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jeka94
3 years ago
13

Please i need help with the answers this is due later on today

Mathematics
2 answers:
Nutka1998 [239]3 years ago
8 0

Answer:

384, 216, 290, 192, 384

1446, 1 roll

Step-by-step explanation:

For rectangular boxes, calculate the sum of each side, then multiply it by two.

Box 1: 2(18 x 5) + 2(18 x 4) + 2(5 x 4) = 364

Box 3: 2(11 x 8) + 2(8 x 3) + 2(3 x 11) = 290

Box 5 is a cube (all sides equal), so you can find 1 side's area and multiply it by 6.

Box 5: 6(8 x 8) = 384

For triangular boxes, calculate the edges, then find the triangular area using area = 0.5(base x height).

Box 2: (15 x 3) + (9 x 3) + (12 x 3) + 2(0.5)(9 x 12) = 216

Box 4: 2(13 x 2) + (10 x 2) + 2(0.5)(10 x 12) = 192

Total: 364 + 290 + 384 + 216 + 192 = 1446

Rolls of wrapping paper:

Area of 1 roll = 30 x 60 = 1800

Since 1446 is less than 1800, you only need 1 roll of wrapping paper.

VLD [36.1K]3 years ago
4 0
The person above me got everything correct so brainliest them
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\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{(8+2x)^2}}\cdot dx\impliedby \textit{now, let's do some substitution}\\\\
-------------------------------\\\\
u=8+2x\implies \cfrac{du}{dx}=2\implies \cfrac{du}{2}=dx\\\\
-------------------------------\\\\

\bf \displaystyle \int\limits_{0}^{28}\ \cfrac{1}{\sqrt[3]{u^2}}\cdot \cfrac{du}{2}\implies \cfrac{1}{2}\int\limits_{0}^{28}\ u^{-\frac{2}{3}}\cdot du\impliedby 
\begin{array}{llll}
\textit{now let's change the bounds}\\
\textit{by using } u(x)
\end{array}\\\\
-------------------------------\\\\
u(0)=8+2(0)\implies u(0)=8
\\\\\\
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\bf \\\\
-------------------------------\\\\
\displaystyle  \cfrac{1}{2}\int\limits_{8}^{64}\ u^{-\frac{2}{3}}\cdot du\implies \cfrac{1}{2}\cdot \cfrac{u^{\frac{1}{3}}}{\frac{1}{3}}\implies \left. \cfrac{3\sqrt[3]{u}}{2} \right]_8^{64}
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3 years ago
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