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fredd [130]
3 years ago
7

use the vertical line test to determine if the relation {(-6-2), (-2,6), (03), (3,5)} is a function. explain your response.

Mathematics
1 answer:
spin [16.1K]3 years ago
7 0

It is not a function :)

hope this helped.

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When factoring an algebraic expression, what property are we "undoing"?
DerKrebs [107]

Answer:

The Distributive Property

4 0
3 years ago
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
2 years ago
Read 2 more answers
Complete the steps to solve for x.
mylen [45]
Answer: 11

Explanation:

(4/5)x - (1/4)x = 11
(16/20)x - (5/20)x = 11
(11/20)x = 11
7 0
2 years ago
Read 2 more answers
Solve for x. 3x + 5 = 23​
Lunna [17]

[|] Answer [|]

\boxed{X \ = \ 6}

[|] Explanation [|]

3x + 5 = 23

_________

_________

Subtract 5 From Both Sides:

3x + 5 - 5 = 23 - 5

Simplify:

3x = 18

Divide Both Sides By 3:

\frac{3x}{3} \ = \ \frac{18}{3}

Simplify:

X = 6

_________

_________

- Check Your Work -

Substitute 6 For X:

3 * 6 + 5 = 23

Parenthesis

Exponents

Multiply

Divide

Add

Subtract

3 * 6 = 18

18 + 5 = 23

\boxed{[|] \ Eclipsed \ [|]}

8 0
3 years ago
Read 2 more answers
Can someone help me for this question please? I just keep getting wrong answer! Pls someone explain to me step by step! ASAP!!!
Kobotan [32]

Answer:

  84 feet

Step-by-step explanation:

This problem involves several steps. The first step is to realize that the given figure does not show the required number of vertical stringers. It shows 5, but there will be 7 of them. The given diagram is helpful in that it shows a vertical stringer on the centerline of the arches.

The second step is to write a function that will tell you how long the stringer will be. I find it convenient to write the equation for an arch shape such as this using the parent function h(x) = 1-x^2. This parent function gives an arch of height 1 and width 1 from center (a total width of 2). You want an arch that is 16 ft high and 40 ft wide (one side from center), so you must scale this parent function both horizontally (by 40) and vertically (by 16). It becomes ...

  H(x) = 16(1 -(x/40)^2)

The taller arch is twice this height, so the length of a vertical stringer at position x is

  vertical length = 2H(x) -H(x) = H(x)

That is, the function H(x) we defined can be used to find the length of the stringers.

The third step is to find the stringer lengths. It can save some energy if you realize that the problem is symmetrical, so that the stringer at x=-30 is the same length as the one at x=30. We need to find stringer lengths every 10 feet from -40 feet to +40 feet. Of course, the ones at ±40 feet are zero length, because that is where the two arches meet.

  H(-30) = 16(1 -(3/4)^2) = 7

  H(-20) = 16(1 -(1/2)^2) = 12

  H(-10) = 16(1 -(1/4)^2) = 15

  H(0) = 16

Then the fourth step is to add up the stringer lengths, rounding the result as required.

  7 +12 +15 +16 +15 +12 +7 = 16 + 2(34) = 84 . . . . feet (no rounding needed)

Finally, you need to answer the question asked:

  The sum of vertical stringer lengths is 84 feet.

5 0
3 years ago
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