A company is designing boxes to ship their product to stores. The design team decides that the width of the box should be five feet shorter than the length, and the height of the box should be three feet longer than the width. Due to shipping constraints, the length of the box can be no greater than six feet. The volume of the box, V(x), can be modeled by a polynomial function, where x is the length of the box. Which of the following correctly models the situation above and gives the correct domain?

  see below

Step-by-step explanation:

You can work this based only on the domain. You don't need to figure the volume, though you can if you want to check the answer further.

x represents the length of the box, which has a restriction that it can be no greater than 6 ft. This tells you x ≤ 6. However, the width is 5 ft shorter than the width, so its value is x-5. But we know the width must be greater than zero:

  0 < x -5

  5 < x

So, the constraints on the domain of x are ...

  5 < x ≤ 6 . . . . . . only matches the last choice, (5, 6]

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Checking the volume

The height is 3 more than the width, so the volume is ...

  V = LWH = (x)(x -5)(x -5 +3) = x(x -5)(x -2) = x(x^2 -7x +10)

  V = x^3 -7x^2 +10x

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.