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Jobisdone [24]
3 years ago
13

A snowstorm began on Monday evening. It snowed steadily until 6:00 am on Tuesday morning when the snow was 12 inches deep. Kevin

wrote the equation y= 3t - 6 to model the depth, y, in inches, of the snow on Tuesday morning, t hours after midnight. Does Kevin's equation correctly predict the total snowfall of 12 inches at 7:00 am on Tuesday morning?
Mathematics
1 answer:
poizon [28]3 years ago
3 0

Answer:

what is expected at 7am is 15 inches deep snow but what we have is 12 inches deep snow. The equation has failed in its prediction.

Step-by-step explanation:

In this question, we are asked to calculate if the prediction made by an equation modeled is correct.

Firstly let’s look at the equation in question;

y = 3t - 6

where y is the snow depth and t is the number of hours after midnight.

now we are looking at 7am, that’s 7 hours past 12am, meaning 7 hours after midnight.

let’s plug the value of t as 7 into the equation

y = 3(7) - 6

y = 21-6

y = 15 inches

according to the equation by Kevin, what is expected is 15 inches deep snow but what we have is 12 inches deep snow. The equation has failed in its prediction.

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Given the sequence in the table below, determine the sigma notation of the sum for term 4 through term 15. N an 1 4 2 −12 3 36 t
Rzqust [24]

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

<h3>What is sequence ?</h3>

Sequence is collection of  numbers with some pattern .

Given sequence

a_{1}=5\\\\a_{2}=-10\\\\\\a_{3}=20

We can see that

\frac{a_1}{a_2}=\frac{-10}{5}=-2\\

and

\frac{a_2}{a_3}=\frac{20}{-10}=-2\\

Hence we can say that given sequence is Geometric progression whose first term is 5 and common ratio is -2

Now n^{th}  term of this Geometric progression can be written as

T_{n}= 5\times(-2)^{n-1}

So summation of 15 terms can be written as

\sum_{n=4}^{15} T_{n}\\\\$\\$\sum_{n=4}^{15} 5(-2)^{n-1}$$

By applying basic property of Geometric progression we can say that sum of 15 terms of a sequence whose first three terms are 5, -10 and 2 is                    \sum_{n=4}^{15} 5(-2)^{n-1}$$  

To learn more about Geometric progression visit : brainly.com/question/14320920

8 0
3 years ago
All I need is help on these two so plz help me finish !!
aivan3 [116]

Answer:5.86 9.26

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A cement pathway of uniform width is to be poured around a rectangular swimming
Nookie1986 [14]

Answer:

A (2) =4*2 +882

Step-by-step explanation:

8 0
3 years ago
I need help answering this question please help
MArishka [77]
40

Or at least that what my calculator says.
8 0
3 years ago
How do you solve this
AlladinOne [14]

Answer:

  • domain: all terms are defined for <em>all real numbers</em>
  • solution: x = 6

Step-by-step explanation:

Rewrite the equation as a single exponential. After taking the log, the solution becomes obvious.

5^{x-2}-7^{x-3}=7^{x-5}+11\cdot 5^{x-4}\\\\5^{x-2}-11\cdot 5^{x-4}=7^{x-3}+7^{x-5}\\\\5^{x-2}(1-11\cdot 5^{-2})=7^{x-3}(1+7^{-2})\\\\5^{x-2}\dfrac{14}{25}=7^{x-3}\dfrac{50}{49}\\\\1=\dfrac{2\cdot 7^{x-3}5^4}{2\cdot 5^{x-2}7^3}=\left(\dfrac{7}{5}\right)^{x-6}\\\\0=(x-6)\log(1.4)\\\\x=6

6 0
3 years ago
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