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Anvisha [2.4K]
3 years ago
6

What is 251,031 rounded to the nearest thousand

Mathematics
2 answers:
Sloan [31]3 years ago
7 0
The answer is <span>251,000</span><span />
insens350 [35]3 years ago
3 0

Answer:

251,000.

Step-by-step explanation:

We have been given a number 251,031.

First of all, we will find the place value of our given number as:

Ones: 1,

Tens: 3

Hundreds: 0

Thousands: 1

Ten thousands: 5

Hundred thousands: 2

We can see that the thousands digit is 1 and hundreds digit is 0. Since the value of hundreds digit is less than 5, so we will round down our number to nearest thousand that is 251,000.

Therefore, our required number would be 251,000.

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Factorising the denominators of both functions,

Factorising the denominator of f(x),

\begin{gathered} f(x)=\frac{x+12}{x^2+4x-12}=\frac{x+12}{x^2+6x-2x-12}=\frac{x+12}{x(x+6)-2(x+6)}=\frac{x+12}{(x-2)(x+6)} \\ f(x)=\frac{x+12}{(x-2)(x+6)} \end{gathered}

Factorising the denominator of g(x),

\begin{gathered} g(x)=\frac{4x^2-16x+16}{4x+48}=\frac{4(x^2-4x+4)}{4(x+12)} \\ \text{Cancel out 4 from both numerator and denominator} \\ g(x)=\frac{x^2-4x+4}{x+12}=\frac{x^2-2x-2x+4}{x+12}=\frac{x(x-2)-2(x-2)}{x+12}=\frac{(x-2)^2}{x+12} \\ g(x)=\frac{(x-2)^2}{x+12} \end{gathered}

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undefined

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