Hurry I need help ( -0.08(6.2)

Mathematics

2 answers:

Karolina [17]3 years ago

8 0

= -0.496 hope this helps a little doing know if it’s right or not

boyakko [2]3 years ago

7 0

The answer to this requires a simple calculation. The final answer would be negative as you're multiplying a positive by a negative. Multiply -.08 by 6.2. The answer would be - 0.496. Hope this helped :)

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1. The most massive planet in the solar system, Jupiter, has a mass of about 1,898,000,000,000,000,000,000,000,000 kg. Estimate

UNO [17]

Answer:

D. $2*10^{27} kg$

Step-by-step explanation:

Mass of Jupiter is given as 1,898,000,000,000,000,000,000,000,000 kg

Estimating this to the nearest whole number would be = 2,000,000,000,000,000,000,000,000,000 kg (rounding up the the "1,898")

To express this in scientific notation form, simply count how many zeros we have to the nearest whole number, 2.

We have 27 zeros there.

Therefore, we would multiply the whole number, 2, by 10 raise to the power of 27.

An estimate of the mass of Jupiter = $2*10^{27} kg$

6 0

3 years ago

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