Because the number is decreasing in size (i.e. less than one) which means you are making it smaller. Try dividing a big number with any fraction. You'll see that it decreases. This is because a fraction is smaller than one and so it will decrease. Another way of looking at it is if you multiply a number by 1, you get the same number. Multiply by 2 and you get twice the number. Therefore, multiplying a number by less than one and your bound to get a smaller number.
(x+y)^2+z^2=x^2+y^2+z^2+2xy or ((x+y)^2+z)^2= (x^2+y^2+2xy+z)^2= x^4+y^4+z^2+6x^2y^2+4x^3y+2x^2z^2+4xy^3+4xyz^2+2z^2y^2
I'm not sure if you're asking what the derivative is for f(x) = 1/[(x^2)+2] or for f(x) = [1/(x^2)]+2 so I'm gonna do the first one for now. Lemme know if you want the other one too! :)**Note: I'm not sure which terminology you're most familiar with using in class but just know that y is the same thing as/another way or writing y(x) or f(x) on the left side of the equation. Similarly, y' = y'(x) = f'(x) = d/dx(y) = dy/dx = d/dx[y(x)] = d/dx[f(x)].**So we know the First Principles from the Fundamental Theorem of Calculus defines the derivative as a limit:f'(x) = lim(h→0)⁡ [f(x+h)-f(x)]/hOur function is:f(x) = 1/[(x^2)+2]Similarly:f(x+h) = 1/[((x+h)^2)+2]Simplify:f(x+h) = 1/[(x+h)(x+h)+2]f(x+h) = 1/[(x^2+2xh+h^2+2]Then we plug it into the limit and simplify:f'(x) = lim(h→0)⁡ 〖[1/(((x+h)^2)+2)] - [1/((x^2)+2))]〗/hf'(x) = lim(h→0)⁡ 〖[1/(((x+h)^2)+2)] - [1/((x^2)+2))]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[1/(((x+h)^2)+2)]*[(x^2)+2]/[(x^2)+2]- [1/((x^2)+2))]*[((x+h)^2)+2]/[((x+h)^2)+2]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[[(x^2)+2]-[((x+h)^2)+2]]/[[(x^2)+2]*[((x+h)^2)+2]]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[[(x^2)+2]-[(x^2+2xh+h^2)+2]]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[(x^2)+2-(x^2)-2xh-(h^2)-2]/[[(x^2)+2]*[((x^2+2xh+h^2)+2]]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[(x^2)+2-(x^2)-2xh-(h^2)-2]/[(x^4)+2(x^3)h+(x^2)(h^2)+2(x^2)+2(x^2)+4xh+2(h^2)+4]〗*(1/h)f'(x) = lim(h→0)⁡ 〖[-2xh-(h^2)]/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗*(1/h)f'(x) = lim(h→0)⁡ (2h)〖(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗*(1/h)f'(x) = lim(h→0)⁡ (2)〖(-x-h)/[(x^4)+2(x^3)h+(x^2)(h^2)+4(x^2)+4xh+2(h^2)+4]〗Finally, plug in 0 for h:f'(x) = lim(h→0)⁡ (2)〖(-x-(0))/[(x^4)+2(x^3)(0)+(x^2)((0)^2)+4(x^2)+4x(0)+2((0)^2)+4]〗 f'(x) = (2)[-x/[(x^4)+4(x^2)+4]Factor the denominator:f'(x) = (2)[-x/[(x^2)+2)^2]Final answer: f'(x) = -2x/[((x^2)+2)^2]Hope this helped!! ~Casey
When an electron goes from a higher state to a lower state, it gives up energy equal to the difference of energy levels of the two states. This energy is in the form of a photon. If it goes directly from n=3 to n=1, then 1 photon is emitted. If it transitions from n=3 to n=2, then from n=2 to n=1, two (2) photons are emitted. Energy level of n=3 for Hydrogen is -1.511 eV (electron volts) Energy level of n=2 for Hydrogen is -3.4 eV (electron volts) Energy level of n=3 for Hydrogen is -13.6 eV (electron volts) The energy levels are 'more negative' at lower levels because the electron becomes more bound to the atom. From n=3 to 1 (gives up 12.089 eV, or a photon with wavelength 102.518 nm - ultraviolet light) From n=3 to 2 (gives up 1.889 eV, or a photon with wavelength 656.112 nm - red light) From n=2 to 1 (gives up 10.2 eV, or a photon with wavelength 121.5 nm - ultraviolet) See related link post.
-(4*log(2*cos(4*x)-4*cos(2*x)+3)-3*log(2*cos(4*x)+2)-2*log(2*cos(2*x)+2))/12
It is 5400.
The positive integer factors of 446 are: 1, 2, 223, 446
2 x 223 = 446
Any of its factors which are: 1, 2, 223 and itself 446
446.3333
To 4 DP, 62.1935 To 3 DP, 62.194 To 2 DP, 62.19 To 1 DP, 62.2
63.7143
The height, is 6.3 metres (to 2 dp).The height, is 6.3 metres (to 2 dp).The height, is 6.3 metres (to 2 dp).The height, is 6.3 metres (to 2 dp).
2-5=85+446
To 1 dp it is 1.0; to 2 dp it is 1.00; to 3 dp it is 1.000
They’re the ‘real value’ of a rounded number. Upper and Lower Bounds are concerned with accuracy. Any measurement must be given to a degree of accuracy, e.g. 'to 1 d.p.', or ' 2 s.f.', etc. Once you know the degree to which a measurement has been rounded, you can then find the Upper and Lower Bounds of that measurement. Phrases such as the 'least Upper Bound' and the 'greatest Lower Bound' can be a bit confusing, so remember them like this: the Upper Bound is the biggest possible value the measurement could have been before it was rounded down; while the Lower Bound is the smallest possible value the measurement could have been before it was rounded up.
It is: 5.58