All categories

3 years ago

HELP

Mathematics

1 answer:

GREYUIT [131]3 years ago

3 0

Answer:

srry think its to late

Step-by-step explanation:

You might be interested in

Please help me i really appreciate it

Morgarella [4.7K]

Answer:

Slope= 1/3

Step-by-step explanation:

find any 2 random points. Find the height difference and put it as the numerator of a fraction. Find the distance horizontally and put it as the denominator of the same fraction.

Let's use (10, 60) and (40, 70)

40-10=30 (denominator

70-60=10 (numerator)

10/30 (simplify)

1/3

4 0

2 years ago

Of 60 randomly chosen students from a school surveyed,16 chose aquarium as their favorite fiel trip.There are 720 students in th

erik [133]

Answer:

192 students

Step-by-step explanation:

60 randomly students surveyed

16 chose aquarium as their favorite field trip.

720 students in total.

$\frac{16}{60}$ = $\frac{?}{720}$ =

720 × 16 ÷ 60 =

192 students

5 0

3 years ago

1.12- 2 3/5 QUICK!!!

Rzqust [24]

Answer:

47/5 or 9 2/5

Step-by-step explanation:

12/1-13/5

60-13/5

=47/5

7 0

2 years ago

Read 2 more answers

Use the grouping method to factor the polynomial below completely 3x^3+9x^2+5x+15

Tomtit [17]

Answer:

(x+3)( 3x^2 +5)

Step-by-step explanation:

3x^3+9x^2+5x+15

Split into two groups

3x^3+9x^2 +5x+15

Factor 3x^2 out of the first group and 5 out of the second group

3x^2 ( x+3) + 5(x+3)

Factor out x+3

(x+3)( 3x^2 +5)

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