Need Math Help?

[`Jammin' Jobus]

i also need help with differential equations.

1) Find the matrix exponential e^At. Hint: use the block diagonal structure of A.

x prime is equal to

1 1 0 0
3 -1 0 0
0 0 0 2
0 0 -2 0

xprime is equal to that matrix multipled by x.

 

[vericz]

i also need help with differential equations.

1) Find the matrix exponential e^At. Hint: use the block diagonal structure of A.

x prime is equal to

1 1 0 0
3 -1 0 0
0 0 0 2
0 0 -2 0

xprime is equal to that matrix multipled by x.

I'll be taking this next semester

 

[Megavitamins]

Hopin you guys can help me out lol

What value of c will make the roots real, rational, and equal?

x^2+6x+c=0

thanks!

 

[`Jammin' Jobus]

I'll be taking this next semester

bad course.


Vector calculus is so much better. I dunno Im only first semester of second year so I still have a lot to take. I heardadvanced algebra (rings and fields) and Complex and Real analysis are worse tho.

 

[`Jammin' Jobus]

Hopin you guys can help me out lol

What value of c will make the roots real, rational, and equal?

x^2+6x+c=0

thanks!

Do you need a really thorough solution...? C should probably be 9 just based on observation. That would give it a repeated root at -3 which is real and rational.

 

[starmatrix]

An inflection point is when the gradient of the gradient is zero. Meaning that the rate at which the gradient curve of the original function is 0. It's a little weird to explain, its the point where the gradient moves from positive to negative or from negative to positive. You can see this on a graph by looking at the point where the curve 'changes direction'.

Thanks for the help, I think I understand it better.. with the help of the textbook too.

bad course.


Vector calculus is so much better. I dunno Im only first semester of second year so I still have a lot to take. I heardadvanced algebra (rings and fields) and Complex and Real analysis are worse tho.

Just started first year calculus.. I don't wanna know what's coming up in upper years.

 

[Signia]

Ok, so my physics prof puts up homework online and the students get to have fun putting up with an extremely unforgiving system that may or may not take our answers, and when we get it wrong, there's absolutely NO explanation. At least we get 20 tries to balance things out a bit, but... it wasn't too helpful with this one that I just officially got WRONG.

Oh jeez... I had use that same website for my high school physics homework a few years ago. Sooo discouraging seeing the red X over and over again after thinking you've solved a page long problem. I would start seeing the green check in my dreams..

 

[Corpsecreate]

Hopin you guys can help me out lol

What value of c will make the roots real, rational, and equal?

x^2+6x+c=0

thanks!

Take out the good old Quadratic Formula:

x = [ -b +- Sqrt ( b² - 4ac) ] / 2a

so x = -6 +- Sqrt (36-4c) / 2

for non real roots, c must be greater than 9
for rational solutions we want (-6 +- (36-4c)) / 2 to be an integer or any number expressable as a fraction (there are infinite solutions).
for equal you want sqrt (36-4c) to be 0 so c = 9

 

[`Jammin' Jobus]

this computer pseudo notation is soo ugly i don't even wanna read it.

 

[Megavitamins]

Do you need a really thorough solution...? C should probably be 9 just based on observation. That would give it a repeated root at -3 which is real and rational.

Edit: thanks corp!

 

[starmatrix]

How's this question?

Consider f(x) = x^3 + 3x^2 + k. If f has 3 roots, find the possible values for k.

 

[Corpsecreate]

I dont like cubics so i'll let someone else tackle that one

Oh fine i'll give it a try.

f(x) = x^3 + 3x^2 + k

A = 1
B = 3
C = 0
D = k

Let Q = (1/3)*C - (1/9)*B^2
Q = -1
Let R = (1/6)*(C*B*-3D)-(1/27)*B^3
R = (-k/2) - 1

For 3 real roots, Q^3 + R^2 < 0

so -1 + [ (-k/2) - 1 ]^2 < 0
(k²/4) + k < 0

-4 < k < 0

Those are your values of k.

Not entirely sure if thats right, should be though

Now someone wanna try my question yet?

If you have a circle of Radius 2 and centre (4,3) then what is the smallest angle of a line needed to only JUST touch the circle?

 

[Death]

A bomber, diving at an angle of 53.0 degrees with the vertical, releases a bomb at an altitude of 730 m. The bomb hits the ground 5.0 s after being released.
a) What was the velocity of the bomber?
b) How far did the bomb travel horizontally during its flight?

A driver skids on a horizontal bridge and crashes through the guard railing and lands in the river 20.0 m below the level of the bridge. The police find that the car is 53.6 m beyond the bridge (horizontally).
a) Deter mine the speed of the car before the crash, in km/h.
b) What assumptions did you make in your solution?

ANY HELP!?

 

[SoldierRift]

What is 5 + 7

Its incredibly hard....

 

[Corpsecreate]

A bomber, diving at an angle of 53.0 degrees with the vertical, releases a bomb at an altitude of 730 m. The bomb hits the ground 5.0 s after being released.
a) What was the velocity of the bomber?
b) How far did the bomb travel horizontally during its flight?

A driver skids on a horizontal bridge and crashes through the guard railing and lands in the river 20.0 m below the level of the bridge. The police find that the car is 53.6 m beyond the bridge (horizontally).
a) Deter mine the speed of the car before the crash, in km/h.
b) What assumptions did you make in your solution?

ANY HELP!?

I'll try the 2nd one, not sure on how to do the 1st.

Assume gravity is 9.8m/s, then it would have taken the car 20/9.8 seconds to hit the water. 20/9.8 is 2.041 seconds

in 2.041 seconds the car moved 53.6m horizontally and so 53.6 / 2.041 will give you the cars horizontal speed per second which is 26.264 m/s.

Convert 26.264 m/s to km/hr and you get 94.55 km/hr.

The assumption that was made was that hitting the guard rail did not at all slow down his speed and that gravity pulls down at a speed of 9.8 m/s

 

[Death]

Ohhh, so much simpler than what I was doing. Thanks a lot.
Oh my god, there is one question that I haven't been able to solve for a long time now. I can't even find information on it on the internet.

A cannon is fired at 30.0 degrees above the horizontal with a velocity of 200 m/s from the edge of a cliff 125 m high. Calculate how far out the cannonball lands on the level plain below. (3740 m is answer)

Any ideas?

 

[JustNoOne]

Ohhh, so much simpler than what I was doing. Thanks a lot.
Oh my god, there is one question that I haven't been able to solve for a long time now. I can't even find information on it on the internet.

A cannon is fired at 30.0 degrees above the horizontal with a velocity of 200 m/s from the edge of a cliff 125 m high. Calculate how far out the cannonball lands on the level plain below. (3740 m is answer)

Any ideas?

The distance the cannonball lands can be described as distance = velocity*time

We need to find time and velocity.


Horizontal velocity is not affected by gravity so horizontal velocity is constant until it hits the ground.


Now for finding time...

To find time, we can use this formula: d=V1t+1/2at^2 where 'd' is the total distance (vertically), 't' is time and 'a' is acceleration (vertically, in this case, gravity).

If you break up the velocity into two parts (horizontal and vertical), it should look like a right angle triangle where 200 m/s is the hypotenuse with the angle of elevation being 30 degrees.

Now we can find the vertical and horizontal velocity of the cannonball.

200sin30=100

200cos30=173.2 (rounded to the nearest tenth)

Now we know that the initial vertical velocity of the cannonball (100m/s) and the horizontal velocity of the cannonball (173.2m/s)

Now pluging in the vertical velocity we just found into that equation as well as the rest of the values:

let down be positive

125= -100t + 1/2(9.81m/s^2)t^2
125= -100t +4.905t^2
0 = 4.905t^2 - 100t - 125m

Now we get a quadratic equation, to solve for t, we can put it into the quadratic formula (I assume you know this so I'll skip to the two answers)

You get 21.56888585 and -1.18246671, but we are solving for t which is time and time cannot be negative or else the cannonball would be behind the cannon, so 21.6 is the answer.


Now you know the time and the horizontal velocity, now we can find the distance the cannonball traveled.

d=v*t
d=173.2m/s*21.6s
d=3741.12m


The cannonbal traveled 3741.12m until it touched the ground.

 

[Fly_Amanita]

What is 5 + 7

Its incredibly hard....

That's something you should ask an engineer. Math people can't do arithmetic.

 

[vericz]

i also have that online physics assignment. It's on wiley plus and i HATE it

 

[ConnorTheKid]

can anyone help me with this stupid problem? It's chemistry but math related.

8. Write each of the following numbers in standard scientific notation.

a. 1/0.00032

b. 10 to the power of 3/10 to the power of -3

c. 10 to the power of 3/10 to the power of 3

d. 1/55,000

e. (10 to the power of 5)(10 to the power of 4)(10 to the power of -4)/(10 to the power of -2)

f. 43.2/(4.32 x 10 to the power of -5)

g. (4.32 x 10 to the power of -5)/432

h. 1/(10 to the power of 5)(10 to the power of 6)

My chemistry teacher is incredibly dumb and never taught my class how to put numbers into scientific notation like this. She only taught us how to put whole numbers in scientific notation (i.e: 4200 = 4.2 x 10 to the power of 3).

Can you guys help me and explain to me how to do these kinds of problems?

 

[Gosu_Engineer]

can anyone help me with this stupid problem? It's chemistry but math related.

8. Write each of the following numbers in standard scientific notation.

a. 1/0.00032

b. 10 to the power of 3/10 to the power of -3

c. 10 to the power of 3/10 to the power of 3

d. 1/55,000

e. (10 to the power of 5)(10 to the power of 4)(10 to the power of -4)/(10 to the power of -2)

f. 43.2/(4.32 x 10 to the power of -5)

g. (4.32 x 10 to the power of -5)/432

h. 1/(10 to the power of 5)(10 to the power of 6)

My chemistry teacher is incredibly dumb and never taught my class how to put numbers into scientific notation like this. She only taught us how to put whole numbers in scientific notation (i.e: 4200 = 4.2 x 10 to the power of 3).

Can you guys help me and explain to me how to do these kinds of problems?

Your teacher probably neglected to teach how to multiply/divide and add/subtract with numbers in scientific notation

To multiply/divide:
Given two numbers in scientific notation,
x = a*10^b
y = c*10^d

x*y = (a*c)*10^(b+d)
x/y = (a/c)*10^(b-d)

Adding/subtracting is a bit different since it requires the numbers to have the same power of 10. just change the power of 10 to be the same then just add the two numbers in front

so as for your problems you need to recognize that these are multiplication/division of two numbers that need to be put into scientific notation then apply the formula:

a.(1/0.00032) = (3.2*10-4)^-1 = (10^0)/(3.2*10^-4)

applying the formula: x = a*10^b (a=1 b=0) y = c*10^d (c=3.2 d=-4)

so x/y = x/y = (a/c)*10^(b-d) = (1/3.2)*10^(0 - -4)
or (.3125)*10^4

from here you can use what you learned to put into standard scientific notation
a. 3.125*10^3

if you want to do the rest on your own I'll put the rest in black text. if you want to check, all my answers requiring deciamls are to three decimal places

b 1.258*10^-1

c 7.943*10^0 or just 7.943

d 1.818*10^-5

e 1*10^7 or just 10^7

f 1*10^6 or just 10^6

g 1*10^-7 or just 10^-7

h 1*10^-11 or just 10^-11

if you want to do it the easy way or short on time just do these in a calculator set to scientific notation mode

 

[Dark Ryu]

Ok I need help.

1) What is the remainder when x^101-x^4+2 is divided by X+1?

2) Find a polynomial with integer coefficients that has a degree of 4, and zeros (1-2i) and 1. With 1 has a multiplicity of 2.

 

[Sinz]

http://img136.imageshack.us/img136/7884/capturedj.png

And here is the problem I am having trouble with.
So far I have drawn a graph, and the areas i got from left to right are.
2,-4.5,20,and -4. So my final answer was 13.5 which was definitely wrong. :/

 

[Corpsecreate]

Ok I need help.

2) Find a polynomial with integer coefficients that has a degree of 4, and zeros (1-2i) and 1. With 1 has a multiplicity of 2.

Looked into it a bit and I got it! A polynomial with degree 4 has the sum of all of its roots as -b/a and product e/a

f(x) = ax^4 + bx^3 + cx^2 + dx + e

A quartic will have 4 roots but you have only been given 3 so you know that the polynomial can be written as...

(X-1) * (X-1) * (X-(1-2i)) * (X-??)

You also know that -b/a = 1 + 1 + (1-2i) + ??
-b/a = 3 - 2i + ??

if b and a are both real numbers then that means that the ??? must be a complex number written as (Y + 2i) because the +2i must cancel out the -2i to give a real number and the value of Y we dont know because we dont know what b and a is.

So we can write the polynomial as...

(X-1) * (X-1) * (X+2i-1) * (X-2i-Y)

Expand that disgusting thing and you get:

X^4 + (-Y-3)*X^3 + (7+3Y + (2-2Y)*i)*X^2+(-3Y-9+(4Y-4)*i)*x + 4+Y+(2-2Y)i

Its horrible I know. If you look at it, you need Y = 1 for all coefficients to be real numbers, so Plug Y = 1 into it and you get...

f(x) = x^4 - 4x^3 + 10x^2 -12x + 5

I checked it in my calculator and I got the 4 roots, 1 (multiplicity 2), 1-2i and 1+2i.

Maybe theres an easier way, im not sure, but thats how I did it. Hope it helps!

 

[Fly_Amanita]

Keep in mind that if a+bi is a root of a polynomial with real coefficients, where a and b are real, then a-bi is also a root of said polynomial. That simplifies the above problem a great deal.

 

[Corpsecreate]

Really? Didn't know that. I guess it makes sense when you think about it though.

 

[Fly_Amanita]

Yep, this is known as the complex conjugate root theorem.

 

[2.72]

Keep in mind that if a+bi is a root of a polynomial with real coefficients, where a and b are real, then a-bi is also a root of said polynomial. That simplifies the above problem a great deal.

Fixed. It's real, not complex, coefficients.

 

[Fly_Amanita]

wtf, I'm stupid. Yeah, there obvious counterexamples to what I said.

 

[Black Waltz]

 

[JustNoOne]

The length should be 1 since:

- the inverse of sine cannot be greater than 1, so the maximum value of x can only be 1 or less
- the value of x cannot be negative since if x was negative number, the result of the root of x minus x squared will always equal a negative number and it is not possible to root a negative number

I think that's the answer to the question.

 

[Corpsecreate]

Here you go!

 

[moogle]

I typed out my solution, then saw Corpsecreate beat me to it. I also get 2.

Spoiler

The domain of both sqrt(x - x^2) and arcsin(sqrt(x)) is the closed interval [0,1], so sqrt(x - x^2) + arcsin(sqrt(x)) can only take x values from 0 to 1, inclusive.

arc length formula:
L = integral from a to b of sqrt(1+(dy/dx)^2))dx

We'll integrate from 0 to 1, since that's our domain.

Let's find dy/dx first:
y = sqrt(x - x^2) + arcsin(sqrt(x))
dy/dx = (1-2x)/(2*sqrt(x-x^2)) + (1/(2*sqrt(x)))/sqrt(1-x)
dy/dx = (1-2x)/(2*sqrt(x-x^2)) + 1/(2*sqrt(x-x^2))
dy/dx = (1-x)/sqrt(x-x^2)

Now plug it into the arc length formula:

L = integral from 0 to 1 of sqrt(1+(1-2x+x^2)/(x-x^2))dx
L = integral from 0 to 1 of sqrt((x-x^2+1-2x+x^2)/(x-x^2))dx
L = integral from 0 to 1 of sqrt((1-x)/(x-x^2))dx
L = integral from 0 to 1 of sqrt(1/x)dx
L = integral from 0 to 1 of x^(-1/2)dx
L = 2*x^(1/2) evaluated from 0 to 1
L = 2*1 - 2*0 = 2

If you want me to elaborate on any step, let me know. The notation gets kinda messy.

 

[Corpsecreate]

I discovered along the way that the area under Coshx from any point A to any point B is equal to the Arc Length of Coshx from point A to point B. Interesting how that works

Where Coshx = (1/2)*(e^x+e^-x)

 

[Black Waltz]

Thanks for the help guys!

 

[DUB]

Dear god, I am definitely posting my next test on here.

 

[Dark Ryu]

Need help and hopefully this can be answered quickly.

To attract customers during the Thanksgiving holiday, Wendy's may offer Turkey Burgers instead of regular hamburgers. Wendy's will make up your burger any way you want. You can choose from 8 different toppings (ketchup, mustard, pickle, etc.)

1. How many different burgers can be made if you choose exactly 2 toppings of the 8? Be careful, a burger with ketchup and mustard is the same as a burger with mustard and ketchup. Is it 28?

2. How many different burgers can be made if you choose exactly 5 toppings of the 8?

3. How many different burgers can be made if you choose all 8 of the toppings? Is it 1?

That is the problem. Hopefully someone can help me out.

 

[Corpsecreate]

Need help and hopefully this can be answered quickly.

To attract customers during the Thanksgiving holiday, Wendy's may offer Turkey Burgers instead of regular hamburgers. Wendy's will make up your burger any way you want. You can choose from 8 different toppings (ketchup, mustard, pickle, etc.)

1. How many different burgers can be made if you choose exactly 2 toppings of the 8? Be careful, a burger with ketchup and mustard is the same as a burger with mustard and ketchup. Is it 28?

2. How many different burgers can be made if you choose exactly 5 toppings of the 8?

3. How many different burgers can be made if you choose all 8 of the toppings? Is it 1?

That is the problem. Hopefully someone can help me out.

Since order doesnt matter, your using Combinations and not Permutations.

1. You have 8 options and are choosing 2, this is 8C2 which is 8!/(2!(8-2)!) = 28
2. You have 8 options and are choosing 5, this is 8C5 which is 8!(5!(8-5)!) = 56
3. 8C8 = NCN = N!/(N!*0!) = N!/N! = 1

Hopefully I got it for you in time

 

[Gates]

I'll help with this thread too. I'm a math tutor and would like to sharpen my skills.

 

[Corpsecreate]

I'll help with this thread too. I'm a math tutor and would like to sharpen my skills.

Ok how about this question for some fun!?

If you have a circle of Radius 2 and centre (4,3) then what is the smallest angle of a line needed to only JUST touch the circle in radians?