I like this thread.

Solve the following equations:

[img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image107.gif[/img]


Simplify( complex number)

(5 + 2i)(- 3 -5i)

At last i saw this thread, i'd be right back.

bawomolo:

Solve the following equations:

[img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image107.gif[/img]

(2x^1/3 - 3)(x^1/3+3) = 0

x^1/3 = 3/2

x = 27/8


Are you sure about that complex no question? I don't see a way to remove the i. Anyway, Ill get back to that later

bawomolo:

Simplify( complex number)

(5 + 2i)(- 3 -5i)

i think the answer is -31i - 5

tRoOE:

okay I just need to cross check if my answer is right, so can someone help me. My question goes as follow

Find an explicit formula for the squence defined by a₁=3, a_n=3a_n-1+1.

I miss this thread. I was banned by Seun for my expressing my heterodox views on his forum.

Find an explicit formula for the squence defined by a₁=3, a_n=3a_n-1+1.

I take it this is an arithmetic sequence that abides by the formula a_n = a₁+ (n-1)d, where d is the common difference between two consecutive terms.

Given: a₁ = 3, and a_n=3a_n-1+1
Find a₂

a₂ = 3a₁ + 1
a₂ = 3(3) + 1 = 10
d = a₂ - a₁ = 10 - 3 = 7

Therefore, a_n = a_n = a₁+ (n-1)d = 3 + (n-1)7 = 7n-4

4 Him:

i think the answer is -31i - 5

Yes, that's correct.

(5+2i)(-3-5i) = -3(5+2i)-5(5+2i) =
-15-6i-25i-10i²
keep in mind i² = -1
-simplifies to
-5-31i

or the imaginary number simplify the following and express your answer in the form of a + bi where a & b are real numbers.

[img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image117.gif[/img]

[img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image116.gif[/img]

change 3+4i to polar coordinates

bawomolo:

or the imaginary number simplify the following and express your answer in the form of a + bi where a & b are real numbers.

1. [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image117.gif[/img]

2. [img]http://www.mc.maricopa.edu/dept/d25/math/AssetReview/COLST/Image116.gif[/img]

3. change 3+4i to polar coordinates

Ah - SS1 Further Maths! Roots of complex nos nearly made me fight my teacher back then.

1. 2i/(i-2i) = 2i/-i = -2

2. 5+i/i³ = 5+i/-i = -5/i - 1 = 5(-1/i) - 1 = 5i - 1

3. r = sqrt(3²+4²) = sqrt(25) = 5

Theta = tan^-1(4/3) or c0s(4/3)/sin(4/3) = 53^o

Therefore polar form is 5(c0s(53) + isin(53))

Find A and B such that COS(100X-45degrees) = ACOS(100X) + BSIN(100X)

MyTempID:

Find A and B such that COS(100X-45degrees) = ACOS(100X) + BSIN(100X)

Should we find X as well?

There's no need to find X, because it is most likely that your coefficients A and B will be valid for all X.

pls do anybody have any info about Aptech or NIIT. i just want to know which is best and which is worth a life career. do they really teach? are there job offers? is it worth the money they are demanding? do Aptech or NIIT students have lots of job opportunities? pls help.

solve the system

2x - 2y - z = -1
x + 2y + 2z = 9
x + y - z = 8


ln x + ln(2x + 1) = 0

bawomolo:

solve the system

ln x + ln(2x + 1) = 0

lnx(2x+1) = 0
e^lnx(2x+1) = e⁰
x(2x+1) = 1
2x² +x - 1 = 0

(2x - 1)(x+1) = 0

x = -1; 1/2

bawomolo:

solve the system

2x - 2y - z = -1
x + 2y + 2z = 9
x + y - z = 8

x + 2y +2z = 9
2x + 2y - 2z = 16

3x + 4y = 25
x-3y = -9

13y = 42
y = 4

x-12 = -9
x = 3

x + 2y + 2z = 9
3 +8 -9 = -2z
2 = -2z
z = -1


x,y,z = 3,4,-1

ln x + ln(2x + 1) = 0

MyTempID:

lnx(2x+1) = 0
e^lnx(2x+1) = e⁰
x(2x+1) = 1
2x² +x - 1 = 0

(2x - 1)(x+1) = 0

x = -1; 1/2

I am not sure about your method . . . . you shouldn't get 2 values for x . . . .

ln x + ln(2x + 1) = 0

ln(2x + 1) = -lnx

e^lnx(2x+1) = e^-lnx

2x+1 = x

x = -1

That's incorrect.  You lost one solution.  Check very well and you will see why.

ln x + ln(2x + 1) = 0

In fact, this is a good reason why checking your solution is important.
According to you, x = -1 . . .
checking x = -1
ln (-1 ) is undefined.

Oh I get it. Thanks.

Solve these:

A man walked for 5 hours, first along a level road, then up a hill, and then he turned around and walked back to the starting point along the same path. He walks 4 mph on the level, 3 mph uphill, and 6 mph downhill. Find the distance he walked.



Uptown is 20 miles frm Ekene's home in Ajegunle. He Drives 60 mph going to a meeting (he is almost late) but coming home the weather is bad and drives 30 mph. What is his average speed for the time he is on the road?

MyTempID:

Solve these:
Uptown is 20 miles frm Ekene's home in Ajegunle. He Drives 60 mph going to a meeting (he is almost late) but coming home the weather is bad and drives 30 mph. What is his average speed for the time he is on the road?

Total time to get Uptown (doing 20 miles at 60 mph) - 1/3 hrs
Total time to get back downtown (doing 20 miles at 30 mph) - 2/3 hrs

Total distance covered - 40 miles
Total time taken - 1/3 hrs + 2/3 hrs = 1 hr

therefore, average speed - 40 mph

MyTempID:

Find A and B such that COS(100X-45degrees) = ACOS(100X) + BSIN(100X)

Cos(100X-45)=Acos100X + Bsin100X

cos100Xcos45 + sin100Xsin45=Acos100X +Bsin100X

cos45cos100X +sin45sin100x=Acos100X + Bsin100X

from the above equation

A=cos45 and B=sin45

but cos45=sin45=0.7071

therefor A=B=0.7071

MyTempID, I will come to your question later. It requires a lot of solving.

Anyway, this question is specifically for Pmann or anyone who wishes to take on the challenge. It involves Calculus:

An open-ended rectangular water tank of diameter 3.1m contains water which flows out through a tap at the bottom. The taphole has a diameter of 0.075m. The height of the water is 3.5m at time t₀ and take the density of the water as 7100Kgm^-3. If the velocity at the taphole is V = sqrt(2gh) where h(t) is the instantaneous height and gravity is 9.81ms^-2, find the time taken to empty the container.

Hint: You need to integrate dh/dt

Ibime:

MyTempID, I will come to your question later. It requires a lot of solving.

Anyway, this question is specifically for Pmann or anyone who wishes to take on the challenge. It involves Calculus:

An open-ended rectangular water tank of diameter 3.1m contains water which flows out through a tap at the bottom. The taphole has a diameter of 0.075m. The height of the water is 3.5m at time t₀ and take the density of the water as 7100Kgm^-3. If the velocity at the taphole is V = sqrt(2gh) where h(t) is the instantaneous height and gravity is 9.81ms^-2, find the time taken to empty the container.

Hint: You need to integrate dh/dt

this would be much easier if we knew the flow rate or relevant equations

solve

ln(x^2 + x - 20) - ln(x+5) = 1
logx + log(2x+5) - log 7

bawomolo:

this would be much easier if we knew the flow rate or relevant equations

OK:

The relevant equation is the continuity equation:

rate of mass accumulation = rate of mass in - rate of mass out

Rate of mass in is zero

Rate of mass out = density x velocity through taphole x Area of taphole


Therefore, this is the equation you start with:

the time differential of mass = 0 - (density x velocity through taphole x Area of taphole)

remember: mass = density x volume

Every station on the Port-Harcourt to Lagos railroad sells tickets to every other station on that route. When the railroad added some new stations, 46 additional sets of tickets had to be printed. How many new stations were added to the railroad and how many stations existed previously?

bawomolo:

this would be much easier if we knew the flow rate or relevant equations

solve

ln(x^2 + x - 20) - ln(x+5) = 1
logx + log(2x+5) - log 7

x²+x-20 = (x+5)(x-4)
lnA - lnB = ln A/B

ln[(x+5)(x-4)/(x+5)] = 1
ln(x-4) = 1

e^ln(x-4) = e
(x-4) = e
x = e+4

Ah

what happened?why did it suddenly stop?

4a + 6b = 12
2a + 5b= 8
solve
lol,u guys should come back pls.this thread was so so refreshing and educating.

ifyalways:

what happened?why did it suddenly stop?

4a + 6b = 12
2a + 5b= 8
solve
lol,u guys should come back please.this thread was so so refreshing and educating.

a = 3/2
b = 1

log₃ x=5

what's x?