1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ohaa [14]
3 years ago
10

10.25 : -0.5 = 0 -20.5 20.5 -0.05 0.05

Mathematics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

20.5

Step-by-step explanation:

hope it helps you that is all I could do.

You might be interested in
I need help finding out what AC is!!
iragen [17]
I think it’s b.5. Hope this helps. :)
5 0
3 years ago
Read 2 more answers
There was only 5/12 of a pizza left. Carl then ate 1/6 of the remaining pizza. How much of the pizza was left?
eimsori [14]
\frac{5}{12} - \frac{1}{6} = \frac{5}{12} -  \frac{2}{12} =  \frac{3}{12} =   \frac{1}{4}

\frac{1}{4} of the pizza was left
4 0
3 years ago
Read 2 more answers
Subtract this pair of fractions<br> 1/3x-5/3x
babunello [35]

Answer:

-7x/6

Step-by-step explanation:

x/2-5x/3

Taking LCM

3x/6-10x/6

(3x-10x)/6

-7x/6

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
I don't know how to do this. I have the formula but I'm confused. Help?
Rom4ik [11]
For the lateral area, don't include the base area. For the surface area, you should include it.
3 0
3 years ago
Other questions:
  • What is the opposite of 4
    7·1 answer
  • What is the correct proportion for corresponding sides? (Last question)
    15·1 answer
  • What does radius mean
    10·2 answers
  • Adam has the scatterplot shown for his data set, where snowfall is measured in inches, and he is trying to describe the line of
    15·1 answer
  • Great! You will purchase snacks for the painting class. You have a budget of $40. You want to buy fruit and granola bars.
    9·1 answer
  • 372-199= What’s the answer
    15·2 answers
  • The list below show test scores for 1st period on a recent test. Finding the mean deviation. 62 63 64 71 96 97 98 99 100 100
    10·1 answer
  • What's the answer to this?​
    9·1 answer
  • If the measure lf 2 is 83, what is the measure of 3?
    14·1 answer
  • Between which two numbers will you find 5.634 x 10^2? A. 0.0562 and 0.0563B. 0.0563 and 0.0564C. 562 and 563D. 563 and 564
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!