1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
natita [175]
3 years ago
8

Convert 11 years to hours

Mathematics
2 answers:
yan [13]3 years ago
5 0

Answer:

96360

Step-by-step explanation:

search hours a month then multiply that by 12

ki77a [65]3 years ago
5 0
The answer is 96,360hrs


If you have to round it it’s is 96,000
You might be interested in
A sample of n=8 scores has a mean of m=12. What is the value of ∑x for this sample
Andrews [41]

A sample of n=8 scores has a mean of m=12. What is the value of ∑x for this sample

Answer: We are given the mean of 8 scores is 12.

We are required to find the sum of these 8 observation's, \sum x

We know that:

\bar{x}=\frac{\sum x}{n}

We are given:

\bar{x} = 12, n=8

\therefore 12=\frac{\sum x}{8}

      \sum x = 12 \times 8

      \sum x=96

Hence, sum of 8 observation's, \sum x =96

7 0
3 years ago
1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

3 0
3 years ago
Which sentence describes why polygon MNOP is congruent to polygon JKLP?
svetlana [45]
The segment MN is equal in size to the segment JK, the segment NO is equal in size to the segment KL, the segment OP is equal in size to the segment LP.
7 0
3 years ago
What is the value of 3 in number 234
Nonamiya [84]
30 because it is in the tenths place
7 0
3 years ago
Read 2 more answers
Find the missing side lengths. Leave your answers as radicals in simplest form
konstantin123 [22]
X=y=29 divided by square root of 2.
Hope this helped!!!
Sorry I could not give you the exact value.
I guess it is 20.506
4 0
4 years ago
Other questions:
  • What is the value of x? 3x−8=8x+4/2
    11·2 answers
  • Out of 60 people surveyed at the movies, 45 people said they bought popcorn.If there were 420 people at the movies how many woul
    7·1 answer
  • What is 2 to the 50th power
    15·2 answers
  • What set of reflections would carry hexagon ABCDEF onto itself?a- x-axis, y=x, x-axis, y=xb- y-axis, x-axis, y-axisc- x-axis, y-
    8·1 answer
  • How do you determine whether to use substitution or elimination. give an example of a system that you would solve using each met
    5·1 answer
  • Draw a shape that uses 5 squares and has a perimeter of 10 units.
    14·1 answer
  • Can someone help me with this problem please. ( I need it as fast as possible).
    11·1 answer
  • HELP ASAP PLS <br> Find the volume
    12·2 answers
  • Si f(x)=2x-6 , entonces f(3)=?
    15·2 answers
  • HELP IT DUE NOW HURRY!
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!