The location of a dolphin in relation to the surface of the sea, h(x), over time, x, in seconds, for 5 seconds can be modeled by a cubic function. Each of the following functions is a different form of the cubic model for the situation given above. Which form would be the most helpful if attempting to determine the time it takes for the dolphin to re-enter the sea after leaping out of the water? h(x) = 2x2(x - 11) + 4(17x - 12) h(x) = 2x(x2 - 11x + 34) - 48 h(x) = 2(x - 1)(x - 4)(x - 6) h(x) = 2x3 - 22x2 + 68x - 48

The most helpful function in an attempt to determine the time it takes for the dolphin to re-enter is h(x) = 2·(x - 1)·(x - 4)·(x - 6)

Step-by-step explanation:

For 2·x²·(x - 11) + 4·(17·x - 12)

h(5) = 2×5^2×(5 - 11) + 4×(17×5 - 12) = -8

For the function h(x) = 2·x·(x² - 11·x + 34) -48 we have;

h(5) = 2×5×(5^2 - 11×5 + 34) -48 = -8

For the function h(x) = 2·(x - 1)·(x - 4)·(x - 6) we have;

h(5) = 2×(5 - 1)×(5 - 4)×(5 - 6) = -8

For the function h(x) = 2·x³ - 22·x² + 68·x -48 we have;

h(5) = 2×5^3 - 22×5^2 + 68× 5  - 48 = -8

Given that the values of the function are all equal at x = 5, the function that will be most helpful in determining the time it takes for the dolphin to re-enter the sea after leaping out of the water is the function that is already factorized

Thereby where the value of the function h(x) at which the dolphin re-enters the the sea is h(x) = 0, we have the function h(x) = 2·(x - 1)·(x - 4)·(x - 6), readily gives the time values, x, as x = 1 second or 4 second or 6 second, therefore, the most helpful function is h(x) = 2·(x - 1)·(x - 4)·(x - 6).

what's that

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)