Surface area of a triangular prism formula?

Mathematics

2 answers:

madreJ [45]3 years ago

7 0

V=1/2 a *c * h
hope this helps

Murrr4er [49]3 years ago

3 0

<span>SA=bh+(s1+s2+s3)H

s1,s2,s3 are sides of prism b is base h is height</span>

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You have two lines. Line A measures 12/16, and line B measures 3/4.

hram777 [196]

Answer:

Step-by-step explanation:

Let's compare the two values of 12/16 and 3/4.

Notice that 12/16 isn't a simplified fraction; both the numerator and denominator are divisible by 4. So divide the top and bottom by 4:

12 / 4 = 3

16 / 4 = 4

We are left with:

12/16 = 3/4

Now, we see that Line A and Line B are equal in length, so the answer is B.

<em>~ an aesthetics lover</em>

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3 years ago

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krok68 [10]

remember: a filled dot means it includes that point and an empty dot means it doesn't

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B. has filled dot at x=0 so x=0 is a solution

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3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$