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hammer [34]
3 years ago
15

8 by the power of -4

Mathematics
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

1/4096

Step-by-step explanation:

Note that if there is a negative in the power, simply put the term and power under 1.

8^(-4) = 1/(8^4)

Simplify:

1/(8 * 8 * 8 * 8) = 1/(4096)

1/4096 is your answer.

~

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Answer: i'm not really sure how to answer this one but i think you're wanting an answer to something. the equation for it would be 5 + 5x meaning the 5 is already there, and x is for the minutes that are they filling it up. so if the question asks how much after 5 minutes, you simply just put the 5 in place for the x.

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What is 7 / 2/3 in simplest form
galina1969 [7]

To do this we have to do 7 divided by 2/3

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7x2/3

Now we want to flip the second fraction over or make it a reciprocal

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7x 3/2 which is equal to 21/2

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4 0
2 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
How to simplify (-12)-7×(-12)7
Strike441 [17]
(-12)1 because of how you break it down
3 0
3 years ago
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vampirchik [111]

Answer:

  • Solution of equation ( x ) = <u>7</u>

Step-by-step explanation:

In this question we have given with an equation that is <u>4</u><u> </u><u>(</u><u> </u><u>5</u><u>x</u><u> </u><u>-</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u> </u><u>(</u><u> </u><u>9</u><u>x</u><u> </u><u>+</u><u> </u><u>3 </u><u>)</u><u>.</u> And we are asked to solve this equation that means we have to find the value of <u>x</u><u>.</u><u> </u>

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

<u>\quad \longrightarrow \: 4 ( 5x - 2 ) = 2 ( 9x + 3 )</u>

<u>Step </u><u>1</u><u> </u><u>:</u> Removing parenthesis :

\quad \longrightarrow \:20x - 8 = 18x + 6

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 8 from both sides :

\quad \longrightarrow \:20x - \cancel{ 8 }+  \cancel{8} = 18x + 6 + 8

On further calculations we get :

\quad \longrightarrow \:20x = 18x + 14

<u>Step </u><u>3 </u><u>:</u> Subtracting 18 from both sides :

\quad \longrightarrow \:20x - 18x =  \cancel{18x }+ 14 -  \cancel{18}

On further calculations we get :

\quad \longrightarrow \:2x = 14

<u>Step </u><u>4</u><u> </u><u>:</u> Dividing with 2 on both sides :

\quad \longrightarrow \: \frac{ \cancel{ 2}x}{ \cancel{2}}  =   \cancel{\frac{14}{2} }

On further calculations we get :

\quad \longrightarrow \:  \blue{\boxed{ \frak{x = 7}}}

  • <u>Therefore</u><u>,</u><u> </u><u>solution</u><u> </u><u>of </u><u>this </u><u>equation</u><u> </u><u>is </u><u>7</u><u> </u><u>or </u><u>we </u><u>can </u><u>say </u><u>that </u><u>value </u><u>of </u><u>this </u><u>equation</u><u> </u><u>is </u><u>7</u><u> </u><u>.</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

We are verifying our answer by substituting value of x in given equation. So ,

  • 4 ( 5x - 2 ) = 2 ( 9x + 3 )

  • 4 [ 5 ( 7 ) - 2 ] = 2 [ 9 ( 7 ) + 3 ]

  • 4 ( 35 - 2 ) = 2 ( 63 + 3 )

  • 4 ( 33 ) = 2 ( 66 )

  • 132 = 132

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> </u><u>our </u><u>value</u><u> for</u><u> x</u><u> is</u><u> </u><u>correct </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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2 years ago
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