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fomenos
3 years ago
9

MZ2= m23= m24= Help due today!!!

Mathematics
2 answers:
denpristay [2]3 years ago
8 0
There is no picture or instructions. what do u need
marysya [2.9K]3 years ago
5 0
There isn’t in pictures? how are we supposed to help? what’s the question??
You might be interested in
How do I solve this?
Minchanka [31]

The maximum happens at x = -b/2a

x = -48/2(-16) = 3/2

Now replace x in the equation and solve for y:

y = 48(3/2) - 16(3/2)^2

y =  -72 - 36

y = 36

The maximum height is 36 feet.

4 0
4 years ago
12 M cubed minus 22 M squared minus 70 m
Andru [333]
I'm guessing you needed this factored and apologies for the lateness, this popped up after I answered another question. 

12m³ - 22m² - 70m

2m(6m² - 11m - 35)
2m(6m² + 10m - 21m - 35)
2m(2m(3m + 5) - 7(3m + 5))

Your factored answer is:

2m(2m - 7)(3m + 5)
3 0
4 years ago
What is the prime factorization of 88?
Harlamova29_29 [7]
Find the factors of 88

88/11 = 8
8/2 = 4
4/2 = 2

11 x 2 x 2 is the prime factorization of 88

hope this helps
7 0
3 years ago
Read 2 more answers
Describe how to locate (3,-1)
Harrizon [31]

Answer:

its A because you go 3 to the right and you go 1 down because its a negative

4 0
3 years ago
Read 2 more answers
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
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