width = x
length = 2x+8
area = l x w
x<span>(2x+8)</span>=120
<span><span>2<span>x^2</span>+8x−120=0 </span>
</span>
<span><span><span>x^2</span>+4−60=0 </span></span>
<span><span><span>(x+10)</span><span>(x−6)</span>=0</span>
</span>
<span><span>x=−10 and x=6 </span></span>
<span><span> width has to be a positive number</span></span>
Width = <span>6
</span> inches.
Answer:
looks like D...............
Answer:
0, for q ≠ 0 and q ≠ 1
Step-by-step explanation:
Assuming q ≠ 0, you want to find the value of x such that ...
q^x = 1
This is solved using logarithms.
__
x·log(q) = log(1) = 0
The zero product rule tells us this will have two solutions:
x = 0
log(q) = 0 ⇒ q = 1
If q is not 0 or 1, then its value is 1 when raised to the 0 power. If q is 1, then its value will be 1 when raised to <em>any</em> power.
_____
<em>Additional comment</em>
The applicable rule of logarithms is ...
log(a^b) = b·log(a)
Sector area= theta/360 x πr^2
Radius= 24/2= 12
Sector area= 120/360 x π(12)^2
Final answer :
Sector area = 48π
