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3 years ago

Find the maclaurin series for f(x=sin^2(x

1 answer:

4 0

Recall that $\sin^2x=\dfrac{1-\cos2x}2$ .

If you know the Maclaurin series for $\cos x$ , finding this expansion should be easy.

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Find the sum of the first ten prime numbers. Find the sum of the first ten prime numbers .

natali 33 [55]

The sum of the first ten prime numbers is 129.

<h3>What is a prime number?</h3>

The prime numbers are those numbers that have only one factor namely 1 and the number itself.

The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.

Their sum is

2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29

= 129

Hence, the sum of the first ten prime numbers is 129.

Learn more about prime numbers;

brainly.com/question/4184435

#SPJ1

5 0

2 years ago

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to <em>x</em> approaching 2)

Since both the numerator and denominator are continuous at <em>t</em> = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at <em>t</em> = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$

6 0

3 years ago

PLZ HELP ASAP

ICE Princess25 [194]

Seems way to hard for anyone to help you XD

8 0

3 years ago

What is the measure of Angle ABC A. 120 degrees B. 125 degrees C. 60 degrees D. 55 degrees

Finger [1]

We have the following:

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8 0

1 year ago

I need help identifying if this is a 90,180,270 degree transformation

Arte-miy333 [17]

$\begin{gathered} Choosing\text{ one point} \\ K(-1,-4) \\ K^{\prime}(1,4) \\ It\text{ means} \\ K^{\prime}=K(-x,-y) \\ Hence \\ This\text{ is a 180\degree transformation} \end{gathered}$

6 0

1 year ago