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kolezko [41]
3 years ago
14

What is the concept of rounding decimals? Please be certain to leave an example

Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0
Look at the digit to the right of the required number. If it is a 5 or greater, add 1 to previous digit.
If it is less than 5, leave the previous digit unchanged. In either case, drop all the remaining digits. Example: Round off this decimal 7.253896 to three decimal digits.
Answer. 7.253896approximately7.254
(The wavy equal sign approximately means "is approximately equal to.")
To round off to three decimal digits, we must look at the digit in the fourth place. The digit in the fourth place is 8 (greater than 5). Therefore, we add 1 to the previous digit 3.
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Please help!!!!! It's urgent
Evgesh-ka [11]
       2b                     2a
----------------- +  -----------------
   (b+a)^2            (b^2 - a^2)

            2b                     2a
= ----------------- +  -------------------
      (b+a)(b+a)         (b+a)(b-a)

         2b(b - a) +  2a(b + a)
= ------------------------------------
           (b+a)(b+a)(b-a)


         2b^2 - 2ab  +  2ab + 2a^2
= ---------------------------------------
           (b+a)(b+a)(b-a)

         2b^2 + 2a^2
= ------------------------
        (b+a)(b+a)(b-a)

         2(b^2 + a^2)
= ------------------------
        (b+a)^2 (b-a)

Answer:

Numerator:        2(b^2 + a^2)
Denominator:    (b+a)^2 (b-a)
7 0
3 years ago
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Musya8 [376]

Answer:

ok

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
-4x+16=12 what is the answer
ale4655 [162]

Answer:

1

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
If you need $8000 to travel to Europe after you graduate in 4 year. How much would your monthly deposits need to be if your acco
lina2011 [118]
\bf \qquad \qquad \textit{Future Value of an ordinary annuity}\\
\left. \qquad \qquad \right.(\textit{payments at the end of the period})
\\\\
A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right]

\bf \qquad 
\begin{cases}
A=
\begin{array}{llll}
\textit{accumulated amount}\\
\end{array}\to &
\begin{array}{llll}
8000
\end{array}\\
pymnt=\textit{periodic payments}\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12\\
t=years\to &4
\end{cases}

\bf 8000=pymnt\left[ \cfrac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]
\\\\\\
\cfrac{8000}{\left[ \frac{\left( 1+\frac{0.05}{12} \right)^{12\cdot 4}-1}{\frac{0.05}{12}} \right]}=pymnt\implies \cfrac{8000}{\frac{\left( \frac{241}{240} \right)^{48}-1}{\frac{1}{240}}}=pymnt
\\\\\\
\cfrac{8000}{53.0148852}\approx pymnt\implies 150.9010152318\approx pymnt
6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B400%20%5Ctimes%20400%20-%2024%20%5Ctimes%20400%20-%20360%20%5Ctimes%20400%20-%202
max2010maxim [7]

Answer:

\sqrt{400 \times 400 - 24 \times 400 - 360 \times 400 - 200} \\  \\  = \sqrt{160000 - 9600 - 144000 - 200} \\  \\  = \sqrt{160000 - 9600 - 144000 - 200} \\  \\ = \sqrt{160000 - 153,800} \\  \\  = \sqrt{6200} \\  \\  = 78.74 \\  \\

7 0
3 years ago
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