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lesya [120]
3 years ago
15

This 7-digit number is 8,920,000 when rounded to the nearest ten thousand. the digits to the tens and hundreds places are the le

ast and same value. the value of the thousands digit is double that of the ten thousands digit. the sum of all its digits is 24.
Mathematics
1 answer:
Aliun [14]3 years ago
3 0
The 7-digit number we are looking for is 8,92a,bcd



"The value of the thousands digit is double that of the ten thousands digit."

The value of the ten thousands digit is 2. Thus the value of the thousands digit is 4.



The 7-digit number we are looking for is 8,924,bcd



"The sum of all its digits is 24."

8+9+2+4+b+c+d = 23 + b+ c+ d

Thus b+c+d = 1



The digits to the tens and hundreds places are the least and same value.

These two digits (b and c) must be 0, because b+c+d = 1

That means that the digit "d" must be equal to 1.



The 7-digit number we are looking for is 8,924,001



This number is the only number that satisfies all the conditions stated in the question.
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Answer:

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B. The factored form of the quadratic expression x^2+9=(x-3i)(x+3i)

Step-by-step explanation:

A. To find the solutions to the quadratic equation x^2+9=0 you must:

\mathrm{Subtract\:}9\mathrm{\:from\:both\:sides}\\\\x^2+9-9=0-9\\\\\mathrm{Simplify}\\\\x^2=-9\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-9},\:x=-\sqrt{-9}

x=\sqrt{-9} = \sqrt{-1}\sqrt{9}=\sqrt{9}i=3i\\\\x=-\sqrt{-9}=-\sqrt{-1}\sqrt{9}=-\sqrt{9}i=-3i

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x=3i,\:x=-3i

B. Two expressions are equivalent to each other if they represent the same value no matter which values we choose for the variables.

To factor x^2+9:

First, multiply the constant in the polynomial by i^2 where i^2 is equal to -1.

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Since both terms are perfect squares, factor using the difference of squares formula

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x^2+9=x^2+9i^2=\left(-3i+x\right)\left(3i+x\right)

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