The answer of the First one is 76
Answer:
∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° , ∠KLJ = 86°
Step-by-step explanation:
Here, given In ΔJLK and ΔMLP
Here, JK II ML, LM = MP
∠JLM = 22° and ∠LMP = 36°
Now, As angles opposite to equal sides are equal.
⇒ ∠MLP = ∠MPL = x°
Now, in ΔMLP
By <u>ANGLE SUM PROPERTY</u>: ∠MLP + ∠MPL + ∠LMP = 180°
⇒ x° + x° + 36° = 180°
⇒ 2 x = 180 - 36 = 144
or, x = 72°
⇒ ∠MLP = ∠MPL = 72°
Now,as JK II ML
⇒ ∠LJK = ∠JLM = 22° ( Alternate pair of angles)
Now, by the measure of straight angle:
∠MLP + ∠JLM + ∠JLK = 180° ( Straight angle)
⇒ 72° + 22° + ∠JLK = 180°
or, ∠JLK = 86°
In , in ΔJLK
By <u>ANGLE SUM PROPERTY</u>: ∠JKL + ∠JLK + ∠LJK = 180°
⇒ ∠JKL + 86° + 22° = 180°
⇒ ∠JKL = 180 - 108 = 72 , or ∠JKL = 72°
Hence, from above proof , ∠MLP = 72° , ∠LJK = 22° , ∠JKL = 72° ,
∠KLJ = 86°
Answer:
2/8 4/16
2/12 3/18
Step-by-step explanation:
Answer:
A = 25pi
Step-by-step explanation:
The area of the circle is given by
A = pi r^2
The radius is 5
A = pi 5^2
A = 25pi
Answer: If the number of bills counted is odd, the ones digit is 5; if the number of bills is even, the ones digit is 0.
Explanation
1) Simulate the counting of the bills:
number of bills counted value ones digit
1 5 5
2 5 + 5 = 10 0
3 10 + 5 = 15 5
4 15 + 5 = 20 0
5 20 + 5 = 25 5
6 25 + 5 = 30 0
2) Pattern: you can see that the ones digit alreternate: 5, 0, 5, 0, 5, 0, ...
If the number of bills counted is odd the ones digit is 5, if the number of bills is even the ones digit is 0.
