Answer:
Step-by-step explanation:
<u>Quadrant I examples:</u>
<u>Quadrant II examples:</u>
<u>Quadrant III examples:</u>
<u>Quadrant IV examples:</u>
<h2>7x + 19 = 208</h2><h2>11x - 89 = 208</h2><h2>--------------------------------------</h2>
<u>Step-by-step explanation:</u>
let third angle be x
60° + 60° + x = 180° (Δ sum property)
x = 60°
so, this is a equilateral triangle (all Δ equal)
<h2>-----------------------------------------</h2>
in equilateral triangle --> all sides are equal
so, therefore =>>
7x + 19 = 11x - 89
19 + 89 = 11x - 7x
108 = 4x
108 ÷ 4 = x
27 = x
<h3>x = 27</h3><h2>-----------------------------------------</h2>
so variable
1.) 7x + 19
= 7(27) + 19
= 189 + 19
= 208
2.) 11x - 89
= 11(27) - 89
= 297 - 89
= 208
Lmk if this helped if not im sorry
Answer:
The absolute minimum value is "
" and the absolute maximum value is "
".
Step-by-step explanation:
Given:
on,
![[0,5]](https://tex.z-dn.net/?f=%5B0%2C5%5D)
By differentiating it, we get
⇒ 
Set 
then,
⇒ 
(Critical point)
When x=0,
⇒ 
When ,
⇒ 
(Absolute minimum)
When 
⇒ 
(Absolute maximum)
Answer:
6 cm
Step-by-step explanation:
If the original side length is x, then the modified square has an area of ...
A = LW
32 = (x +2)(x -2) = x^2 -4
36 = x^2 . . . . . . . . add 4
6 = x . . . . . . . . . . take the square root
The original figure has a side length of 6 cm.
__
<em>Check</em>
The modified figure is 8 cm by 4 cm = 32 cm^2.
