Math HELP!!!?

Question

Thanks!!
^= exponent

1. 0= 10x^3 - 15x^2
A. The solution set is (0,3)
B. THe solution set is (0)
c. The Solution set is (0, 3/2)
D. The solution set is (0, -3/2)

2. u^5 -13u^3 + 36u =0
a. (0,2,3)
b (-3, -2,0)
c. (-3,-2,2,3)
d. (0,-2,2,-3,3)

3. (2t-5)(t-1) = 2
a. (1/2, 3)
b. (2,3)
c. (1/2, 2)
d. (-1/2, -3)

Thanks So much!

THEN DONT LOOK AT MY QUESTION JUST TO ***** AT ME TO DO MY HOMEWORK OKAY!

Answer 1

FIrst one: Common factor. 5x^2(2x - 3)= 0
x = 0 or 2x - 3 = 0
x = 3/2
So the answer is C.

Second one: Common factor first: u(u^4 - 13u^2 +36)=0
Simple trinomial factor: u(u^2 -9)(u^2 - 4) =0
Difference of Square factor: u(u-3)(u+3)(u -2)(u+2) = 0
THis gives u = 0, 3, -3, 2, -2
ANswer is D

Third one: Expand and get in right form first.
2t^2 - 2t - 5t + 5 - 2 =0
2t^2 - 7t + 3 =0
Complex Trinomial factoring gets you to:
(2t - 1)(t - 3) =0
2t -1=0 or t - 3 =0
t =1/2 or t = 3
Answer is A

Answer 2

basically you are trying to solve polynomial equations, or, equivalently, find their x-intercepts if you were to graph them.

1. 10x^3 - 15x^2 = 0
so first you factor out the x^2 as it is common to both terms:

x^2 * (10x - 15) = 0

set each terms equal to 0 and solve for x:

x^2 = 0 and 10x - 15 = 0

so x=0 and x= 15/10 = 3/2

2. u^5 - 13u^3 + 36u = 0
factor out u which is common to all terms

u * (u^4 -13u^2 +36) = 0

to factor u^4 - 13u^2 + 36 = 0 may seem difficult, but basically, the easy way to think about this is to determine if their is any set of numbers that multiply to equal and 36 and add/subtract to equal 13 because the answer has to be whole numbers given your choices so the u^4 term must be able to be factored. the numbers 9 and 4 have this property so now you just have to determine the positivity or negativity of the numbers. and -9 and a -4 will multiply to be +36 and added together will be -13 so:

u^4 - 13u^2 + 36 = (u^2 - 9)(u^2 - 4) = 0

the full equation is now:

u(u^2 - 9)(u^2 - 4) = 0

so u = 0, u^2 - 9 = 0, and u^2 - 4 = 0

so u = 0 u^2 = 9, and u^2 = 4

so u = 0, +3, -3, +2, -2

3. (2t - 5)(t - 1) = 0

this is straightforward because the equation is given to you in a factored form so set each term equal to 0 and solve

2t - 5 = 0, and t- 1 = 0

so t= 5/2 and 1

there you go, hope this helps with your math problems.

Answer 3

1. 0= 10x^3 - 15x^2
5x^2(2x-3)=0
there are actually 3 solutions
0, 0, 3/2
c. The Solution set is (0, 3/2) is correct

A. The solution set is (0,3)
B. THe solution set is (0)
c. The Solution set is (0, 3/2)
D. The solution set is (0, -3/2)

2. u^5 -13u^3 + 36u =0
u(u^4-13u^2+36u)=0
u(u^2-4)(u^2-9)=0
u=0,+/-2, +/-3
d. (0,-2,2,-3,3) is correct

a. (0,2,3)
b (-3, -2,0)
c. (-3,-2,2,3)
d. (0,-2,2,-3,3)

3. (2t-5)(t-1) = 2 use Foil
2t^2-2t-5t+5=2
2t^2-7t+3=0
(2t-1)(t-3)=0
t=1/2, 3
a. (1/2, 3) is correct

a. (1/2, 3)
b. (2,3)
c. (1/2, 2)
d. (-1/2, -3)

Answer 4

1. c because 10x^3 - 15x^2 = 5x^2(2x-3) = 0
2. d because u^5 -13u^3 + 36u = u(u^2-4)(u^2-9) = 0
3. a because (2t-5)(t-1) - 2 = (2t-1)(t-3) = 0

Answer 5

just look it up some where else
i mean it is probably your homework that you are too lazy to do
if you dont do it yourself then you wont learn

Answer 6

u should do ur homework by urself

Answer 7

1>) B
2.)A
3.)B