The last side would be 10
Problem 215
Check out the attached image. Line up the letters ABC and CDE so the first set of letters are over the second set of letters. Order is important here. Note how A corresponds to C (second C), B corresponds to D, and (the first) C corresponds to E.
We have this mapping:
A <--> C
B <--> D
C <--> E
This means...
Angle BAC corresponds to Angle DCE (red angles)
Angle ABC corresponds to Angle CDE (blue angles)
Angle BCA corresponds to Angle DEC (green angles)
It also means...
Side BC corresponds to Side DE (red sides)
Side AC corresponds to Side CE (blue sides)
Side AB corresponds to Side CD (green sides)
Check out the attached image which I hope clears up any confusion you may have. Often I think it helps to represent stuff like this in a visual way.
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Problem 216
Answer: True
Since triangle ABC is congruent to triangle DEF, this means that side AB is congruent to side ED. These are corresponding sides. Since AB = 6, this means ED = 6 as well.
Answer:
One
General Formulas and Concepts:
<u>Algebra I</u>
- Reading a coordinate plane
- Solving systems of equations by graphing
Step-by-step explanation:
If 2 lines are parallel, they will have no solution.
If 2 lines are the same, they will have infinite amount of solutions.
We see from the graph that the 2 lines intersect at one point, near x = 1.5.
∴ our systems has 1 solution.
(√3 - <em>i </em>) / (√3 + <em>i</em> ) × (√3 - <em>i</em> ) / (√3 - <em>i</em> ) = (√3 - <em>i</em> )² / ((√3)² - <em>i</em> ²)
… = ((√3)² - 2√3 <em>i</em> + <em>i</em> ²) / (3 - <em>i</em> ²)
… = (3 - 2√3 <em>i</em> - 1) / (3 - (-1))
… = (2 - 2√3 <em>i</em> ) / 4
… = 1/2 - √3/2 <em>i</em>
… = √((1/2)² + (-√3/2)²) exp(<em>i</em> arctan((-√3/2)/(1/2))
… = exp(<em>i</em> arctan(-√3))
… = exp(-<em>i</em> arctan(√3))
… = exp(-<em>iπ</em>/3)
By DeMoivre's theorem,
[(√3 - <em>i </em>) / (√3 + <em>i</em> )]⁶ = exp(-6<em>iπ</em>/3) = exp(-2<em>iπ</em>) = 1
Answer: 100
Step-by-step explanation:
Given : The lodhl diner offers a meal combination consisting of an appetizer, a soup, a main course, and a dessert.
There are 5 appetizers, 5 soups, 4 main courses, and 5 desserts.
Also, a dessert and a appetizer are not allowed to take together.
By Fundamental counting principal ,
Number of three-course meals with dessert and without appetizer :
(1)
Number of three-course meals with appetizer and without dessert :
(2)
Now, the number of meals with either dessert or appetizer :-
[Add (1) and (2)]