5x^3+8x^2/3x^4-16x^2=x^2(5x+8)/x^2(3x^2-16)=5x+8/3x^2-16
lim x--->0 5x+8/3x^2-16=8/-16=-1/2
Answer:
x = 33
Step-by-step explanation:
sin 29° = 16/x
x = 33
Answer:
8/50
Step-by-step explanation:
2/10 x 4/5 = 8/50
Hope that helps!
<span>The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides </span>⇒
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10/12 = (3x+2)/(5x-0.4)
10(5x-0.4) = 12(3x+2)
50x - 4 = 36x + 24
50x - 36x = 24 + 4
14x = 28
x = 28/14
x = 2
QS = 3x + 2 = 3*2 + 2 = 6 + 2 = 8
SR = 5x - 0.4 = 5*2 - 0.4 = 10 - 0.4 = 9.6
Answer is QS=8, SR=9.6
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Answer: Area of Δ DUO = 12.0 square units.
Step-by-step explanation:
From the diagram, Δ DPA is a right angled triangle and right angled at P.
Therefore ∠D will be
Tan ∅° = PA/DP ie, opposite side all over the adjacent.
= 4.5/3.75
Tan∅° = 1.2
to calculate ∅°, we know find the inverse of Tan 1.2
∅ = Tan^-1 1 .2 from your log tables or calculator
∅° = 50.20°.
= 50°
Since line DR is ⊥ to line OP
∠ADR = 90° - 50°
= 40°.
From the diagram,
∠ADR = ∠UDR = 40°
Therefore,
∠ODU = 180 - ( 40 + 40 + 90 ) { Angle on a straight line }
= 180 - 170
= 10°
From Δ UDM , line MU is he height of the required Δ DUO whose area is to be determined.
Now find the height MU
Tan10.0° = MU/10, where MU is the opposite side and 10.0 is the adjacent from the diagram given.
MU = Tan10.0 x 10.0
= 0.1763 x 10.0
= 1.763
Therefore to calculate the area of Δ DUO
= 1/2 x base x height
= 1/2 x line OD x line MU
= 1/2 x 14.0 x 1.763
= 7 x 1.763
= 12.341
= 12.0 square units.
=
