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Angelina_Jolie [31]
3 years ago
5

Complete the ratio table to convert the units of measure from centimeters to inches or inches to centimeters.

Mathematics
2 answers:
djverab [1.8K]3 years ago
8 0

Answer:

Step-by-step explanation:

2.45 m in inches = 96.457 in. 2.45 meters to inches = 96.457″. To convert 2.45 meters to inches you have to divide the length expressed in the base unit of length in the International System (SI) of Units by 0.0254 .25.4Centimeters

Evgen [1.6K]3 years ago
6 0

Answer:

5.08 centimeters equals 2 inches

10 inches equals 25.4 centimeters

Step-by-step explanation:

we have to complete the following table.

According to the information contained in table 2.54 centimeters it is equal to 1 inch

The first information we have to complete is how many inches equals 5.08 centimeters

to solve it we apply a simple rule of three

5.08 centimeters equals 2 inches

The second information that we have to complete is how many centimeters equals 10 inches

10 inches equals 25.4 centimeters

Step-by-step explanation:

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Answer: The measure of <b is 26 degrees.

Step-by-step explanation:  

Complementary angles are angles that add up to 90 degrees. I remember this because my teacher told me this: "Complements are always RIGHT!" Of course, this is a reference to complementary angles combining to make right angles, which are 90 degrees.

To go back on your question, if  <a and <b are truly complementary, then m<a +m<b = 90 degrees. Substitute the measure of <a for 64 degrees, then you get 64 degrees + m<b = 90 degrees. Simplify. Then you'd get that the measure of <b is 26 degrees.

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<u>Question 4</u>

1) \overline{BD} bisects \angle ABC, \overline{EF} \perp \overline{AB}, and \overline{EG} \perp \overline{BC} (given)

2) \angle FBE \cong \angle GBE (an angle bisector splits an angle into two congruent parts)

3) \angle BFE and \angle BGE are right angles (perpendicular lines form right angles)

4) \triangle BFE and \triangle BGE are right triangles (a triangle with a right angle is a right triangle)

5) \overline{BE} \cong \overline{BE} (reflexive property)

6) \triangle BFE \cong \triangle BGE (HA)

<u>Question 5</u>

1) \angle AXO and \angle BYO are right angles, \angle A \cong \angle B, O is the midpoint of \overline{AB} (given)

2) \triangle AXO and \triangle BYO are right triangles (a triangle with a right angle is a right triangle)

3) \overline{AO} \cong \overline{OB} (a midpoint splits a segment into two congruent parts)

4) \triangle AXO \cong \triangle BYO (HA)

5) \overline{OX} \cong  \overline{OY} (CPCTC)

<u>Question 6</u>

1) \angle B and \angle D are right angles, \overline{AC} bisects \angle BAD (given)

2) \overline{AC} \cong \overline{AC} (reflexive property)

3) \angle BAC \cong \angle CAD (an angle bisector splits an angle into two congruent parts)

4) \triangle BAC and \triangle CAD are right triangles (a triangle with a right angle is a right triangle)

5) \triangle BAC \cong \triangle DCA (HA)

6) \angle BCA \cong \angle DCA (CPCTC)

7) \overline{CA} bisects \angle ACD (if a segment splits an angle into two congruent parts, it is an angle bisector)

<u>Question 7</u>

1) \angle B and \angle C are right angles, \angle 4 \cong \angle 1 (given)

2) \triangle BAD and \triangle CAD are right triangles (definition of a right triangle)

3) \angle 1 \cong \angle 3 (vertical angles are congruent)

4) \angle 4 \cong \angle 3 (transitive property of congruence)

5) \overline{AD} \cong \overline{AD} (reflexive property)

6) \therefore \triangle BAD \cong \triangle CAD (HA theorem)

7) \angle BDA \cong \angle CDA (CPCTC)

8) \therefore \vec{DA} bisects \angle BDC (definition of bisector of an angle)

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