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Mariulka [41]
1 year ago
6

What is the domain, range, and function? {(-3, 3), (1, 1), (0, -2), (1,4), (5, -1)}

Mathematics
1 answer:
wolverine [178]1 year ago
8 0

The domain are all the inputs, that is; the x-values

Domain = { -3, 0, 1, 5}

The range are all the output, that is the y-values

Range = { 3, 1, -2, 4, -1}

This is just a relation and not a function, as we have more than 2 same value of x

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CAN SOMEONE PLEASE PLEASE PLEASE HELP ME, YOU’LL GET FREE EASY POINTS IF YOU GIVE ME THE RIGHT ANSWER !!
bekas [8.4K]

Answer:

  1. reflection across BC
  2. the image of a vertex will coincide with its corresponding vertex
  3. SSS: AB≅GB, AC≅GC, BC≅BC.

Step-by-step explanation:

We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.

__

<h3>1.</h3>

Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.

__

<h3>2.</h3>

In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.

__

<h3>3.</h3>

The markings on the figure show the corresponding parts to be ...

  • side AB and side GB
  • side AC and side GC
  • angle ABC and angle GBC
  • angle BAC and angle BGC

And the reflexive property of congruence tells us BC corresponds to itself:

  • side BC and side BC

There are four available congruence theorems applicable to triangles that are not right triangles

  • SSS -- three pairs of corresponding sides
  • SAS -- two corresponding sides and the angle between
  • ASA -- two corresponding angles and the side between
  • AAS -- two corresponding angles and the side not between

We don't know which of these are in your notes, but we do know that all of them can be used. AAS can be used with two different sides. SAS can be used with two different angles.

SSS

  Corresponding sides are listed above. Here, we list them again:

  AB and GB; AC and GC; BC and BC

SAS

  One use is with AB, BC, and angle ABC corresponding to GB, BC, and angle GBC.

  Another use is with BA, AC, and angle BAC corresponding to BG, GC, and angle BGC.

ASA

  Angles CAB and CBA, side AB corresponding to angles CGB and CBG, side GB.

AAS

  One use is with angles CBA and CAB, side CB corresponding to angles CBG and CGB, side CB.

  Another use is with angles CBA and CAB, side CA corresponding to angles CBG and CGB, side CG.

3 0
1 year ago
Read 2 more answers
Solve 12−5x−4kx=y for x.
IgorC [24]

Answer:

x = \frac{12-y}{4k+5}

Step-by-step explanation:

12-5x-4kx=y

1. Subtract 12 from both sides.

2.Distrubutive property (factor) a(b+c)=ab+ac

x(-5-4k)=y-12

  • Divide by -5-4k

x = \frac{y-12}{-5-4k}

= \frac{12-y}{4k+5}

5 0
3 years ago
What is the solution to 4(x-1)-2(3x+5)=-3x-1
Margarita [4]

To solve this problem, we need to get the variable x alone on one side of the equation. To begin, we are going to use the distributive property twice on the left side of the equation to expand the multiplication and get rid of the parentheses.

4(x-1) - 2(3x + 5) = -3x -1

4x - 4 -6x - 10 = -3x - 1

Next, we should combine like terms on the left side of the equation. This means we should add/subtract the variable terms and the constant terms in order to simplify this equation further.

-2x - 14 = -3x - 1

Then, we have to add 3x to both sides of the equation to get the variable terms all on the left side of the equation.

x - 14 = -1

After that, we should add 14 to both sides of the equation to get the variable x alone one the left side of the equation.

x = 13

Therefore, the answer is 13.

Hope this helps!

4 0
3 years ago
Read 2 more answers
Solve for x please help me!!
lyudmila [28]

Answer:

it should be 64 ....

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Which property tells us to multiply the powers to simplify
MaRussiya [10]
There are a couple of operations you can do on powers
We can multiply powers with the same base
x4⋅x2=(x⋅x⋅x⋅x)⋅(x⋅x)=x6
xa⋅xb=xa+b
This is an example of the product of powers property tells us that when you multiply powers with the same base you just have to add the exponents
8 0
3 years ago
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