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natta225 [31]
2 years ago
6

Don't know this one, please help

Mathematics
2 answers:
vekshin12 years ago
3 0

Answer:

D) 45/n 11

Step-by-step explanation:

(6n^-5(3n^-3)

= 6n^-5×9n^-6

= 54n ^-11

= 54/n 11

hope it helps..

Paladinen [302]2 years ago
3 0

Answer:

D. 54/n¹¹

Step-by-step explanation:

(6n⁻⁵)(3n⁻³)²

<em>Note that (ax)ᵇ=aᵇxᵇ</em>

(6n⁻⁵)(3²)(n⁻³)²

(6)(n⁻⁵)(9)(n⁻³)²

54(n⁻⁵)(n⁻³)²

<em>Note that (xᵃ)ᵇ=xᵃᵇ</em>

54(n⁻⁵)(n⁻³⁽²⁾)

54(n⁻⁵)(n⁻⁶)

54(n⁻⁵)(n⁻⁶)

54(n⁻⁵)(n⁻⁶)

<em>Note that (xᵃ)(xᵇ)=xᵃ⁺ᵇ</em>

54(n⁻⁵⁻⁶)

54n⁻¹¹

<em>Note that x⁻ᵃ=1/xᵃ</em>

54/n¹¹

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Find the number b such that the line y = b divides the region bounded by the curves y = 36x2 and y = 25 into two regions with eq
Gemiola [76]

Answer:

b = 15.75

Step-by-step explanation:

Lets find the interception points of the curves

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x² = 25/36 = 0.69444

|x| = √(25/36) = 5/6

thus the interception points are 5/6 and -5/6. By evaluating in 0, we can conclude that the curve y=25 is above the other curve and b should be between 0 and 25 (note that 0 is the smallest value of 36 x²).

The area of the bounded region is given by the integral

\int\limits^{5/6}_{-5/6} {(25-36 \, x^2)} \, dx = (25x - 12 \, x^3)\, |_{x=-5/6}^{x=5/6} = 25*5/6 - 12*(5/6)^3 - (25*(-5/6) - 12*(-5/6)^3) = 250/9

The whole region has an area of 250/9. We need b such as the area of the region below the curve y =b and above y=36x^2 is 125/9. The region would be bounded by the points z and -z, for certain z (this is for the symmetry). Also for the symmetry, this region can be splitted into 2 regions with equal area: between -z and 0, and between 0 and z. The area between 0 and z should be 125/18. Note that 36 z² = b, then z = √b/6.

125/18 = \int\limits^{\sqrt{b}/6}_0 {(b - 36 \, x^2)} \, dx = (bx - 12 \, x^3)\, |_{x = 0}^{x=\sqrt{b}/6} = b^{1.5}/6 - b^{1.5}/18 = b^{1.5}/9

125/18 = b^{1.5}/9

b = (62.5²)^{1/3} = 15.75

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3 years ago
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Irina18 [472]

Answer:

Step-by-step explanation:

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Taya2010 [7]

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Step-by-step explanation:

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7 0
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