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2 years ago

What is the reciprocal of 5 !!!help !!

Mathematics

2 answers:

enot [183]2 years ago

7 0

Answer:

1/5

Step-by-step explanation:

slega [8]2 years ago

5 0

The answer is 1/5 because 5= 5/1 as fraction then flip

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Marcella bought a 25-ounce bottle of olive for $5.88. She used 60% of the olive in two weeks. Which of the following is the cost

sergij07 [2.7K]

Aprox.$3.52. if you ever need to find the percentage of something you just times it by the fraction. e.g. 60% of 5 = 5x0.6. so all you need to do to figure this out is to find 60% of 5.88 which is apporxamitely 3.52

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3 years ago

What does 100,000 • 1,470 • 300,890 • 69 =

Mazyrski [523]

If you multiply them all you get 3.0519273e+15 as your answer.

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3 years ago

Translate the phrase to an algebraic expression.

mario62 [17]

Answer:

Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an equivalence relation.

4 0

3 years ago

5x3 + 14x2 + 9x help

Cerrena [4.2K]

<h3>Answer: x(x+1)(5x+9) </h3>

===================================================

Work Shown:

5x^3 + 14x^2 + 9x

x( 5x^2 + 14x + 9 )

To factor 5x^2 + 14x + 9, we could use the AC method and guess and check our way to getting the correct result.

A better way in my opinion is to solve 5x^2 + 14x + 9 = 0 through the quadratic formula

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(14)\pm\sqrt{(14)^2-4(5)(9)}}{2(5)}\\\\x = \frac{-14\pm\sqrt{16}}{10}\\\\x = \frac{-14\pm4}{10}\\\\x = \frac{-14+4}{10} \ \text{ or } \ x = \frac{-14-4}{10}\\\\x = \frac{-10}{10} \ \text{ or } \ x = \frac{-18}{10}\\\\x = -1 \ \text{ or } \ x = \frac{-9}{5}\\\\$

Then use those two solutions to find the factorization

x = -1 or x = -9/5

x+1 = 0 or 5x = -9

x+1 = 0 or 5x+9 = 0

(x+1)(5x+9) = 0

So we have shown that 5x^2 + 14x + 9 factors to (x+1)(5x+9)

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Overall,

5x^3 + 14x^2 + 9x

factors to

x(x+1)(5x+9)

6 0

3 years ago