Answer:
(p ⋅ 10) − (p ⋅ 2)
Step-by-step explanation:
Try to substitute "p" with any number.
Let's say p = 2
So
p ⋅ (10 − 2)
2 ⋅ (10 − 2)
2 ⋅ (8)
= 16
Now for the other equations
(p ⋅ 10) − (p ⋅ 2)
(2 ⋅ 10) − (2 ⋅ 2)
(20) - (4)
= 16
(10 ⋅ 2) − p
(10 ⋅ 2) − 2
(20 ) − 2
= 18
10 ⋅ 2 − p
10 ⋅ 2 − 2
20 - 2
= 18
(10 ⋅ 2) ⋅ p
(10 ⋅ 2) ⋅ 2
(20) ⋅ 2
= 40
Only (p ⋅ 10) − (p ⋅ 2) has the same result as p ⋅ (10 − 2).
Cheers
√(1/121) = √1 / √121 = 1 / 11 = <em>11⁻¹</em>
Answer:
I think the answer should be A
Answer:
Correct option is
B
0
D
−1
sinx+sin2x+sin3x
=sin(2x−x)+sin2x+sin(2x+x)
=2sin2xcosx+sin2x [ by using sin(A+B)=sinAcosB+sinBcosA and sin(A−B)=sinAcosB−sinBcosA ]
=sin2x(2cosx+1)........(i)
cosx+cos2x+cos3x
=cos(2x−x)+cos2x+cos(2x+x)
=2cos2xcosx+cos2x [By using cos(a−b)=cosa⋅cosb+sina⋅sinb and cos(a+b)=cosa⋅cosb−sina⋅sinb]
=cos2x(2cosx+1).....(ii)
∴(sinx+sin2x+sin3x)
2
+(cosx+cos2x+cos3x)
2
=1
sin
2
2x(2cosx+1)
2
+cos
2
2x(2cosx+1)
2
=1.......[From(i)(ii)]
⇒(2cosx+1)
2
=1
⇒2cosx+1=±1
∴cosx=0or−1
