Answer:
(a)18.85 Cubic Inches
(b)The box with dimensions of 4 in. x 3in. x 2 in.
Step-by-step explanation:
<u>Part A</u>
Height =2 Inches
Diameter=1 Inch
Radius=Diameter/2=1/2=0.5 Inch
Volume of a cylinder
Volume of 1 Container
Volume of the 12 Containers
<u>Part B</u>
Next, we determine the container that should be used. Let us first determine the volume of each of the boxes available for easy comparison.
Volume of the boxes
- Volume of box with dimension 3 in. x 2 in. x 5 in.=3X2X5=30 Cubic Inches
- Volume of box with dimension 4 in. x 3in. x 2 in.=4X3X2=24 Cubic Inches
- Volume of box with dimension 5 in. x 6 in. x 5 in.=5X6X5=150 Cubic Inches
- Volume of box with dimension 3 in. x 2 in. x 3 in. =3X2X3=18 Cubic Inches
The volume of the 12 film containers <u>is greater than 18 but less than 24</u>. Therefore,the box that should be used with the least amount of wasted space is the box with a volume of 24 Cubic inches.
Answer:
15 divided by 3 -\(0-0)/-
Step-by-step explanation:
My guess is 9 hours, because since pipe a takes 12 hours and with pipe b it takes 9, it's 12-9=3
I think.
Check the picture below.
so the perimeter of the kite is x+x+y+y, namely 2x + 2y.
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=\stackrel{hypotenuse}{x}\\ a=\stackrel{adjacent}{8}\\ b=\stackrel{opposite}{15}\\ \end{cases} \\\\\\ x=\sqrt{8^2+15^2}\implies x=\sqrt{289}\implies \boxed{x=17} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3D%5Cstackrel%7Bhypotenuse%7D%7Bx%7D%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B8%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B15%7D%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20x%3D%5Csqrt%7B8%5E2%2B15%5E2%7D%5Cimplies%20x%3D%5Csqrt%7B289%7D%5Cimplies%20%5Cboxed%7Bx%3D17%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \begin{cases} c=\stackrel{hypotenuse}{y}\\ a=\stackrel{adjacent}{8}\\ b=\stackrel{opposite}{4}\\ \end{cases}\implies y=\sqrt{8^2+4^2}\implies y=\sqrt{80}\implies \boxed{y\approx 8.94} \\\\[-0.35em] ~\dotfill\\\\ 2x+2y\implies 2(17)+2(8.94)\implies 51.88\implies \stackrel{\textit{rounded up more}}{51.9}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20c%3D%5Cstackrel%7Bhypotenuse%7D%7By%7D%5C%5C%20a%3D%5Cstackrel%7Badjacent%7D%7B8%7D%5C%5C%20b%3D%5Cstackrel%7Bopposite%7D%7B4%7D%5C%5C%20%5Cend%7Bcases%7D%5Cimplies%20y%3D%5Csqrt%7B8%5E2%2B4%5E2%7D%5Cimplies%20y%3D%5Csqrt%7B80%7D%5Cimplies%20%5Cboxed%7By%5Capprox%208.94%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%202x%2B2y%5Cimplies%202%2817%29%2B2%288.94%29%5Cimplies%2051.88%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Brounded%20up%20more%7D%7D%7B51.9%7D)
Answer:
Step-by-step explanation:
help me with this and the. i will