Use subtraction the answer is 44
Answer:
y^ ⁻⁹/⁴ * z^⁽⁻³⁾
Step-by-step explanation:
(Y^3 z^4)^(-3/4) = (y^3)^(-3/4) * (z^4)^(-3/4)
= y^3* ⁽⁻³/⁴⁾ * z^4 *⁽⁻³/⁴⁾
=y^ ⁻⁹/⁴ * z^⁽⁻³⁾
Answer:
BC < ED ⇒ answer A
Step-by-step explanation:
* Lets revise some facts in the triangle
- If one side of a triangle is longer than another side, then the angle
opposite the longer side will be larger than the angle opposite the
shorter side
- If one angle in a triangle is larger than another angle in a triangle,
then the side opposite the larger angle will be longer than the side
opposite the smaller angle
* Lets solve the problem
- In the two triangles BCD and DEB
∵ CD = 8 and BE = 8
∴ CD = BE
∵ Side BD is a common side in the two triangles
- The third side in Δ BCD is BC and the third side in DEB is DE
∵ BC is the opposite side to the angle of measure 24°
∵ ED is the opposite side to the angle of measure 30°
∵ The measure 24° < the measure 30°
∴ The side opposite to the angle of measure 24° < the side opposite
to the angle of measure 30°
∵ The other two sides of the 2 triangles BCD and DEB are equal
∴ We can compare between the 3rd sides in the Δ BCD and Δ DEB
∴ BC < ED
tan( 20 ) + 4 Sin( 20 ) =
( Sin( 20 ) / Cos( 20 ) ) + 4 Sin( 20 ) =
Sin( 20 ) + 4 Sin( 20 ).Cos( 20 ) / Cos( 20 ) =
Sin( 20 ) + 2 × 2 Sin(20).Cos(20)/ Cos(20) =
Sin( 20 ) + 2 × Sin( 40 ) / Cos( 20 ) =
Sin( 20 ) + 2Sin( 40 ) / Cos( 20 ) =
Sin( 20 ) + 2Cos( 50 ) / Cos ( 20 ) =
Sin( 20 ) + 2Cos( 20 + 30 ) / Cos( 20 ) =
________________________________
2 × Cos( 30 + 20 ) =
2 × [ Cos(30).Cos(20) - Sin(30).Sin(20) ] =
2 × [ √3/2 × Cos(20) - 1/2 × Sin(20) ] =
√3 Cos(20) - Sin(20)
_________________________________
Sin( 20 ) + 2Cos ( 20 + 30 ) / Cos( 20 ) =
Sin( 20 ) + √3 Cos(20) - Sin(20) / Cos(20) =
Sin(20) - Sin(20) + √3 Cos(20) / Cos(20) =
0 + √3 Cos(20) / Cos(20) =
√3 Cos(20) / Cos(20) =
Cos(20) simplifies from the numerator and denominator of fraction
√3 × 1 / 1 =
√3
And we're done ....
