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3 years ago

Simplify 12^4 times 30^8 times 125^2 divided by 128^4 times 625^3 timed 27^6

Mathematics

1 answer:

natta225 [31]3 years ago

7 0

Answer:

25/2985984

Step-by-step explanation:

12^4*30^8*125^2/(128^4*625^3*27^6)

(2^2*3)^4*(2*3*5)^8*(5^3)^2/(2^7)^4*(5^4)^3*(3^3)^6)

(2^8*3^4*2^8*3^8*5^8*5^6)/(2^28*5^12*3^18)

2^(8+8-28)*3^(4+8-18)*5^(8+6-12)

2^-12*3^-6*5^2

25/(2^12*3^6)

25/2985984

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(b) The force, F, between two objects is inversely proportional to the square of the distance, d, between

LiRa [457]

Answer:

F is reduced by 31%

Step-by-step explanation:

F = 1/(d²)

now, we increase the distance by 20% (multiply by 1.2).

F new = 1/(1.2×d)² = 1/(1.44×d²) = (1/1.44) × (1/(d²)) =

= 1/1.44 × old F = 0.69 × old F

100 - 69 = 31%

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3 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

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3 years ago

Find the quotient of 812.30 divided by 83. Round to the nearest tenth.

elena-14-01-66 [18.8K]

812.30/83 = 9.78674698.....
Nearest tenth means one number under decimal place so...

812.30/83 ~ 9.8

Explanation: The 9.7... rounded up to 9.8 because the number after the tenths place (in this case the number after the 7) is greater or equal to 5.

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3 years ago

Eumin pays $3.56 for 4 juice boxes.

Alborosie

Answer:

No, You divide 3.56/4 and you get 0.89 and then after that you multiply it by 7 so you do0.89x7=6.23

Step-by-step explanation:

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3 years ago

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Leona [35]

Answer:

It would be

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3 years ago

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