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alekssr [168]
3 years ago
14

Find four consecutive even integers such that the sum of the second and fourth and integer is one half times the sum of the firs

t and third integers
Mathematics
1 answer:
Artyom0805 [142]3 years ago
8 0
Okay lets create an eqn from that information

A first int, B second int, C third int, D fourth int.

B = A + 2
C = A + 4
D = A + 6
A is the smallest integer

B + D = 0.5 (A + C) 

Now lets substitute

(A + 2) + (A + 6) = 0.5(A + (A + 4))
now lets dist

2a + 8 = 0.5(2a +4)
2a + 8 = a + 2
a + 8 = 2
a = -6

B = -6 +2
B = -4
C = -6 + 4
C = -2
D = -6 + 6
D = 0

Now using B + D = 0.5(A + C)
-4 + 0 = 0.5(-6 + (-2))
-4 = 0.5 (-8)
-4 = -4

Correct 
Therefore, First integer is -6, second integer is -4, third integer is -2 and fourth integer is 0
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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
Simplify the expression.<br> -2.7f + 0.8f - 16 - 4
Mazyrski [523]

Answer:

-1.9f-20

Step-by-step explanation:

-2.7f+0.8f-16-4

<em>Collect similar items</em>

-2.7f+0.8f=-1.9f

=-1.9f-16-4

<em>Subtract the numbers</em>

-16-4=-20

=-1.9f-20

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2 years ago
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Increase 620ml by 35%
taurus [48]
620*1.35=837 so 837ml is the answer
6 0
3 years ago
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The length of a rectangular room is 8 feet more than twice the width. If takes perimeter of the room is 124 feet, what are its d
Vlad1618 [11]

Hey there! I'm happy to help!

Let's call the length and width L and W respectively.

L=2W+8

2W+2L=124

We plug our value of L into the second equation and solve for W.

2W+2(2W+8)=124

We undo the parentheses with the distributive property.

2W+4W+16=124

Combine like terms.

6W+16=124

Subtract 16 from both sides.

6W=108

Divide both sides by 6.

W=18

We plug this W value into the first equation to solve for L.

L=2(18)+8

L=36+8

L=44

So, the length is 44 feet and the width is 18 feet.

Have a wonderful day! :D

8 0
3 years ago
A Scientist had a bottle that contains 56.6 mL of solution.She used 3.2 mL of the solution for an experiment.She poured half of
mario62 [17]
56.6 - 3.2 = 53.4
53.4 / 2 = 26.7
26.7 - 6 = 20.7ml
For the final answer she has 20.7ml of solution left.

Please mark as brainliest if you can
7 0
3 years ago
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