Suppose you need to build a box with a surface area of 90 square feet. If the length is 5 feet and the height is 2 feet, what is the width of the box?

Hi there!

Your answer is:

W=17.5feet

Step-by-step explanation:

We use the surface area form for a cuboid:

2h(l+w)=SA

Fill in the known values!

2(2)* (5+ W) = 90

Solve for W!

4*(5+W)=90

20+4w=90

-20

4w=70

/4

W=17.5feet

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.

idk lol

step-by-step explanation: