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coldgirl [10]
3 years ago
10

Simplify. Show your work.

Mathematics
2 answers:
Shtirlitz [24]3 years ago
5 0

Answer:

5. 12√3

6. 46 - 14√5

Step-by-step explanation:

5.

We want to simplify these radicals. Look at the first term: -3√3. Well, since 3 is a prime number, we cannot square root any of the factors of 3 further, so this is already in its simplest form.

Look at the next term: 3√12. Remember that 12 = 3 * 4, and 4 = 2², so we can actually write 3√12 as 3 * √3 * √4 = 3 * √3 * 2 = 6√3.

Look at the final term: 3√27. Remember that 27 = 3 * 9 and 9 = 3², so we can write 3√27 as 3 * √3 * √9 = 3 * √3 * 3 = 9√3.

Our new expression is -3√3 + 6√3 + 9√3. We can now combine them because they're like terms: 12√3.

6.

Let's use FOIL to expand this. FOIL is first, outer, inner, and last.

-3 and -2 are the first terms, so multiply them: (-3) * (-2) = 6.

-3 and 2√5 are the outer terms, so multiply them: (-3) * (2√5) = -6√5.

4√5 and -2 are the inner terms, so multiply them: (4√5) * (-2) = -8√5.

4√5 and 2√5 are the last terms, so multiply them: (4√5) * (2√5) = 8 * √25 = 8 * 5 = 40.

Now, add all these up:

6 + (-6√5) + (-8√5) + 40 = 46 - 14√5.

Naily [24]3 years ago
3 0

Answer:

5. 12√3

6. 46 - 14√5

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Ana's manager is trying to figure out how much to charge for a chair that just arrived. If the wholesale price of the chair is $
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3 years ago
Which coefficient matrix represents a system of linear equations that has a unique solution ?
finlep [7]

Answer:

Option C

Step-by-step explanation:

We are given a coefficient matrix along and not the solution matrix

Since solution matrix is not given we cannot check for infinity solutions.

But we can check whether coefficient matrix is 0 or not

If coefficient matrix is zero, the system is inconsistent and hence no solution.

Option A)

|A|=\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0

since II row is a multiple of I row

Hence no solution or infinite

OPtion B

|B|=\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0

Hence no solution or infinite

Option C

\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68

Hence there will be a unique solution

Option D

\left[\begin{array}{ccc}5&10&5\\4&1&4\\-1&-2&-1\end{array}\right] \\=2(10)-2(10)=0=0

(since I row is -5 times III row)

Hence there will be no or infinite solution

Option C is the correct answer



4 0
3 years ago
Read 2 more answers
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