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3 years ago

I will give u a five dollar gift card plz help!!!

Mathematics

2 answers:

juin [17]3 years ago

6 0

Answer:

1. The measure of ∠WOV is 60°. You would use complementary angles that are adjacent (∠WOV, and ∠XOW)

2. The measure of ∠YOZ is 60°. You would use the vertical angles that are non-adjacent (∠WOV, and ∠YOZ). These two angles are congruent so they would have the same measure. These angles combined also create supplementary angles

3. Another way to find the measure of ∠YOZ would be to make/write an equation and solve for x. For example, (3x+30)°=60°. x would equal 10 because 10x3=30+30=60°

Step-by-step explanation:

1. Since a complementary angle would equal 90°, simply subtract 30° from 90° resulting in 60°.

2. Because vertical angles are congruent and (∠WOV, and ∠YOZ) are a pair of them, they equal the same as each other so they're both 60°.

3. You can make any equation with x included as long as it equals 60° mine was just an example you can make your own like 10x+10=60 or 4x+20=60. Also to create your equation you also need to use the angle fact of the vertical angles

NikAS [45]3 years ago

6 0

Answer:

1. The measure of ∠WOV is 60°. You would use complementary angles that are adjacent (∠WOV, and ∠XOW)

3. Another way to find the measure of ∠YOZ would be to make/write an equation and solve for x. For example, (3x+30)°=60°. x would equal 10 because 10x3=30+30=60°

Step-by-step explanation:

1. Since a complementary angle would equal 90°, simply subtract 30° from 90° resulting in 60°.

2. Because vertical angles are congruent and (∠WOV, and ∠YOZ) are a pair of them, they equal the same as each other so they're both 60°.

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If k = −1 then the system has no solutions.

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Step-by-step explanation:

We have the following system of equations

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$\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]$

The last row determines, if there are solutions or not. To be consistent, we must have k such that

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If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.

Case k = 2:

$\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]$

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