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Featured 3 months ago

Suppose you know that:

#log 2 ~~ 0.30103#

#log 3 ~~ 0.47712#

Then note that:

#43 = 129/3 ~~ 128/3 = 2^7/3#

So

#log 43 ~~ log(2^7/3) = 7 log 2 - log 3 ~~ 7*0.30103-0.47712 = 1.63009#

We know that the error is approximately:

#log (129/128) = log 1.0078125 = (ln 1.0078125) / (ln 10) ~~ 0.0078/2.3 = 0.0034#

So we can confidently give the approximation:

#log 43 ~~ 1.633#

A calculator tells me:

#log 43 ~~ 1.63346845558#

Featured 3 months ago

The general term of a geometric series can be represented by the formula:

#a_n = a r^(n-1)#

where

We find:

#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a + color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N#

#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1 - r^N)#

Dividing both ends by

#sum_(n=1)^N ar^(n-1) = (a(1 - r^N))/(1-r)#

If

#sum_(n=1)^oo ar^(n-1) = lim_(N->oo) (a(1 - r^N))/(1-r) = a/(1-r)#

In the given example,

#sum_(n=1)^oo 6*(1/2)^(n-1) = 6/(1-1/2) = 12#

Featured 2 months ago

The remainder is

The remainder theorem states that when a polynomial

Here,

and

The remainder is

We now perform the synthetic division

The remainder is also

The quotient is

Featured 2 months ago

The vertex is

The focus is

The directrix is

Let's rearrange the equation by completing the squares

This is the equation of a parabola.

We compare this equation to

The vertex is

The focus is

The directrix is

graph{((y-2)^2-x/2)(y-1000(x+1/8))((x-1/8)^2+(y-2)^2-0.001)=0 [-5.222, 5.874, -0.687, 4.86]}

Featured 1 month ago

Let

Also,

Hence, using

**Soln.**

**Enjoy Maths.!**

Featured 6 days ago

**Axis** or the **Real Axis.**

This meas that, under the given condition,

Hence, all such **Axis,** or, the **Real Axis** in

the **Argand's Diagram.**

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