Hey, I can answer in the comments but you need to have the boxes down in the picture if you can tell me the multiple degrees it has in the boxes then I can help thx :)
Answer:
C. f has a relative maximum at x = 1.
Step-by-step explanation:
A. False. f(x) is concave down when f"(x) is negative. f"(x) is the tangent slope of the graph, f'(x). So f(x) is concave down between x = -1.5 and x = 1.5.
B. False. f(x) is decreasing when f'(x) is negative. So f(x) is decreasing in the intervals x < -3 and 1 < x < 2.
C. True. f(x) has a relative maximum where f'(x) = 0 and changes from + to -.
Answer:
Step-by-step explanation:
2x+x+3(2x)=63
3x+6x=63
9x=63
9x/9=63/9
X=7
2x=14
3(2x)=42
Surface area of box=1200 cm²
<span>Volume of box=s²h </span>
<span>s = side of square base </span>
<span>h = height of box </span>
<span>S.A. = s² + 4sh </span>
<span>S.A. = surface area or 1200 cm², s²
= the square base, and 4sh = the four 'walls' of the box. </span>
<span>1200 = s² + 4sh </span>
<span>1200 - s² = 4sh </span>
<span>(1200 - s²)/(4s) = h </span>
<span>v(s) = s²((1200 - s²)/(4s)) </span>
<span>v(s) = s(1200 - s²)/4 . </span>
<span>v(s) = 300s - (1/4)s^3</span>
by derivating
<span>v'(s) = 300 - (3/4)s² </span>
<span>0 = 300 - (3/4)s² </span>
<span>-300 = (-3/4)s² </span>
<span>400 = s² </span>
<span>s = -20 and 20. </span>
again derivating
<span>v"(s) = -(3/2)s </span>
<span>v"(-20) = -(3/2)(-20) </span>
<span>v"(-20) = 30 </span>
<span>v"(20) = -(3/2)(20) </span>
<span>v"(20) = -30 </span>
<span>v(s) = 300s - (1/4)s^3 </span>
<span>v(s) = 300(20) - (1/4)(20)^3 </span>
<span>v(s) = 6000 - (1/4)(8000) </span>
<span>v = 6000 - 2000
v=4000</span>
Answer:
5/16m long.
Step-by-step explanation:
- 5/8 ÷ 2/1
- 5/8 × 1/2
- =5/16m long.
