STEP
1
:
y
Simplify —
3
Equation at the end of step
1
:
y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((24•(x6))•—)•x)•y)
3
STEP
2
:
Equation at the end of step
2
:
y
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((((23•3x6)•—)•x)•y)
3
STEP
3
:
Canceling Out:
3.1 Canceling out 3 as it appears on both sides of the fraction line
Equation at the end of step
3
:
(((18•(x5))•(y3))+((6•(x2))•(y4)))+((8x6y•x)•y)
STEP
4
:
Equation at the end of step
4
:
(((18•(x5))•(y3))+((2•3x2)•y4))+8x7y2
STEP
5
:
Equation at the end of step
5
:
(((2•32x5) • y3) + (2•3x2y4)) + 8x7y2
STEP
6
:
STEP
7
:
Pulling out like terms
7.1 Pull out like factors :
8x7y2 + 18x5y3 + 6x2y4 = 2x2y2 • (4x5 + 9x3y + 3y2)
Trying to factor a multi variable polynomial :
7.2 Factoring 4x5 + 9x3y + 3y2
Try to factor this multi-variable trinomial using trial and error
Factorization fails
Final result :
2x2y2 • (4x5 + 9x3y + 3y2)
Answer:
7889/6114
Step-by-step explanation:
Answer:
First one
Step-by-step explanation:
Answer:
h = f(t) = -15cos((π/5)t) +20
Step-by-step explanation:
If you like, you can make a little table of positions:
(t, h) = (0, 5), (5, 35)
Since the wheel is at an extreme position at t=0, a cosine function is an appropriate model:
h = Acos(kt) +C
The amplitude A of the function is half the difference between the t=0 value and the t=5 value:
A = (1/2)(5 -35) = -15
The midline value C is the average of the maximum and minimum:
C = (1/2)(5 + 35) = 20
The factor k satisfies the relation ...
k = 2π/period = 2π/10 = π/5
So, the function can be written as ...
h = f(t) = -15cos((π/5)t) +20
Answer:
x=24
Step-by-step explanation:
HE= 1/3(HG)
x-3=1/3(4x-15)
3x-9=4x-15
-x=-24
x=24