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s2008m [1.1K]
3 years ago
6

Simplify each of the following:

Mathematics
2 answers:
Zepler [3.9K]3 years ago
7 0

Answer:

a)

  • √25 + √50 - √24 + √49 =
  • 5 + 5√2 - 2√6 + 7 =
  • 12 + 5√2 - 2√6

b)

  • √2(2√8 – 3√32 + 4√50) =
  • 2√16 - 3√64 + 4√100 =
  • 2*4 - 3*8 + 4*10 =
  • 8 - 24 + 40 =
  • 24
Novay_Z [31]3 years ago
6 0

Answer:

a.) 12 + 5√2 - 2√6

b.) 24

Step-by-step explanation:

a) √25 + √50 - √24 + √49

  • √25 + √50 - √24 + √49

Calculate the square root .

  • 5 + √50 - √24 + 7

Simplify the radical expression.

  • 5 + 5√2 - 2√6 + 7

Combine like terms.

  • 5 + 7 + 5√2 - 2√6
  • 12 + 5√2 - 2√6

b.) √2 ( 2√8 - 3√32 + 4√50 )

  • √2 ( 2√8 - 3√32 + 4√50 )

Simplify the radical expression.

  • √2 ( 4√2 - 3 × 2²√2 + 20√2)

Evaluate the power.

  • √2 ( 4√2 - 3 × 4√2 + 20√2)

Calculate the products.

  • √2 ( 4√2 -12√2 + 20√2)

Combine like terms.

  • √2 × (4 - 12 + 20 )√2
  • √2 × 12 √ 2

Multiply.

  • 2 × 12
  • 24
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Rewrite the system of linear equations as a matrix equation AX = B.
iren2701 [21]

Answer:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

Step-by-step explanation:

Let's find the answer.

Because we have 3 equations and 3 variables (x1, x2, x3) a 3x3 matrix (A) can be constructed by using their respectively coefficients.

Equations:

Eq. 1 : x1 + 2x2 + 5x3 = 5

Eq. 2 : x1 + x2 + x3 = 6

E1. 3 : 4x1 + 6x2 + 5x3 = 7

Coefficients for x1 ; x2 ; x3

From eq. 1 : 1 ; 2 ; 5

From eq. 2 : 1 ; 1 ; 1

From eq. 3 : 4 ; 6 ; 5

So matrix A is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]

And the vector of vriables (X) is:

\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]

Now we can find the resulting vector (B) using the 'resulting values' from each equation:

\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

In conclusion, AX=B is:

\left[\begin{array}{ccc}1&2&5\\1&1&1\\4&6&5\end{array}\right]*\left[\begin{array}{ccc}x1\\x2\\x3\end{array}\right]=\left[\begin{array}{ccc}5\\6\\7\end{array}\right]

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