Answer:
30
Step-by-step explanation:
<span>Let b= speed of boat in still water
Let c= speed of current
Upstream
Speed is ; b-c
d u=(b-c)* t u
Downstream
speed is ; b+c
d d=(b+c)*t d
given:
d u=18
t u=3
d d=24
t d=2
------------------
(1) 18=(b-c)*3
(2) 24=(b+c)*2
and
(1) 18=3b-3c
(2) 24=2b+c
Multiply both sides of (1) by 2
Multiply both sides of (2) by 3
(1)
(2)
Add (1) and (2)
And, from (1),
18=3*9-3c
3c=27-18
3c=9
c=3
The speed of the boat in still water is 9 mi/hr
The speed of the current is 3 mi/hr
check:
(1) 18=(9-3)*3
(2) 24=(9+3)*2
and
(1) 18=16*3
18=18
(2) 24=24
</span>
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The answer would be 4+y=-36
Solution:
LHS = (cosA+cosB)²+(sinA+sinB)²
= (cos²A+2cosAcosB+cos²B)+
(sin²A+2sinAsinB+sin²B)
/* By algebraic identity:
(x+y)² = x²+2xy+y² */
Rearranging the terms, we get
= (cos²A+sin²A)+(cos²B+sin²B)+2(cosAcosB+sinAsinB)
= 1+1+2cos(A-B)
________________________
we know that,
i) cos²x + sin²x = 1
ii) cosxcosy+sinxsiny = sin(x-y)
________________________
= 2+2cos(A-B)
= 2[1+cos(A-B)]
=
=
=$RHS$
Therefore,
(cosA+cosB)²+(sinA+sinB)²
=
••••
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