So 6 tons = 12,000 pounds
12,000 - 7,860 pounds (the original weight) = 4140 pounds which would be the weight of his items
m = 90 - 6d is the equation to determine m, the number of measures Harita still needs to memorize, as a function of d, the number of days of practice since she began learning the piece
<em><u>Solution:</u></em>
Given that Harita must memorize 90 measures of music for her cello solo at a concert
To find: equation to determine m, the number of measures Harita still needs to memorize, as a function of d, the number of days of practice since she began learning the piece
Let "m" be the number of measures Harita still needs to memorize
Let "d" be the number of days of practice since she began learning the piece
Given that She plans on memorizing 18 new measures for every 3 days of practice
<em><u>Rate per day is given as:</u></em>

Therefore she memorises 6 per day
<em><u>Therefore, the equation that relates 'm' to 'd' is: </u></em>
m = total measures of music she must memorise - (number of measures she memorises per day x d)
m = 90 - 6(d)
m = 90 - 6d
Thus the required equation is found
3/5
2-(-3)=5
7-4=3
slope should be 0.6 (3/5)
2/x + 1/7 = 4/x
2/x + 4/x = -1/7 move the terms
Transform the expression
-2/x =-1/7 change the signs
2/x = 1/7 cross multiply
Final answer X= 14
Answer:
sin 3 θ = 3 sin θ - 4 sin³θ
Step-by-step explanation:
<u><em>Step(i):</em></u>-
Given sin 3 θ
= sin ( 2θ + θ )
apply trigonometric formula
<em> sin ( A + B) = sin A cos B + cos A sin B </em>
<em> sin 2 A = 2 sin A cos B</em>
<em> Cos 2 A = 1 - 2 sin² A </em>
<em> cos ² A - sin ² A = 1</em>
<u><em>Step(ii):</em></u>-
sin 3 θ = sin ( 2θ + θ )
= sin 2θ cosθ + cos2θ sin θ
= 2 sin θ cos θ cos θ +( 1 - 2 sin² θ )sin θ
= 2 sin θ (cos² θ ) + sin θ- 2 sin³ θ
= 2 sin θ ( 1- sin²θ) + sin θ- 2 sin³ θ
= 2 sin θ - 2sin³θ + sin θ- 2 sin³ θ
= 3 sin θ - 4 sin³θ
<u><em>Final answer</em></u> :-
sin 3 θ = 3 sin θ - 4 sin³θ