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Aleksandr [31]
3 years ago
8

Why 2+2=4 and not 2+2=22 ?

Mathematics
1 answer:
Lunna [17]3 years ago
8 0

2 + 2 = 4 because 4 is a sum of 1-digit numbers

2 + 2 ≠ 22 because 22 is the sum of 1-digit and 2-digit numbers

Step-by-step explanation:

Lets explain some important notes:

  • The 1-digit number consists of one digit, 4 is a one-digit number
  • The 2-digit number consists of two digits, ones digit and tens digit, 23 is a 2-digit number, the ones digit is 3 and the tens digit is 2, so its value is 20 + 3
  • The 3-digit number consists of ones digit, tens digit and the hundreds digit, 123 is a 3-digit number, the ones digit is 3, the tens digit is 2 and the hundreds digit is 1, so its value is 100 + 20 + 3

∵ 2 + 2 is the addition problem of two 1-digit numbers

∴ There sum will be 4 which is also 1-digit number

∴ 2 + 2 must be 4

∵ 22 is a two digit number

∴ Its value is 20 + 2 ⇒ not 2 + 2

∴ 2 + 2 can not be 22

2 + 2 = 4 because 4 is a sum of 1-digit numbers

2 + 2 ≠ 22 because 22 is the sum of 1-digit and 2-digit numbers

Learn more:

You can learn more about the place value of numbers in brainly.com/question/13174282

#LearnwithBrainly

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We know, that the <span>area of the surface generated by revolving the curve y about the x-axis is given by:

\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}

In this case a = 0, b = 15, y=\dfrac{x^3}{15} and:

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A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\&#10;\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\

=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\

\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=&#10;\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=&#10;\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}

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