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3 years ago

Can someone help me with this please?

Mathematics

2 answers:

alisha [4.7K]3 years ago

7 0

Answer:

Step-by-step explanation:

After finding the median, 49, you must find the median from 71-98, which is 72. P.S. the median is the number in the middle.

Blizzard [7]3 years ago

4 0

Answer:

Step-by-step explanation:

You might be interested in

Find the correct classisication irratinal number rational number whole number intergers

elena-s [515]

Step-by-step explanation:

Real numbers include:

Rational numbers include

Fractions, Integers

Integers include

Negative Integers, Whole numbers

Whole numbers include

Zero, Natural number

Irrational numbers

6 0

2 years ago

What is the area of the sector? Either enter an exact answer in terms of \piπpi or use 3.143.143, point, 14 for \piπpi and enter

n200080 [17]

Question:

What is the area of the sector? Either enter an exact answer in terms of π or use 3.14 and enter your answer as a decimal rounded to the nearest hundredth.

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete as the values of radius and central angle are not given.

However, I'll answer the question using the attached figure.

From the attached figure, the radius is 3 unit and the central angle is 120 degrees

The area of a sector is calculated as thus;

$Area = \frac{\alpha }{360} * \pi r^2$

Where $\alpha$ represents the central angle and r represents the radius

By substituting $\alpha = 120$ and r = 3

$Area = \frac{\alpha }{360} * \pi r^2$ becomes

$Area = \frac{120}{360} * \pi * 3^2$

$Area = \frac{1}{3} * \pi * 9$

$Area = \pi * 3$

$Area = 3\pi$ square units

Solving further to leave answer as a decimal; we have to substitute 3.14 for $\pi$

So, $Area = 3\pi$ becomes

$Area = 3 * 3.14$

$Area = 9.42$ square units

Hence, the area of the sector in the attached figure is $3\pi$ or 9.42 square units

8 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