Can you help me solve this please :/

Mathematics

2 answers:

alex41 [277]3 years ago

7 0

Answer:

2√3

Step-by-step explanation:

In a special right triangle called the 30-60-90 triangle, the ratios of the side lengths are x-x√3- 2x respectively. This means that the side opposite to the 60 degrees is x√3 and the side opposite to the 30 degrees is x.

Since the side opposite to the 60 degrees is 6, x√3=6

x√3=6

x=6/√3

rationalize the denominator:

6√3/3

simlify

2√3

so the answer is the third one. 2√3

yawa3891 [41]3 years ago

5 0

It is 6/3 trust me or don’t

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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