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Ber [7]
2 years ago
11

Devon's father is installing new wood flooring. He bought boards that are 10 ft long, 3/4 in. thick, and 11/24 in. wide. Determi

ne the number of boards Devon's father will need for a 10 ft by 16 1/2 ft room. Plz plz help and the way to solve it plz!!!
Mathematics
1 answer:
kvv77 [185]2 years ago
5 0

Answer:

36 boards

Step-by-step explanation:

The boards are as long as the room, so I only had to worry about the width. If the boards were 1/2 ft wide, I would need 2 boards for every foot of width. For 16 ft, I would need 32 boards.

Instead of dividing, I wrote the equivalent multiplication by multiplying by the reciprocal of 11/24 (the reciprocal of 11/24 is 24/11)

I simplified by dividing both 2 and 24 by 2, and both 11 and 33 by 11.

Therefore he needs 36 boards

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Answer:

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Step-by-step explanation:

As they said, I will be using 3 as pi.

3 (pi) * (\frac{26}{2})^2 (radius squared)

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A skier is trying to decide whether or not to buy a season ski pass. A daily pass costs ​$69. A season ski pass costs ​$350. The
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2 years ago
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

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