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77julia77 [94]
3 years ago
6

Which inequality is shown in this graph? (03) 6 5 (2, -3)

Mathematics
1 answer:
Flura [38]3 years ago
8 0
The inequality is 5 explanation
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Evaluate<br> A) 34 <br> B) 46 <br> C) 40 <br> D) 7, 10, 13, 16
mezya [45]
The answer is B.

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8 0
4 years ago
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Help! Will give brainly! Explain why the value x=4 is not a solution to the inequality 2x&gt;10
melamori03 [73]

Answer

8 is not greater than 10

Step-by-step explanation:

2x>10

2(4)>10

8<10

x=4 can not be the solution to this equation because 8 is less than 10

*just as a side note think of the > sign as an alligator, the mouth always is open towards the larger number

7 0
3 years ago
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Which of the following is the inverse of f(x)=3-x/5?​
pshichka [43]

Answer:

-5x+3 OR 5x-3

Step-by-step explanation:

You find the inverse by replacing the x with the y, and solving for y. Doing this actually gives you -5x+3 but in this case,  the closest answer would be 5x-3.

UPDATE: -5x + 3 can be written as 3 - 5x lol

5 0
3 years ago
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Two parallel lines are intersected by a transversal. Prove: Angle bisectors of the same side interior angles are perpendicular.
coldgirl [10]

Answer:

The answer is given below

Step-by-step explanation:

From the diagram below,Let the line AB and CD be parallel line. Let transversal line EF cut AB at Y and transversal line EF cut CD at Z.

The bisector of ∠BYZ and ∠DZY meet at O. Therefore ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

∠BYZ and ∠DZY are interior angles.

∠BYZ + ∠DZY = 180 (sum of consecutive interior angles)

∠BYZ/2 + ∠DZY/2 = 180/2

∠BYZ/2 + ∠DZY/2 = 90°

In ΔOYZ:

∠YZO + ∠ZYO + ∠YOZ = 180 (sum of angles on a straight line).

But  ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

∠DZY/2 + ∠BYZ/2 + ∠YOZ = 180

90 + ∠YOZ = 180

∠YOZ = 180 - 90

∠YOZ = 90°

Therefore Angle bisectors of the same side interior angles are perpendicular.

7 0
3 years ago
Melissa needs 3.87 feet of string to use as a border for her poster. Melissa finds a piece of string that is 4.25 feet long. How
Tasya [4]
[tex] 4.25 - 3.87 = .38ft [tex]
7 0
4 years ago
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