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Novosadov [1.4K]
3 years ago
9

Match the reasons with the statements in the proof if the last line of the proof would be 6. ∠1 and ∠7 are supplementary by defi

nition. Given: s || t Prove: 1, 7 are supplementary 1. s||t Given 2. ∠5 and ∠7 are supplementary. Exterior sides in opposite rays. 3. m∠5 + m∠7 = 180° Definition of supplementary angles. 4. m∠1 = m∠5 Substitution 5. m∠1 + m∠7 = 180° If lines are ||, corresponding angles are equal.
Mathematics
2 answers:
nataly862011 [7]3 years ago
7 0

Answer:

Step-by-step explanation:

Given: s II t

Substitution: m<1 + m<7 = 180*

If lines are II, corresponding angles are equal: m<1 = m<5

<5 and <7 are supplementary: definition of supplementary angles

m<5 + m<7 = 180* : exterior sides in opposite rays

Alecsey [184]3 years ago
4 0

Answer:

Given : s||t

Exterior sides in opposite rays : ∠5 and ∠7 are supplementary

Definition of supplementary angles : m∠5 + m∠7 = 180°

If lines are ||, corresponding angles are equal : m∠1 = m∠5

Substitution : m∠1 + m∠7 = 180°

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a student walks 50m on a bearing 0.25 degrees and then 200m due east how far is she from her starting point.​
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I'm going to use Physics here for this concept of vectors. Here are some stipulations I have set for the problem (aka rules I set and then followed throughout the problem):

** I am counting the 50 m as 2 significant digits even though it is only 1, and I am counting 200 as 3 significant digits even though it is only 1. 1 sig dig doesn't really give us enough accuracy, in my opinion.

** A bearing of .25 degrees is measured from the North and goes clockwise; that means that measured from the x axis, the angle is 89.75 degrees. This is the angle that is used in place of the bearing of .25 degrees.

** Due east has an angle measure of 0 degrees

Now let's begin.

We need to find the x and y components of both of these vectors. I am going to call the first vector A and the second B, while the resultant vector will be C. Starting with the x components of A and B:

A_x=50cos(89.75) so

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B_x=200cos(0) so

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Now for the y components:

A_y=50sin(89.75) so

A_y=50 (which I'm counting as 2 sig fig's)

B_y=200sin(0) so

B_y=0 and we need to add those results together.

C_y=50

Now for the resultant magnitude:

C_{mag}=\sqrt{(200)^2+(50)^2}  and that gives us a final magnitude of

C_{mag}=206 m

Now for the angle:

Since both the x and y components of the resultant vector are in quadrant 1, we don't need to add anything to the angle to get it right, so

tan^{-1}(\frac{50}{200})=14

The girl is 206 meters from her starting point at an angle of 14 degrees

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