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jekas [21]
3 years ago
14

Solve by completing the square.

Mathematics
1 answer:
enot [183]3 years ago
8 0

Answer:

x^2 -8x -5 = -3

x^2 -8x -2 = 0

We complete the square by:

1) Moving the "non X" term to the right:

x^2 -8x  = 2

2) Dividing the equation by the coefficient of X²

The coefficient of x is 1 so we don't do anything

3) Now here's the "completing the square" stage in which we:

      • take the coefficient of X

that is -8

     • divide it by 2

-8 ÷ 2 = -4

     • square that number

-4*-4 = 16

     • then add it to both sides of the  equation.

x^2 -8x +16 = 2 +16

That becomes

(x -4)^2 = 18

we take the square root of both sides:

(x -4) = sqr root (18)

x1 = sqr root (18) +4

AND

(x+4) = sqr root (18) -4

x1 =  sqr root (18) +4 = 4.2426406871  + 4 = 8.2426406871

x2 = sqr root (18) -4 =  = 4.2426406871  - 4 = .2426406871

       

Step-by-step explanation:

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Answer:

the degree of the monomial is 11

Step-by-step explanation:

x = 1

y^3 = 3

z^7 = 7

add the exponents 1+3+7 = 11

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Your brother has $2000 saved for a vacation. His airplane ticket is $637. Write and solve an inequality to find how much he can
lapo4ka [179]

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$1,363

Step-by-step explanation:

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The slope formula can be used to prove that a triangle has:
lisabon 2012 [21]

Answer:

a right angle

Step-by-step explanation:

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8 0
3 years ago
What is 10 times 3/5 ??​
ASHA 777 [7]

Answer:

6

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
Use matrices and elementary row to solve the following system:
LiRa [457]

I assume the first equation is supposed to be

5x-3y+2z=13

and not

5x-3x+2x=4x=13

As an augmented matrix, this system is given by

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]

Multiply through row 3 by 1/2:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]

Add -1(row 2) to row 3:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]

Multiply through row 3 by 1/5:

\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -3(row 2) to row 1:

\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Multiply through row 1 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]

Add -2(row 1) to row 2:

\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]

Multipy through row 2 by -1:

\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]

The solution to the system is then

\boxed{x=1,y=-2,z=1}

5 0
3 years ago
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