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3 years ago

How do you write 67.24 million in scientific notation

Mathematics

2 answers:

Mekhanik [1.2K]3 years ago

8 0

6.724 * 10^7 because it has to be in the best form and the 7 is how many decimal places.

netineya [11]3 years ago

7 0

I believe the answer would be 6.724

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Look at the pentagonal numbers. Use finite differences to determine which function represents the pattern.

Ahat [919]

Answer:

f(x) = 1.5x - 0.5x

Step-by-step explanation:

The function of the pattern represented by the pentagonal numbers is the sum of three triangular numbers.

The triangular number general formula

x (x + 1) / 2

For example,

The sequence

1, 3, 6, 10

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* * * * * *

* , * *, * * *, * * * *

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The pentagonal numbers

The sequence:

1, 5, 12, 22, 35

As shown in the picture can be divided into three triangles

Triangle 2

x (x + 1) / 2

Triangle 1 and 3 (they are triangles one unit smaller than 2)

n (n + 1) / 2

n= x-1

Replacing n

(x-1) ((x-1) + 1) / 2

(x-1) (x) / 2

(x-1) x / 2

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Function represents the pattern

Triangle 2 + (Triangle 1 + Triangle 3)

Triangle 1 = Triangle 3

So then,

Triangle 2 + 2* Triangle 1

x (x +1) /2 + 2* (x -1) x/2

Rearranging

0.5 x (x +1) + x(x -1)

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1.5 x^2 - 0.5 x

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3 years ago

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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