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charle [14.2K]
2 years ago
5

HELPPP Provide the missing statements and reasons for the following proof:

Mathematics
1 answer:
agasfer [191]2 years ago
5 0

The missing statements and reasons in the proof showing that ∠1 is supplementary to ∠2 are:

S2: ∠2 ≅ ∠3

R2: corresponding angles

S3: m∠2 ≅ m∠3

R3: Definition of congruent angles

R5: Definition of supplementary angles

S7: m∠1 + m∠2 = 180°

S8: ∠1 is supplementary to ∠2

Given the proof to show that ∠1 is supplementary to ∠2, thus:

Since lines m and n are parallel to each other,

∠2 and ∠3 are corresponding angles, so, ∠2 ≅ ∠3 based on the corresponding angles theorem.

Since ∠1 and ∠3 form a linear pair, they are supplementary to each other.

By substitution, we would have m∠1 + m∠2 = 180°.

Therefore, we can conclude that ∠1 is supplementary to ∠2.

In summary, the missing statements and reasons in the proof showing that ∠1 is supplementary to ∠2 are:

S2: ∠2 ≅ ∠3

R2: corresponding angles

S3: m∠2 ≅ m∠3

R3: Definition of congruent angles

R5: Definition of supplementary angles

S7: m∠1 + m∠2 = 180°

S8: ∠1 is supplementary to ∠2

Learn more about supplementary angles on:

brainly.com/question/12919120

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<h3>Answer:  B) 7 units</h3>

===========================================================

Explanation:

The y coordinates of the two points are the same, so we can subtract the x coordinates and apply absolute value

|R - T| = |-6 - 1| = |-7| = 7

Or we can say

|T - R| = |1 - (-6)| = |1 + 6| = |7| = 7

Either way, the two points are 7 units apart.

You could use the distance formula to get the same answer, but that's definitely overkill in my opinion. The trick mentioned above also could work if the x coordinates were the same, but the y coordinates were different. In any other case, you would have to use the distance formula.

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Directions: Calculate the area of a circle using 3.14x the radius
Leokris [45]

\qquad\qquad\huge\underline{{\sf Answer}}♨

As we know ~

Area of the circle is :

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 4.4\div 2

\qquad \sf  \dashrightarrow \:r = 2.2 \: mm

Now find the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.2)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {4.84}^{}

\qquad \sf  \dashrightarrow \:area  \approx 15.2 \:  \: mm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 2</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 3.7 \div 2

\qquad \sf  \dashrightarrow \:r = 1.85 \:  \: cm

Bow, calculate the Area ~

\qquad \sf  \dashrightarrow \: \pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (1.85) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 3.4225 {}^{}

\qquad \sf  \dashrightarrow \:area  \approx 10.75 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 3 </h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (8.3) {}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 68.89

\qquad \sf  \dashrightarrow \:area \approx216.31 \:  \: cm {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>Problem 4</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 5.8 \div 2

\qquad \sf  \dashrightarrow \:r = 2.9 \:  \: yd

now, let's calculate area ~

\qquad \sf  \dashrightarrow \:3.14 \times  {(2.9)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  8.41

\qquad \sf  \dashrightarrow \:area  \approx26.41 \:  \: yd {}^{2}

・ .━━━━━━━†━━━━━━━━━.・

<h3>problem 5</h3>

\qquad \sf  \dashrightarrow \:r = d \div 2

\qquad \sf  \dashrightarrow \:r = 1 \div 2

\qquad \sf  \dashrightarrow \:r = 0.5 \:  \: yd

Now, let's calculate area ~

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times (0.5) {}^{2}

\qquad \sf  \dashrightarrow \:3.14  \times 0.25

\qquad \sf  \dashrightarrow \:area \approx0.785 \:  \: yd {}^{2}

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<h3>problem 6</h3>

\qquad \sf  \dashrightarrow \:\pi {r}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times  {(8)}^{2}

\qquad \sf  \dashrightarrow \:3.14 \times 64

\qquad \sf  \dashrightarrow \:area = 200.96 \:  \: yd {}^{2}

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

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