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3 years ago

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A division problem divides a 3-digit number by a 1 digit number. The quotient is 307 r3. What could the division problem be?

1 answer:

Tanya [424]3 years ago

5 0

Answer:

There is no solution to this problem.

Step-by-step explanation:

The 1 digit number cannot be 1, since there would be no remainder.

The 1 digit number cannot be 2, since the remainder could be 0 or 1.

The 1 digit number cannot be 3, since the remainder could be 0, 1, or 2.

So, the 1 digit number must be greater than 3. Let's start with 4. Since the quotient is 307 (remainder 3), 4 x 307 is already more than 3 digits.

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Answer:

1. {5,-9}

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Step-by-step explanation:

The equation $f(x)=x+\lvert 2x-3 \rvert=12$ can be solved as follows:

$x+\lvert 2x-3 \rvert =12\,\,\,\text{(given equation)}$

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$2x-3=12-x\,\,\text{or}\,\,2x-3=-(12-x)$

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The last equation is $h(x-1)=x-3$ , where $h(x)=\lvert x \rvert -2x+5.$ . To solve this equation we can proceed as follows:

$h(x-1)=x-3$

$\lvert x-1\rvert -2(x-1)+5=x-3$

$\lvert x-1 \rvert -2x+2+5=x-3$

$\lvert x-1\rvert =3x-10$

$x-1=3x-10\,\,\text{or}\,\,x-1=-(3x-10)$

$-2x=-9\,\,\text{or}\,\,4x=11$

$x=9/2\,\,\text{or}\,\,x=11/4$

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Step-by-step explanation:

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