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3 years ago

Real quick question. need help plz

Mathematics

2 answers:

goldenfox [79]3 years ago

7 0

Answer:

Step-by-step explanation:

360-124-96-94=46

11111nata11111 [884]3 years ago

3 0

Add them all up and subtract it from 360

All figures with 4 sides have a total of 360 inside it.

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Evaluate the expression -4x+5y-14 when x=0 and y= 3/5

emmainna [20.7K]

Answer:

-11

Step-by-step explanation:

Plug in 0 for x and 3/5 for y into the expression

-4(0) + 5(3/5) - 14

Multiply the numbers inside the parentheses

0 + 3 - 14

3 - 14 = -11, which is your answer

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3 years ago

Mimi has 16 more bouncy balls than Leah

Mnenie [13.5K]

Let X: Leah
Answer: X+16

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3 years ago

Read 2 more answers

Suppose Larry has 16 books and Pat has x books. (Larry still has at least 3 more than Pat.)

zimovet [89]

Answer:

13 ≥ x

Step-by-step explanation:

Larry = 16
Pat = x

<u>Larry has at least 3 more than Pat</u>

16 ≥ x + 3

<u>Simplified</u>

16-3 ≥ x + 3 - 3
13 ≥ x

4 0

3 years ago

Giovanna has a score of

12345 [234]

Answer:'

-9.7 minus, in 13

Step-by-step explanation:

7 0

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

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

3 0

3 years ago