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lesya [120]
3 years ago
6

Show work and explain with formulas:

Mathematics
1 answer:
Marina86 [1]3 years ago
6 0

17 Answer:  205

<u>Step-by-step explanation:</u>

\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}

18 Answer:  1968

<u>Step-by-step explanation:</u>

a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88

\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}

19 Answer:  -116

<u>Step-by-step explanation:</u>

\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}

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The value of the \lim_{x \to 0} f(x) and \lim_{x \to \frac{\pi }{3} } f(x) are 0 and 1.153 .

<h3></h3><h3>What is the limiting value of a function?</h3>

Limiting Value of a Function. The function's limit is the value of the function as its independent variable, such as x approaches a certain value called the limiting value. For simple equations, this is similar to finding out the value of y when x has a unique value.

Given that,

f(x) = 4x cos x

First to calculate the limit value of the given function at x=0.

\lim_{x \to 0} f(x) = \lim_{x \to 0} 4x cosx

                   = 4×0×1                              (∵ cos0 = 1)

\lim_{x \to 0} f(x) = 0

Similarly,

\lim_{x \to \frac{\pi }{3} } f(x) = \lim_{x \to \frac{\pi }{3} } 4x cosx

                     =  4×\frac{\pi }{3}×cos\frac{\pi }{3}

                     =  4×\frac{\pi }{3}×\frac{1}{2}                          (∵cos60° = \frac{1}{2})

\lim_{x \to \frac{\pi }{3} } f(x)  = 1.153

Hence, The value of the \lim_{x \to 0} f(x) and \lim_{x \to \frac{\pi }{3} } f(x) are 0 and 1.153.

To learn more about the limit of the function from the given link:

brainly.com/question/23935467

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