A-length
b-width
perimeter of rectangular floor: 2a+2b=204
<span>The length is two times the width: a=2b
</span>2a + 2b = 204
2*2b + 2b = 204
6b = 204|:6
b = 34
a=2b=2*34=68
a=68->length
b=34->width
It should be 12 ft away from the wall. Not 4 ft. You do not just subtract the difference because as you pull it away from the wall, the ladder is mostly moving horizontally with only a small vertical change.
You can use Pythagorean Theorem to find the answer. Think of the ladder as the hypotenuse of a right triangle with the wall and the ground as the legs.
a^2 + b^2 = c^2
16^2 + b^2 = 20^2
b^2 = 20^2 - 16^2
b^2 = 144
b = 12 ft
(3x+4) (x+2)
This gives you 3x^2 + 6x + 4x + 8
Which simplifies to 3x^2 + 10x + 8
For 
between 
and 
, we have
is continuous over its domain, so the intermediate value theorem tells us that
is true for 
.
For all , we take into account that is 
-periodic, so the above inequality can be expanded to
where 
is any integer. Equivalently,
To get the corresponding solution set for
simply replace with 
:
The correct answer is B) ASA postulate
when
;
for
; and
for