Answer:
A. EG = √3 × FG
D. EG = √3/2 × EF
E. EF = 2 × FG
Step-by-step explanation:
∵ tan 60 = √3
∵ tan60 = EG/GF
∴ EG/GF = √3
∴ EG = √3 × GF ⇒ A
∵ m∠F = 60°
∵ sin60 = √3/2
∵ sin 60 = EG/EF
∴ √3/2 = EG/EF
∴ EG = √3/2 × EF ⇒ D
∵ cos60 = 1/2
∵ cos60 = GF/EF
∴ GF/EF = 1/2
∴ EF = 2 × GF ⇒ E
Hi. 2=(7+h)/4 ; 2 * 4 = 7+h ; 8=7+h ; 8-7=h ; 1=h. The answer is h=1.
The aides of a right triangle will always satisfy the Pythagorean relationship ...
a² + b²= c²
We can take each of these and try it out:
(1) ... 10² + 24² = 100 + 576 = 676
√676 = <em>26 Yes</em>.
(2) ... 12² + 18² = 144 + 324 = 468
√468 = 21.6, not 20. <em>No.</em>
(3) ... 15² + 26² = 225 + 676 = 901
√901 = 30.017 Awfully close. As an engineer, I'll buy this one. <em>Yes.</em>
(4) ... 40² + 50² = 1,600 + 2,500 = 4,100
√4,100 = 64.03, not 80.<em> No.</em>
Answer:
The number of ways is equal to 
Step-by-step explanation:
The multiplication principle states that If a first experiment can happen in n1 ways, then a second experiment can happen in n2 ways ... and finally a i-experiment can happen in ni ways therefore the total ways in which the whole experiment can occur are
n1 x n2 x ... x ni
Also, given n-elements in which we want to put them in a row, the total ways to do this are n! that is n-factorial.
For example : We want to put 4 different objects in a row.
The total ways to do this are
ways.
Using the multiplication principle and the n-factorial number :
The number of ways to put all 40 in a row for a picture, with all 12 sophomores on the left,all 8 juniors in the middle, and all 20 seniors on the right are : The total ways to put all 12 sophomores in a row multiply by the ways to put the 8 juniors in a row and finally multiply by the total ways to put all 20 senior in a row ⇒ 