4.46 is a fixed number: it has no upper nor lower bound. To 2 dp it is 4.46
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
The lower bound of 5.56 to 2 decimal places is 5.555. This is because the lower bound represents the smallest value that rounds up to 5.56 when rounded to two decimal places. Therefore, any value from 5.555 up to but not including 5.565 will round to 5.56.
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.