The length of a rectangle is 6 inches longer than its width. what are the possible widths if the area of the rectangle is at least 667 square inches

width must be atleast 23 inches.

Step-by-step explanation:

Given that the length of a rectangle is 6 inches longer than its width.

If l is length, then w = l-6

Area =lw

i.e. Area

For this inequality being the product of two numbers is positive if both terms have the same sign.

This is possible only if l is atleast 29 inches (ignoring negative solution for l)

Hence possible widths are atleast 23 inches.

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.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: