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pishuonlain [190]
2 years ago
9

The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of

5.56mm is 12 inches tall. What is the maximun number of crates that can be stacked within regulations?
Mathematics
1 answer:
KIM [24]2 years ago
3 0

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

\frac{height \hspace{0.1cm} of  \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of  \hspace{0.1cm} each \hspace{0.1cm} crate}  = \frac{10 \hspace{0.1cm}  feet}{12 \hspace{0.1cm}  inches}  =  \frac{10 \times 12  \hspace{0.1cm} inches }{12  \hspace{0.1cm} inches}  = 10 \hspace{0.1cm}  crates

where 1 foot  = 12 inches

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stira [4]

The solution is x=-1 and x=-0.5

Step-by-step explanation:

The expression is 2x(x+1.5)=-1

Multiplying the term x+1.5 by 2x, we get,

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Adding both sides by 1,

2x^{2} +3x+1=0

Taking factor, we get,

\begin{array}{r}{x^{2}+2 x+x+1=0} \\{2 x(x+1)+1(x+1)=0} \\{(2 x+1)(x+1)=0}\end{array}

Equating the factors to 0,

\begin{aligned}2 x+1 &=0 \\2 x &=-1 \\x &=-\frac{1}{2}\\x&=-0.5\end{aligned} and \begin{aligned}x+1 &=0 \\x &=-1\end{aligned}

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Step-by-step explanation:

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Solve by elimination -2x+2y=6 and 4x×2y=-5
Valentin [98]

Answer:

x=-11/6, y=7/6. (-11/6, 7/6).

Step-by-step explanation:

-2x+2y=6

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-----------------

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4x+2y=-5

----------------------

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