Find the zero of the graph

How many 1/2 cup servings are in 9 cups of cereal?

Arlecino [84]

Answer:

18

Step-by-step explanation:

9 divided by 1/2 is the same as 9 multiplied by 2, which equals 18

5 0

1 year ago

When he receives his first memory from the old man, what did Jonas experience for the first time?

Nina [5.8K]

Answer:

C. The feeling of snow and of riding a sled down a hill.

Step-by-step explanation

I read the book.

5 0

2 years ago

Read 2 more answers

C=5(y-8) solve for y

Sloan [31]

Answer:

The steps to do this :

$c = 5(y - 8)$

$c = 5y - 40$

$c + 40 = 5y$

$y = \frac{c + 40}{5}$

7 0

3 years ago

Read 2 more answers

Why is the answer 2 in the subtractive equation 1+1?

Yuki888 [10]

Answer:

That's not a subtractive equation, the symbol that you used indicates that it is an addition equation. if the question was 1-1 then the answer wouldn't be 2 but since the question is 1+1 the answer is 2

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

3 years ago