Answer:
h, j2, f, g, j1, i, k, l (ell)
Step-by-step explanation:
The horizontal asymptote is the constant term of the quotient of the numerator and denominator functions. Generally, it it is the coefficient of the ratio of the highest-degree terms (when they have the same degree). It is zero if the denominator has a higher degree (as for function f(x)).
We note there are two functions named j(x). The one appearing second from the top of the list we'll call j1(x); the one third from the bottom we'll call j2(x).
The horizontal asymptotes are ...
- h(x): 16x/(-4x) = -4
- j1(x): 2x^2/x^2 = 2
- i(x): 3x/x = 3
- l(x): 15x/(2x) = 7.5
- g(x): x^2/x^2 = 1
- j2(x): 3x^2/-x^2 = -3
- f(x): 0x^2/(12x^2) = 0
- k(x): 5x^2/x^2 = 5
So, the ordering least-to-greatest is ...
h (-4), j2 (-3), f (0), g (1), j1 (2), i (3), k (5), l (7.5)
180°=triangle
360°=quadrangle
180°*34+180=6300°
Answer:
A. 2·x² + 16·x + 32 ≥ 254
Step-by-step explanation:
The given dimensional relationship between the dimensions of the photo in the center of the cake and the dimensions of the cake are
The width of the cake = The width of the photo at the center of the cake, x + 4 inches
The length of the cake = 2 × The width of the cake
The area of the cake Wanda is working on ≥ 254 in.²
Where 'x' represents the width of the photo (at the center of the cake), let 'W' represent the width of the cake, let 'L' represent the length of the cake, we get;
W = x + 4
L = 2 × W
Area of the cake, A = W × L ≥ 254
∴ A = (x + 4) × 2 × (x + 4) = 2·x² + 16·x + 32 ≥ 254
The inequality representing the solution is therefore;
2·x² + 16·x + 32 ≥ 254
Answer:
Step-by-step explanation:
just replace x into the equation to see if y match with the table.
C is the answer
for example
assume that the height of a rectangular prism is 2, width is 1 and length measures 3
the volume will be 6
(2*1*3)
if we multiply height with the scale factor of 1/2
it becomes 1
so the volume will be 3
(1*1*3)
this situation goes for other examples, too
good luck
