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2 years ago

Help a brother out #2

Mathematics

1 answer:

valentinak56 [21]2 years ago

5 0

Answer:

He is αβ 60° and 30° i need some in my answers 5 brain list answers

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The temperature fell by 15 degrees Fahrenheit today.

katrin2010 [14]

<h3>Answer is -15</h3>

We read this out as "negative 15" or you could write "negative fifteen".

The negative indicates the temperature went down. It doesn't necessarily mean that the final temperature is -15 degrees.

For example, if it starts being 50 degrees F, and drops by 15 degrees, then 50+(-15) = 50-15 = 35 degrees is the temperature after that drop. It might help to use a vertical number line.

8 0

3 years ago

ANSWER QUICK PICTURE ATTACHED

Sonbull [250]

Answer:

Step-by-step explanation:

x+5+x+7x-5=180 degree (being the interior angles of a triangle)

9x=180

x=180/9

x=20 degree

3 0

3 years ago

Read 2 more answers

V=πr^2h/3 solve for h

Aneli [31]

Answer:

h = 3V / πr^2

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

6 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

Write the equation of the line that is perpendicular to the line 3x + y = 7 and passes through the point (6, -1).

ira [324]

Line:
3x+y=7
y=-3x+7

Slope of Perpendicular Line: 1/3x

For the perpendicular line to contain point (6,-1) the y-intercept would be (0,1), thus the equation of the line would be y=1/3x+1

y=1/3x+1

5 0

2 years ago