ago

Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

const2013 [10]

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$