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Roman55 [17]
3 years ago
15

What is the value of x and y

Mathematics
1 answer:
lidiya [134]3 years ago
4 0
If you have given an equation, you see the x and y in it. Just taking second equation and think any common factor between x's variable. With common factor, multiply both and you will find the value of y and put it in any equation, you will find the value of X
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Simplify the following expression:<br> 4x - 3y + 2z - 3x - y + 3z = ?
nordsb [41]
Add all the like terms together so 4x-3x is x or 1x -3y-y is -4y 2z+3z is 5z so put them all together it's x-4y+5z
7 0
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Given: m
marissa [1.9K]

Answer:

m∠LTE = 110

Step-by-step explanation:

1. add up all of the arcs.

2x+3x+4x-8+x-12

2. all of the arcs equal 360

2x+3x+4x-8+x-12=360

3. Find x

10x-20=360, x=38

4. angle LTE is equal to half of the sum of the intercepted arcs.

0.5(arc LE +GF)

5. plug in LE +GF with x

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7 0
3 years ago
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The equation below describes a parabola. If a is negative, which way does the parabola open? x = ay2
Anastasy [175]
This type of parabola opens either to the left or to the right. The negative makes it open to the left.
8 0
3 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 <em>equivalence relation</em> 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.

<em>Reflexive: </em>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.

<em>Symmetric</em>: 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.

<em>Transitive</em>: 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 <em>equivalence relation</em>.

4 0
4 years ago
We would like to estimate the proportion of uf students who owns a scooter to within 1% of the truth, with 95% confidence. We be
Varvara68 [4.7K]

Answer: 3458

i hope this work for you

8 0
2 years ago
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