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snow_tiger [21]
2 years ago
13

Which equation represents the function shown on the graph?

Mathematics
1 answer:
Anna11 [10]2 years ago
4 0

Answer:

x = 0, y = 0

Step-by-step explanation:

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Please help.<br> Algebra.
Talja [164]

Answer:

y=4

Step-by-step explanation:

7 0
2 years ago
Write an expression that represents the height of a tree that begins at 10 feet and increases by 3 feet per year. Let t represen
mestny [16]
  • t represents number of years
  • initial height=10ft
  • Increases per year=3ft

Expression:-

\\ \sf\longmapsto 10+3t

3 0
3 years ago
PLZ HELP IMPORTANT!
ad-work [718]
After you've calibrated the sprayer to know your application rate in gallons per acre, divide that number into the gallons applied with each tankful to find how many acres each tankful covers. Then, multiply the acres per tank by the herbicide needed per acre to find the herbicide to put in each tankful.

Mark me as brainliest please
4 0
2 years ago
Beth solved the system of equations by making a substitution for y in the second equation. What equation did she get as a result
Slav-nsk [51]
Hello there!

The correct answer is option C

3x + 20x - 55 = -9


I am not sure if you wanted some more details but I hope this helps and if you do, just let me and I will edit it.

Good luck with your studies!

Happy New Year!~
3 0
2 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 <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
3 years ago
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