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goldenfox [79]
4 years ago
9

What is the reason for Statement 3 of the two-column proof?

Mathematics
1 answer:
Ivenika [448]4 years ago
8 0

Refer to the attached image.

Given:

The measure of \angle JMK= 52^\circ and \angle KML= 38^\circ.

Also, Three rays ML, MK, and MJ share an endpoint M. Ray MK forms a bisector as shown in the attached image and the bisector divides angle JML into two parts.

To Prove: \angle JML is a right angle.

Proof:

  Statements                                                                                 Reasons

1. m \angle JMK=52^\circ                                            Given

2. m \angle KML=38^\circ                                           Given

3. m \angle JMK+m \angle KML=m \angle JML  

The reason for statement 3 is Angle addition postulate. As angle JML is composed of 2 angles that is angle JMK and angle KML. So by adding the measures of angles JMK and KML, we will get the measure of angle JML which is referred as Angle addition postulate.

4. 52^\circ+38^\circ = m \angle JML   Substitution property of equality

5. 90^\circ = m \angle JML                  Simplification

6. \angleJML is a right angle.      Definition of right angle

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GalinKa [24]
Hello! 

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1 m = 3 ft 3.37 in

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3 years ago
Factor out the GCF from the given polynomial. 10^3 + 16^2 - 18x 10^3 + 16^2 - 18x = (Type your answer in factored form.)​
Vesna [10]

Correct question is;

Factor out the GCF from the given polynomial. 10x³ + 16x² - 18x. (Type your answer in factored form.)​

Answer:

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Step-by-step explanation:

The given polynomial is;

10x³ + 16x² - 18x

First thing is to inspect the coefficients and see the greatest number that all three can be divided by evenly without remainder.

The greatest number is 2.

Secondly let's inspect the variables to see what we can factor out.

They all contain x. Thus, x is a factor.

Thus, we will divide each term by 2x to obtain the GCF of the polynomial.

GCF = (10x³/2x) + (16x²/2x) - (18x/2x)

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tamaranim1 [39]

Let, price of fudge and gum is x and y respectively.

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