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2 years ago

CAN sOME One Help!!!!!!!!!!!!!!

Mathematics

2 answers:

Anon25 [30]2 years ago

7 0

Answer:

it is -1/2x + 3

Step-by-step explanation:

stich3 [128]2 years ago

6 0

Answer:

The answer would be y = -1/2 x + 3

Step-by-step explanation:

So to solve this problem we have to look at the graph.

We have to put it into y = mx + b.

The B would be where the line crosses the y - intercept.

The mx would be how fast is the line going up/down.

If it is going down, it would be negative.

If it is going up, it would be positive.

I hope I helped.

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-3(x – 14) + 9x = 6x + 4<br>

8090 [49]

Answer:

there are no solutions :3

no this is not a joke sjsjsj there are equations who has an answer and ones that are impossible to have a solution. use the no solution sign when you answer it or sum

Step-by-step explanation:

6 0

2 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

2 years ago

Determines whether the line through the points

kotegsom [21]

Find their gradients using the change in y coords divided by the change in x coords. once you have the gradients (or slopes), multiply them by eachother - if the product is (-1) then theyre perpdendicular, if not, they are either parallel or intersect at a point

5 0

2 years ago

Use the givin values to complete each table

Vilka [71]

To find the answers, substitute the variables (letters) in the equation for the number on that line that corresponds to that letter, then solve with the equation.

4 0

2 years ago

The solution set for an inequality is graphed on this number line which numbers below belong to the solution set of the inequali

Rudiy27

Answer:

B, C E and F

Step-by-step explanation:

The inequality is x > -30

the numbers belonging to the solution are all on the heavy line.

Note -30 is not included because of the clear circle on -30

also -44 is not on the heavy line.

7 0

2 years ago