Given , ⊙A ≅ ⊙V
To find:
What congruency statements can you make? Check all that apply.
BC ≅ ZY
∠ DAB ≅ ∠ ZVX
BE ≅ ZX
Solution:
Consider the attached figure while going through the following steps.
If two circles are congruent, then their corresponding chords are also congruent along with the congruent intercepted arcs.
⇒ BC ≅ ZY
and
⇒ BE ≅ ZX
Answer:
Step-by-step explanation:

<h2 /><h2>
<u>Consider</u></h2>

<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>




So, on substituting all these values, we get




<h2>Hence,</h2>

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
Answer:
Part A:




Part B:


and 
Step-by-step explanation:
Part A:
The inicial concentration of the lemonade is 50%, and the volume is 4 quarts, and we will add x quarts of a lemonade with a concentration of 100%, so the total volume will be y, and the concentration will be 0.7, so we have that:


Using the value of y from the first equation in the second one, we have:





Part B:
If he shoots a total of ten targets, we can write the equation:

Each stationary target is 2 points, and each moving target is 3 points, so if the total points is 23, we have:

If we subtract the second equation by two times the first one, we have:



⇒ 