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3 years ago

Please help thank you

Mathematics

2 answers:

True [87]3 years ago

7 0

Hello again
Answer: 1 1\3
~hope i help~

vfiekz [6]3 years ago

3 0

The answer is 1/3. Have a nice day! :)

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in 2018 it will be 8.09 million

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3 years ago

PLEASE HELP, I NEED THIS OR I WILL FAIL MY CLASS PLZZZZ.

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3 years ago

PLZ HELP ILL MARK YOU BRAINLEST ‼️ a) 120° b) 50° c) 70° d) 180°

Alexus [3.1K]

Answer:

120°

Step-by-step explanation:

Add the 2 opposite angles to find the exterior angle.

50° + 70° = 120°

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3 years ago

Read 2 more answers

Last month you went on three hikes. One hike was 2 3/4 miles long, one was 1 1/2 miles long, and the other was 3 3/8 miles long

solniwko [45]

2 6/8 + 1 4/8 + 3 3/8
2 + 1 + 3 = 6
6/8 + 4/8 + 3/8 = 13/8
So 6 13/8 or 7 5/8

8 0

3 years ago

Enter the coefficients of the fifth Taylor polynomial T5(x) for the function f(x) = x5−3x4+2x2+5x−2 based at b=1. T5(x)= + (x−1)

DENIUS [597]

Compute the necessary values/derivatives of $f(x)$ at $x=1$ :

$f(1)=3$

$f'(1)=2$

$f''(1)=-12$

$f'''(1)=-12$

$f^{(4)}(1)=48$

$f^{(5)}(1)=120$

Taylor's theorem then says we can "approximate" (in quotes because the Taylor polynomial for a polynomial is another, exact polynomial) $f(x)$ at $x=1$ by

$T_5(x)=\dfrac3{0!}+\dfrac2{1!}(x-1)-\dfrac{12}{2!}(x-1)^2-\dfrac{12}{3!}(x-1)^3+\dfrac{48}{4!}(x-1)^4+\dfrac{120}{5!}(x-1)^5$

$T_5(x)=3+2(x-1)-6(x-1)^2-2(x-1)^3+2(x-1)^4+(x-1)^5$

###

Another way of doing this would be to solve for the coefficients $a,b,c,d,e,g$ in

$f(x)=a+b(x-1)+c(x-1)^2+d(x-1)^3+e(x-1)^4+g(x-1)^5$

by expanding the right hand side and matching up terms with the same power of $x$ .

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3 years ago