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3 years ago

14

What is the slope of points (4,9) (-8,-6)?

2 answers:

12345 [234]3 years ago

6 0

Answer:

5/4

Step-by-step explanation:

(y₂ - y₁) / (x₂ - x₁)

(4,9) (-8, -6)

plug in

(-6 - 9) / (-8 - 4)

solve within parentheses

-15/-12

simplify

5/4

MrMuchimi3 years ago

3 0

what is the slope of points (4,9) (-8,-6)

Answer:

5/4

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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,

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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

Question in picture solve

Anna007 [38]

I think it’s 3 times because 2*3=6 and 4*3=12 and 6*3=18

8 0

3 years ago

r-ruslan [8.4K]

Answer:

$1.25

September

$30

Step-by-step explanation:

Let's take this a step a time.

First we need to find how much the price of the flowers were in September.

We know that each flower cost $1.50 on October.

The October price was a 20% increase of the September price.

To calculate for the price of the flowers on September, we can solve it like this:

Let x = Price during September

1.2x = 1.50

We used 1.2 because the price of $1.50 is 120% of the original price.

Now we divide both sides by 1.2 to find x.

$\dfrac{1.2x}{1.2}=\dfrac{1.5}{1.2}$

x = 1.25

The price of the flowers during September was $1.25 each.

Now the 7th grade class earned 40% of the selling price of each flower.

40% = 0.40

To find how much they made on each month, we simply multiply the percentage to the price and the number of flowers sold.

September = 0.40 x 1.25 x 900

September = 0.5 x 900

September = $450

Now for October.

October = 0.40 x 1.50 x 700

October = 0.6 x 700

October = $420

The 7th Graders earned more on September.

They earned $30 more on September than October.

5 0

3 years ago