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3 years ago

What’s 16.001 to the nearest one? help me pleaseee

Mathematics

2 answers:

Dahasolnce [82]3 years ago

8 0

Answer: 16
The number directly after the decimal is less than 5, therefore it remains the same

puteri [66]3 years ago

4 0

Answer:

Step-by-step explanation:

I think you're referring to the nearest whole number, which should be 16 because that's after the decimal point

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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

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

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3 years ago

In at least one complete sentence, describe in words which numbers are included in this

ikadub [295]

The numbers included in this interval are from -12/17 and on.

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2 years ago

Lester bought 6.73 pounds of bologna at the deli. Renee bought 1.41 pounds of bologna. How much more bologna did Lester buy than

atroni [7]

Answer:

5.32

Step-by-step explanation:

You would have to subtract 6.73-1.41.

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3 years ago

I don't know how to do this please help explain how to do it aswell as give the answer so i understand

Zinaida [17]

We are given:

The function: y = -16t² + 64

where y is the height from ground, t seconds after falling

Part A:

when the droplet would hit the ground, it's height from the ground will be 0

replacing that in the given function:

0 = -16t² + 64

16t² = 64 [adding 16t² on both sides]

t² = 4 [dividing both sides by 16]

t = 2 seconds [taking square root of both sides]

Part B:

for second droplet,

height from ground = 16 feet

time taken = t seconds

acceleration due to gravity = 10 m/s²

initial velocity = 0 m/s

h = ut + (1/2)at² [second equation of motion]

16 = (0)(t) + (1/2)(10)(t²)

16 = 5t²

t² = 16/5

t = 1.8 seconds (approx)

Therefore, the second droplet takes the least amount time to hit the ground

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3 years ago

Is the following function even, odd, or neither? f(x) = 5x

salantis [7]

Answer:the function is odd

Step-by-step explanation:

because when we put a negative value of x it will give -f(x)

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3 years ago