Answer:
Poster’s perimeter = 150 in
Step-by-step explanation:
Given:
Width of poster = 2.5 ft = 2.5 × 12 = 30 in
Find:
Poster’s perimeter.
Computation:
Height of poster = [30×6]/4
Height of poster = 45 in
Poster’s perimeter = 2 [Width of poster + Height of poster]
Poster’s perimeter = 2 [30+45]
Poster’s perimeter = 2 [75]
Poster’s perimeter = 150 in
Try this option:
1. if 8*x²+b*x+3=8*x²+p*x+q*x+3, then ⇒ b=p+q.
2. p*q = max_value (b²/2), if p=q=0.5*b, and p*q→-oo, if p>0 and q<0 or p>0 q<0.
3. example:
given 8x²+10x+3, the student rewrites it as a) 8x²+5x+5x+3 (5*5=25-max value); b) 8x²+0.01x+9.99x+3 (9.99*0.01=0.0999→0); c) 8x²-20x+30x+3 (p*q=-600).
answer: (-oo;0.5b²)
Answer:
6√5
Step-by-step explanation:
<span>$132 / 16 = $8.25
answer
</span> rate = $8.25 per hour
Set up the equation as follows:
150x represents the dog's speed, 100x represents the squirrel's speed, and 200 represents the distance the squirrel has on the dog. x is the amount of minutes that elapse.
Subtract 100x from both sides.
Divide both sides by 50.
It will take the dog 4 minutes to catch up with the squirrel.