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borishaifa [10]
3 years ago
6

what is the answer of This The next day I got home from school and there was nothing baking in the oven nor any dishes in the si

nk. I went to my room and everything was back to “normal.” On my bed was a box with beautiful wrapping and a handmade bow. I opened the package, saving the bow. I hugged the gorgeous white angora sweater and whispered to the air, “Good-bye, Phyllis and Charlie, and thanks.”
Mathematics
1 answer:
Svetllana [295]3 years ago
7 0
Honestly i have no clue what ur talking abt
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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.

5 0
3 years ago
NEED ASAP:
k0ka [10]

Step-by-step explanation:

the area of this rectangle is

(10 + 2x)(8 + 2x) = 168

80 + 20x + 16x + 4x² = 168

4x² + 36x = 88

because for the whole length you need to add x twice (left and right). the same for the width (top and bottom).

7 0
2 years ago
If z = 3 , what is 5 x ( 6 -z)?
Lerok [7]

Answer:

15

Step-by-step explanation:

Just insert 3 wherever the z is at.

5 * (6 - 3)

5 * 3

= 15


8 0
3 years ago
What is the sum of 835,245 and 34,765
svetoff [14.1K]

835,245+ 34,765=  870,010

8 0
4 years ago
Multiply 6 x 298 using friendly numbers
Leni [432]
298 into 
300 times 6=1800




7 0
3 years ago
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