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hodyreva [135]
3 years ago
8

Plz upload picture this is graphing

Mathematics
1 answer:
harina [27]3 years ago
6 0
G = 18.5. Add 1.5 to both sides, and calculate
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A (blank) angle has the same measure as its arc.
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A (central angle) has the same measure as its arc.
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3 years ago
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Solve the equation. Show the process, not just an answer. 6.EE.7
Natalka [10]

Isolate the variable, note the equal sign, what you do to one side, you do to the other.

1) 4k = 24

Isolate the variable, k. Divide 4 from both sides of the equation:

(4k)/4 = (24)/4

k = 24/4

k = 6

2) 34 + h = 60

Isolate the variable, h. Subtract 34 from both sides of the equation:

34 (-34) + h = 60 (-34)

h = 60 - 34

h = 26

3) 1/5x = 30

Isolate the variable, x. Multiply 5 to both sides of the equation:

(5) * (1/5)x = (30) * (5)

x = 30 * 5

x = 150

4) m - 42 = 85

Isolate the variable, m. Add 42 to both sides of the equation:

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6 0
3 years ago
Find the value of x, with work shown if possible
Furkat [3]

Answer:

Step-by-step explanation:

x / x + 7 = 21/27
cross multiply
27x=21x+147
6x = 147

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I hope this helps

4 0
2 years ago
Describe at least two similarities between constructing a perpendicular line through a point on a line and constructing a perpen
Vitek1552 [10]

The similarities between constructing a perpendicular line through a point on a line and constructing a perpendicular through a point off a line include:

  • Both methods involve making a 90-degree angle between two lines.
  • The methods determine a point equidistant from two equidistant points on the line.

<h3>What are perpendicular lines?</h3>

Perpendicular lines are defined as two lines that meet or intersect each other at right angles.

In this case, both methods involve making a 90-degree angle between two lines and the methods determine a point equidistant from two equidistant points on the line.

Learn more about perpendicular lines on:

brainly.com/question/7098341

#SPJ1

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2 years ago
What is the solution to 0.15x=40multiplied by x .05
galina1969 [7]
800n-3x=0 this is the answer help this helps
4 0
3 years ago
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