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EastWind [94]
3 years ago
11

A Relation Is Shown In The Graph Below. Write Out The Ordered Pairs For The Inverse, And Determine If The Inverse Is A Function

(Picture Attached).
I need the inverse of the points, and where or not the inverse is or isnt a function

Mathematics
2 answers:
Alexus [3.1K]3 years ago
7 0
The 2nd selection is correct.

_____
If your relation has some point (x1, y1), the inverse relation has the point (y1, x1). For example, one of the points of your relation is (-1, 0), so (0, -1) will be a point in the inverse relation.

A relation is a function if it does not contain two or more points with the same x-value.

(The relation shown in the graph is NOT a function, but its inverse relation IS a function.)
scZoUnD [109]3 years ago
3 0

Answer & Step-by-step explanation:

If your relation has some point (x1, y1), the inverse relation has the point (y1, x1). For example, one of the points of your relation is (-1, 0), so (0, -1) will be a point in the inverse relation.

A relation is a function if it does not contain two or more points with the same x-value.

(The relation shown in the graph is NOT a function, but its inverse relation IS a function.

B. (0,-1), (2,1), (-2,1), (3,3), (-3,3); inverse is a function is the answer.

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Help me out with this^^
sergij07 [2.7K]

Answer: 76 m^2

Step-by-step explanation:

A=\frac{4+15}{2}*8 = 76\;m^2

3 0
3 years ago
Area of a triangle = 1/2 • b•h
makvit [3.9K]

Answer:

15

Step-by-step explanation:

5*6=30

1/2*30=15

4 0
3 years ago
What is 56% in simplest form as fraction
muminat

Answer:

14/25

Step-by-step explanation:

Convert 56% to fraction form: 56/100, and then simplify to 14/25

4 0
3 years ago
Read 2 more answers
Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0
3 years ago
Corinne bought 5 bags of chips and 4 jars of dipping sauce for $21.82. At the same prices, Ginger bought 4 bags of chips and 3 j
AURORKA [14]
<h2>$2.98</h2>

Step-by-step explanation:

       5 bags of chips and 4 jars of dipping sauce cost $21.82. Also, 4 bags of chips and 3 jars of dipping sauce cost $16.86.

       This problem can be modeled using linear equations in two variables.

Let c be the cost of a bag of chip and s be the cost of a jar of dipping sauce.

The information yields the equations

       5c+4s=21.82,4c+3s=16.86\\\textrm{Multiply first equation by 4 and second equation by 5}\\20c+16s=87.28,20c+15s=84.3\\\textrm{Subtracting second equation from the first one}\\s=\$2.98

∴ A jar of dipping sauce costs \$2.98.

3 0
3 years ago
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