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nataly862011 [7]
4 years ago
6

Find the product.

Mathematics
1 answer:
lesya [120]4 years ago
4 0
The answer is 9n^2- 60p + 100p^2

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Simplify (5x^3-x+14)-(3x^2-9x+4)
forsale [732]

5x^3 -3x^2 +8x +10 Combine all the like terms and multiply everything in the second parentheses by -1

I hope this help's you :)

7 0
4 years ago
Read 2 more answers
The elevation E, in meters, above sea level at which the boiling point of a certain liquid is t degrees Celsius is given by the
otez555 [7]
Given function is 

E(t) = 1100(80 - t) + 500(80 - t)^2

At t= 79.5 degree.

So we can place t= 79.5 in given equation

E(79.5) = 1100 (80 - 79.5) + 500* (80 - 79.5)^2

                    = 1100 * 0.5 + 500 * (0.5)^2
                     = 550 + 500 * 0.25
                    = 550 + 125 = 625

So elevation equal to 625 meter at t = 79.5 degree.


3 0
3 years ago
Use the Alternating Series Approximation Theorem to find the sum of the series sigma^infinity_n = 1 (-1)^n - 1/n! with less than
DanielleElmas [232]

Answer:

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that |-0.0000248|= 0.0000248 and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118

Step-by-step explanation:

Assuming the following series:

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!}

We want to approximate the value for the series with less than 0.0001 of error.

First we need to ensure that the series converges. If we have a series \sum a_n where a_n = (-1)^n b_n [/tex] or a_n =(-1)^{n-1} b_n where b_n \geq 0 for all n if we satisfy the two conditions given:

1) lim_{n \to \infty} b_n =0

2) {b_n} is a decreasing sequence

Then \sum a_n is convergent. For this case we have that:

lim_{n \to \infty} \frac{1}{n!} =0

And \frac{1}{n!} because \frac{1}{n!} =\frac{1}{n (n-1)!} and \frac{1}{n(n-1)!} < \frac{1}{(n-1)!}

So then we satisfy both conditions and then the series converges. Now in order to find the approximation with the error required we can write the first terms for the series like this:

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198 -0.0000248

For the 7th term we have 3 decimals of approximation but our value is 0.000198 higher than the error required, so we can use the 8th term and we have that |-0.0000248|= 0.0000248 and with this we have 4 decimals of approximation so if we add the first 8 terms we have a good approximation for the series with an error bound lower than 0.0001.

\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n!} = 1-0.5+0.16667-0.04167 +0.00833-0.001389 +0.000198-0.0000248 =0.632118

6 0
4 years ago
what is the equation of the line in standard form through each point? (3,1) &amp; perpendicular to -4c + y - 1= 0
ad-work [718]
C = 1/4 y + -1/4 is the answer
5 0
3 years ago
Read 2 more answers
The pic again i hope less blurry
Karolina [17]
The top and bottom lines a 5 feet. The side lines are 3 feet. You get this answer by:
1) how to find top and bottom lengths: 10-3-2= 5
2) how to find side lengths: 7-2-2= 3
7 0
3 years ago
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