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 #76
avatar+118725 
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So I assume that means that in a day or 2 you will stop answering questions for a few weeks but you intend to start up again when you get back.
Is that right? You might have a hard time finding this post. If you are a formal member it will be easy. It would just be under your posts. But I have a feeling that you haven't formally joined up so it might be harder. (just remember the date of your last post, that should make it easier.)
Thanks for letting me know. Have a great holiday. Do you live in Western Australia, Perth maybe?

That last question was brilliantly answered.

I'll give you another on like that:
--------------------------------------------------------------------------
I want you to copy this question and put the answers after each of my questions. So that the answers stay with the question.

Toni has some in an account
The interest rate is 6.8% per annum
Interest is compounded every 2 months
a) ( think about how many 'nets' of interest there are in a year.) How many times each year is the interest compouned ?
b) What is the 2monthly rate of interest? This is the interest per compounding period, don't forget to put r= ? You forgot the r= last time

If Toni leaves his money in the account for just 2 month then he has left it there for just 1 compounding period. so n=1
what will n be if [Note: Put n = every time ]
c) john leaves the money there for 4 months Ans: n=?
d) for 6 months Ans:
e) for 8 months Ans:
f) for 10 months Ans:
g) for 1 year Ans:
h) for 18 months Ans:
i) for 2 years Ans:
j) for 2 years and 2 months Ans:
g) for 3 years Ans:
(if you are starting to get confused by the big numbers, look back and see how we did it before - we used the number of compounding periods 'nets' in a year to help us.)
h) for 9 years Ans:
i) for 25 years Ans:
j) for 68 years Ans:
k) If Toni leaves his money in this account for 68 years and the bank doesn't change the account in any way, what will r be after 68 years? Please think
Please think about your answers and follow all my instructions.
Jan 10, 2014
 #1
avatar+330 
0
Postby Guest » Thu Jan 09, 2014 8:45 pm
find the height of a cone with a half sphere on top (3-D object) given the volume is 0.65m cubed and a base of both cone and sphere with a diameter of 1 m
************

To solve this will require use of two basic equations.

The volume of a standard cone

1/3Pi*(r 2)*h or ((Pi*d 2)*h)/12 = volume of cone

and the volume of a sphere

4/3Pi*(r 3) or (Pi*d 3)/6 = volume of sphere

Note: The above two equations for each shape are mathematically the same: one uses radius the other diameter and an integer divisor is used instead of a fractional multiplier. This is common in some engineering applications

Note also that a cone has two (2) parameters: diameter and height; and a sphere has only one (1) which is diameter.

The diameter is given in the question and is fixed at 1 meter, so the sphere is a fixed volume. The only variable we can control is the height of the cone.

The total volume (cone volume plus sphere volume) is 0.65 meters cubed (0.65 3).

A one (1) meter diameter sphere has a volume of (Pi*1 3)/6= 0.52 cubic meters.

Because it is a half-sphere take half the volume: 0.26. Subtract this value from the total volume: 0.65-0.26 = 0.39 cubic meters. A cone with a diameter 1 meter and volume of 0.39 Cubic Meters

Rearrange the volume of a cone equation to solve for its height.

((Pi*d 2)*h)/12 = volume of cone

((Pi*d 2)/v)/12 = height = 12v/(Pi(d 2))

h=(12*(0.39m 2))/(Pi(1.00 2)) = ??? meters high.

Don't forget to use units and significant digits

~~D~~
Jan 10, 2014
 #2
avatar+118725 
0
bob:

Solve the system:
y = 2x2 - 4x - 4
y = 2x^2 - 4x - 4 OR y = 2x2 - 4x - 4
x + y = -5
I wrote my second formula the same as you Bob but then I highlighted the 2 and pressed the 'sup' button above the smilies. (sup stands for superscript).

the choices are
a. {-6, 1}
b. no real solution
c. {(1, -6) and (0.5,-5.5)}
d. [{1,6}, {-1,-6}]



Thank you for your solution 1<3 Algebra
Maybe it is correct but I think that Bob intended what I wrote above.
Method 1
Now you can do this like 1<3 Algebra did, and if you do that Algebra's (1<3 is too hard to write every time) way (but with the correct original y formulas) you will get to the correct answer.
It does involve solving a quadratic equation. If you need help solving the quadratic please post again.
Method 2
The other method is by elimination.
You know that x+y=-5
look at what you have -6+1=5 good, 1+-6=-5 good, 0.5+-5,5=-5 good, 1+6 doesn't equal -5 so d is definitely wrong.
So far it could be a,b,or c.
-------------------------------------------------
now the choices are
a. {-6, 1}
b. no real solution
c. {(1, -6) and (0.5,-5.5)}
--------------------------------------------------
You also know that
y = 2x 2 - 4x - 4
I'd change it to f(x) = 2x 2 - 4x - 4 just because the notation is easier for me, but you don't have to.
Now I'll test f(-6) if i get an answer of 1 then I know that (a) is a correct answer otherwise it is not a correct answer. That's part (a)
now for part d) I'd test f(1) and f(0.5) to see if the given points work.
Now you either have a or c or nothing.
If neither a nor b are correct then you still can't say for sure that b is the answer. I didn't think of that before I started writing all this.
So maybe you should do it the first way.
Jan 10, 2014

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