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3 years ago

Which expression is equivalent to √-80? 4√5 -4√5 -4i √ 5 4i √ 5

Mathematics

1 answer:

miskamm [114]3 years ago

6 0

Answer:

4i√5

Step-by-step explanation:

$\sqrt{-80}=\sqrt{(-1)(16)(5)}=\sqrt{-1}\cdot\sqrt{16}\cdot\sqrt{5}=4i\sqrt{5}$

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WINSTONCH [101]

I don’t understand the question can you give more detail

4 0

3 years ago

Read 2 more answers

Out of a class of 150 students, two-fifths opted for German classes and one-third opted for Italian classes. The remaining opted

Yuliya22 [10]

Fraction of students that opted for French = $\frac{4}{15}$

The total number of students = 150

Fraction of students that opted for German = $\frac{2}{5}$

Fraction of students that opted for Italian = $\frac{1}{3}$

Fraction of students that opted for French = 1 - Fraction of students that opted for German - Fraction of students that opted for Italian

Fraction of students that opted for French = $1-\frac{2}{5} - \frac{1}{3}$

Fraction of students that opted for French = $\frac{15-6-5}{15}$

Fraction of students that opted for French = $\frac{4}{15}$

Learn more here: brainly.com/question/18819021

8 0

3 years ago

The area of the square game board is 144 as in what's the length of the sides of the board

Tasya [4]

The length of the sides of the board are 12. (12 times 12 equals 144).

4 0

4 years ago

I need help??? f(x) = 3x + 10x and g(x) = 4x – 2, find (f – g)(x)

goldfiish [28.3K]

F(x) = 3ˣ + 10x and g(x) = 4x – 2, find (f-g)(x)
(f-g)(x) = f(x) - g(x):

(3ˣ + 10x) - (4x - 2) →→→ 3ˣ + 6x - 2. Then:

(f-g)(x) = 3ˣ + 6x - 2

5 0

4 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

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

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

4 years ago