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Iteru [2.4K]
3 years ago
8

A situation for which each member of a set, such as angles of a triangle, can be paired with one and only one member of another

set. - ? Angles paired with one another in a one-to-one correspondence between geometric figures. - ? Sides paired with one another in a one-to-one correspondence between figures. - ? If a one-to-one correspondence between the parts of two figures is such that the corresponding parts are equal, then the figures are congruent. - ? The angle formed by two sides of a triangle. - ? The side of a triangle that is formed by the common side of two angles. - ?
Mathematics
1 answer:
sineoko [7]3 years ago
3 0

Answer: A one-to-one correspondence between the parts of two figures is such that the corresponding parts are equal, then the figures are congruent.


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Answer:

1/3

Step-by-step explanation:

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Peter is reading a 193-page book. He has read three pages more than one-fourth of the number of pages he hasn’t yet read. How ma
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If (angle)1 = 7x and (angle)4 = 3x + 20, what is the value of x?
Semenov [28]
The value of x is 5. Angle one and four are vertical. You can do this by writing 7x = 3x + 20. Isolate the variable by moving 3x to the other side. If you do this, 3x becomes negative, and you end up with 7x-3x = 20 --> 4x = 20.
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7 0
2 years ago
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3х - 30 = у<br> 7y - 6 = 3х
mars1129 [50]

Answer:

Step-by-step explanation:

This one is simple substitution... at least, substitution is the easiest method. The first equation is 3<em>x</em> – 30 = <em>y</em>  and the second is 7<em>y</em> – 6 = 3<em>x</em>

As I look, I see 3 ways to use substitution to solve this:

  1. substitute 7<em>y</em> – 6 for 3<em>x</em>
  2. substitute 3<em>x</em> – 30 for <em>y</em>
  3. solve 3<em>x</em> – 30 = <em>y</em> for 3<em>x</em> and make it equal to 7<em>y</em> – 6

We're going to only use 1 method for the sake of time. Try the other two on your own. Assuming you don't make any mistakes, they will work.

<u>Method 1</u>:

3<em>x</em> – 30 = <em>y</em>

7<em>y</em> – 6 = 3<em>x</em>  — initial system of equations

7<em>y</em> – 6 – 30 = <em>y</em>  — substitute 7<em>y</em> – 6 for 3<em>x</em>

<u>7</u><u><em>y</em></u> – 6 – 30 = <u><em>y</em></u>  — marking like terms, bold for constants, <u>underlined</u> for variables

7<em>y</em> – 36 = <em>y</em>  — combining the constants and simplifying

Here, you could diverge into multiple paths: add 36 to both sides, subtract <em>y</em> from both sides, divide by 6 OR subtract 7<em>y</em> from both sides and divide by –6 . For the sake of time, I'm subtracting 7<em>y</em>, though I don't like dealing with negatives.

7<em>y</em> – 7<em>y</em> – 36 = <em>y</em> – 7<em>y</em>  — subtract 7<em>y</em> from both sides

–36 = –6<em>y</em>  — simplify

–36 ÷ –6 = –6<em>y</em> ÷ –6  — divide by –6 on both sides

<em>y</em> = 6  — simplify

Again, we can diverge here: substitute <em>y</em> into 3<em>x</em> – 30 = <em>y</em> or substitute <em>y</em> into 7<em>y</em> – 6 = 3<em>x</em>

I'm going to choose 3<em>x</em> – 30 = <em>y</em> but it will work either way, should you take the time (if you have it) to chase down every path this problem can take.

3<em>x</em> – 30 = <em>y</em>  — initial equation

3<em>x</em> – 30 = 6  — substitute 6 for <em>y</em>

3<em>x</em> – 30 + 30 = 6 + 30  — add 30 to both sides to isolate 3<em>x</em>

3<em>x</em> = 36  — simplify the expression

3<em>x</em> ÷ 3 = 36 ÷ 3  — divide both sides by 3 to isolate <em>x</em>

<em>x</em> = 12  — simplify

So, we have <em>x</em> = 12 and <em>y</em> = 6 . We know they work for 3<em>x</em> – 30 = <em>y</em>  but not if they work for 7<em>y</em> – 6 = 3<em>x</em> . Let's substitute those in to see if (12, 6) really is the solution point.

7<em>y</em> – 6 = 3<em>x</em>  — original equation

7(6) – 6 ≟ 3(12)  — substitute 6 for <em>y</em> and 12 for <em>x</em>

42 – 6 ≟ 36  — simplify by multiplying

36 = 36 ✔  — simplify by combining like terms on left side

Success! It works! We have found our solution!

I hope this helps increase your understanding of the concept. Have a great day!

8 0
3 years ago
Graph the system of equations. {x−y=64x+y=4
Fantom [35]
X-y=6 Equation 1
x+y=4 Equation 2

To graph the given system of equation, first find x and y-intercept of each equation.
x-y=6
When y=0
x=6  Point is (6,0)
When x=0
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y=-6  Point is (0,-6)

Now x-intercept and y-intercept for equation 2.
x+y=4
When x=0
y=4  Point is (0,4)
When y=0
x=4 Point is (4,0)

Now plot these points on the graph, the lines intersect each other at point (5,-1), which is the solution of the given system.

Answer: (5,-1)

5 0
3 years ago
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