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
vitfil [10]
3 years ago
9

A circle have a diameter of 18 meters. What is its

Mathematics
1 answer:
Sergio [31]3 years ago
3 0

Answer:

A. 56.54867 rounded to 56.56

Step-by-step explanation:

18/2=9

r=9

C=2πr=2·π·

C=2(3.14)(9)

9≈56.54867

You might be interested in
Which equation has infinitely many solutions?
777dan777 [17]

Answer:

B

Step-by-step explanation:

4 0
3 years ago
NEED HELP FAST WILL GIVE BRAILYIST What is the solution to Negative 1 minus 7?<br><br> –8 –6 6 8
kirza4 [7]

Answer:

-8

Step-by-step explanation:

took the test

6 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
2 years ago
What is the degree of 15x^3+63x+27x^4+35?<br> A) 0<br> B) 1<br> C) 3<br> D) 4
levacccp [35]

Answer:

D.) 4

Step-by-step explanation:

Your highest exponent is your degree.

Its easier if you put the polynomial in order:

27x^4+15x^3+63x+35

5 0
2 years ago
Read 2 more answers
Help for 13-15 pls and explain for 15
Pani-rosa [81]
13. 1
14. 1
15. any number and 0 will have a quotient of 0.
for example 7/0=0 .     8/0=0 .    9/0=0
if you try and check your work using multiplication 7*0=0 .    8*0=0 .    9*0=0
then you will realize the rule is right
4 0
3 years ago
Other questions:
  • A dilation has center (0, 0, 0). Find the image of the point (-1, -2, 0) for the scale factor 3.
    13·1 answer
  • Two ways to solve -2(4x+3)=12x-6
    7·1 answer
  • What is -5a + 3b + 8a
    5·1 answer
  • People with bmi less than 18.5 are often classified as underweight. What percent of men aged 20-29 are underweight by this crite
    12·1 answer
  • On average, a server is tipped 95 cents for each customer served. If 18 customers are served, how much money, in dollars, should
    8·1 answer
  • Is the answer is b ? Help
    8·1 answer
  • Use the net to find the surface area of the square pyramid
    9·2 answers
  • Segment &amp; Angle Addition Test
    7·1 answer
  • 5x+9=84 please someone answer this^^!
    7·2 answers
  • I think of a number. I add 3 then divide it by 2. My answer is 12. Form an equation and solve it to find my number.
    15·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!