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2 years ago

30 POINTS!!

Mathematics

1 answer:

Viefleur [7K]2 years ago

6 0

Answer:

400

Step-by-step explanation:

He makes $4 per shirt ($12-$8). $1600/4 = 400. So he woud need to see at least 400 shirts to break even.

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What is the equation of the line through the origin and (-5, 6) I hate try it's .-.

vazorg [7]

Answer:

$y=-\frac{6}{5} x$

Step-by-step explanation:

$y=-6/5x+0$

$y=-6/5x$

6 0

2 years ago

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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

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

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

If angle A = 34°, angle B = (x-5)°, angle C = 4y °, what is the value of y?

nevsk [136]

180-34=146
146-5=141
ratio=1:4
1+4=5
141÷5=28.2
x=28.2
y=112.8

I may be wrong, ask for more opinions
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4 0

3 years ago

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Hehehe hahaha ahahaha help me plz :(

Serggg [28]

5.2 is irrational cause the .222 keeps going

7 0

3 years ago

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A certain right triangle has area 30 in. squared . one leg of the triangle measures 1 in. less than the hypotenuse. let x repres

Brut [27]

A) The length of the longer leg is x-1

b) Based on the area, the other leg is 2*30/(x -1). Based on the Pythagorean theorem, the other leg is √(x^2 -(x -1)^2).

c) Equating the two expressions for the shorter leg, we have
.. 60/(x -1) = √(2x -1)
.. 3600/(x -1)^2 = (2x -1)
.. (2x -1)(x^2 -2x +1) = 3600
.. 2x^3 -5x^2 +4x -3601 = 0

d) There is one positive real root, at x=13. A graphical solution works well.

The three sides of the triangle are 5 in, 12 in, 13 in.

_____
5-12-13 is a well-known Pythagorean triple. It is the next smallest one after 3-4-5.

6 0

3 years ago