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3 years ago

Part D Describe any patterns or trends you noticed when finding the products in part c

Mathematics

1 answer:

maria [59]3 years ago

8 0

Answer:

The middle two terms cancel out each time, and the final product is x2 plus the product of the positive coefficients of i in the original expression. All imaginary parts either cancel out or get replaced with -1, leaving an expression that doesn’t include imaginary numbers.

Step-by-step explanation:

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Please help me with this question

san4es73 [151]

Step-by-step explanation:

Given: $f'(x) = x^2e^{2x^3}$ and $f(0) = 0$

We can solve for f(x) by writing

$\displaystyle f(x) = \int f'(x)dx=\int x^2e^{2x^3}dx$

Let $u = 2x^3$

$\:\:\:\:du=6x^2dx$

Then

$\displaystyle f(x) = \int x^2e^{2x^3}dx = \dfrac{1}{6}\int e^u du$

$\displaystyle \:\:\:\:\:\:\:=\frac{1}{6}e^{2x^3} + k$

We know that f(0) = 0 so we can find the value for k:

$f(0) = \frac{1}{6}(1) + k \Rightarrow k = -\frac{1}{6}$

Therefore,

$\displaystyle f(x) = \frac{1}{6} \left(e^{2x^3} - 1 \right)$

5 0

3 years ago

May someone please help me?

kvv77 [185]

Answer:

the answer is ac line segment

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4 years ago

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Help...anyone plzzzzzzzzz

Arte-miy333 [17]

Can you please take another picture that is closer to the diagram, it's quite blurry.

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3 years ago

Read 2 more answers

The fifth grade students 15/20 went to the book fair of the students who went to the book fair 12/16 bought at least one book wh

grin007 [14]

Answer:

$\frac{9}{16}$

Step-by-step explanation:

Out of a total of 1, $\frac{15}{20}$ went to the fair and $\frac{12}{16}$ OF THOSE, bought atleast one book.

To find the fraction of students who bought at least one book, we will multiply both the fractions. That is the answer. Shown below:

$\frac{15}{20}*\frac{12}{16}\\=\frac{3}{4}*\frac{3}{4}\\=\frac{9}{16}$

Answer is $\frac{9}{16}$

8 0

3 years ago

At the center of espionage in Kznatropsk one is thinking of a new method for sending Morse telegrams. Instead of using the tradi

alexgriva [62]

Answer:

It takes less time sending 5 letters the traditional way with a probability of 36.7%.

Step-by-step explanation:

First we must take into account that:

- The traditional method is distributed X ~ Poisson(L = 1)

- The new method is distributed X ~ Poisson(L = 5)

$P(X=x)=\frac{L^{x}e^{-L}}{x!}$

Where L is the intensity in which the events happen in a time unit and x is the number of events.

To solve the problem we must calculate the probability of events (to send 5 letters) in a unit of time for both methods, so:

- For the traditional method:

$P(X=5)=\frac{1^{5}e^{-1}}{1!}\\\\P(X=5) = 0.367$

- For the new method:

$P(X=5)=\frac{5^{5}e^{-5}}{5!}\\\\P(X=5) = 0.175$

According to this calculations we have a higher probability of sending 5 letters with the traditional method in a unit of time, that is 36.7%. Whereas sending 5 letters with the new method is less probable in a unit of time. In other words, we have more events per unit of time with the traditional method.

7 0

3 years ago