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Vsevolod [243]
3 years ago
12

What time is it 120 minutes after midday?​

Mathematics
1 answer:
coldgirl [10]3 years ago
7 0
Midday is 12pm so you add whatever hours to that. Which is 1:20pm after midday (12pm)
You might be interested in
How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?
Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+...  \ \ \  \ \ --- (1)

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)          

f'(x) = \dfrac{1}{1+x}

f'(0)   = \dfrac{1}{1+0}=1

f ' ' (x)    = \dfrac{1}{(1+x)^2}

f ' ' (x)   = \dfrac{1}{(1+0)^2}=-1

f '  ' '(x)   = \dfrac{2}{(1+x)^3}

f '  ' '(x)    = \dfrac{2}{(1+0)^3} = 2

f ' '  ' '(x)    = \dfrac{6}{(1+x)^4}

f ' '  ' '(x)   = \dfrac{6}{(1+0)^4}=-6

f ' ' ' ' ' (x)    = \dfrac{24}{(1+x)^5} = 24

f ' ' ' ' ' (x)    = \dfrac{24}{(1+0)^5} = 24

Now, the next process is to substitute the above values back into equation (1)

f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \  '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...

In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

To estimate the value of In(1.4), let's replace x with 0.4

In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...

In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...

Therefore, from the above calculations, we will realize that the value of \dfrac{0.4^5}{5}= 0.002048 as well as \dfrac{0.4^6}{6}= 0.00068267 which are less than 0.001

Hence, the estimate of In(1.4) to the term is \dfrac{0.4^5}{5} is said to be enough to justify our claim.

∴

The estimate of In(1.4) is the first five non-zero terms.

8 0
2 years ago
The graph of f(x) = 2x + 1 is shown below. Explain how to find the average rate of change between x = 2 and x = 5. (10 points) G
levacccp [35]
First we need to find f(2) and f(5), which are 5 and 11 respectively; all you have to do is plug in 2 and 5.

Then, we use the following formula:

(f(b)-f(a))/(b-a),

where b and a are the largest and smallest x values respectively.

Finally, we plug our values in:

(11-5)/(5-2)=6/3=2

In fact, the average rate of change of any linear function is just the coefficient of the x term. Hope this helped!
4 0
3 years ago
Points M and P or shown on the number line which expression gives the distance between two points
Korolek [52]

Answer:

A

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
HELPP
LenaWriter [7]
F, D, C
F: 5 x 3 = 15
D: 7 x 3= 21 - 2 = 19
C: 5 x 3 = 15 + 6 = 21
6 0
2 years ago
Read 2 more answers
Marissa wants to buy a DVD player that runs for at least 150 dollars she already saved $80 and plans to save an additional $10 e
SIZIF [17.4K]
Price of DVD player ≥ $150
Saved = $80
Amount needed = $150 - $80 = $70
Saving per week = $10

Let x be the number of weeks needed:
80 + 10x ≥ 150
10x ≥ 150 - 80
10x ≥ 70
x ≥ 70 ÷ 10 
x ≥ 7

Answer: She needs to save for for at least 7 weeks.

3 0
3 years ago
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