Answer:
What are the bases and heights?
It doesnt make sense because i got 104 but thats not one of the options.
Answer:
B. Dodecahedron
Step-by-step explanation:
It has more of a hectagon shape, instead of a triangular one.
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)
9514 1404 393
Answer:
1) f⁻¹(x) = 6 ± 2√(x -1)
3) y = (x +4)² -2
5) y = (x -4)³ -4
Step-by-step explanation:
In general, swap x and y, then solve for y. Quadratics, as in the first problem, do not have an inverse function: the inverse relation is double-valued, unless the domain is restricted. Here, we're just going to consider these to be "solve for ..." problems, without too much concern for domain or range.
__
1) x = f(y)
x = (1/4)(y -6)² +1
4(x -1) = (y-6)² . . . . . . subtract 1, multiply by 4
±2√(x -1) = y -6 . . . . square root
y = 6 ± 2√(x -1) . . . . inverse relation
f⁻¹(x) = 6 ± 2√(x -1) . . . . in functional form
__
3) x = √(y +2) -4
x +4 = √(y +2) . . . . add 4
(x +4)² = y +2 . . . . square both sides
y = (x +4)² -2 . . . . . subtract 2
__
5) x = ∛(y +4) +4
x -4 = ∛(y +4) . . . . . subtract 4
(x -4)³ = y +4 . . . . . cube both sides
y = (x -4)³ -4 . . . . . . subtract 4
