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Answer:

#log 43 ~~ 1.633#

Explanation:

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#

Answer:

#12#

Explanation:

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

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

where #a# is the initial term and #r# the common ratio.

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 #(1-r)# we find:

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

If #abs(r) < 1# then #lim_(N->oo) r^N = 0# and we find:

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

In the given example, #a=6# and #r=1/2#. So #abs(r) < 1# and:

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

Answer:

The remainder is #=75#

Explanation:

The remainder theorem states that when a polynomial #f(x)# is divided by #x-c#

#f(x)=(x-c)q(x)+r(x)#

#f(c)=0+r#

Here,

#f(x)=2x^3-7x^2#

and #(x-5)#

#f(5)=2*125-175=250-175=75#

The remainder is #=75#

We now perform the synthetic division

#color(white)(aaaa)##5##color(white)(aaaa)##|##color(white)(aaaa)##2##color(white)(aaaa)##-7##color(white)(aaaa)##0##color(white)(aaaa)##0#

#color(white)(aaaa)##color(white)(aaaaa)##|##color(white)(aaaaa)##color(white)(aaaa)##10##color(white)(aaaa)##15##color(white)(aaaa)##75#

#color(white)(aaaaaaaaaa)#------------------------------------------------------------

#color(white)(aaaa)##color(white)(aaaaa)##color(white)(aaaaaa)##2##color(white)(aaaa)##3##color(white)(aaaa)##15##color(white)(aaaa)##color(red)(75)#

The remainder is also #=75#

The quotient is #=2x^2+3x+15#

Answer:

The vertex is #V=(0,2)#
The focus is #F=(1/8,2)#
The directrix is #x=-1/8#

Explanation:

Let's rearrange the equation by completing the squares

#2y^2+8=x+8y#

#2y^2-8y=x-8#

#2(y^2-4y)=x-8#

#2(y^2-4y+4)=x-8+8#

#2(y-2)^2=x#

#(y-2)^2=1/2x#

This is the equation of a parabola.

We compare this equation to

#(y-b)^2=2p(x-a)#

#p=1/4#

The vertex is #V=(a,b)=(0,2)#

The focus is #F=(a+p/2,b)=(1/8,2)#

The directrix is #x=-1/8#

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]}

Answer:

#x=0.#

Explanation:

Let #2^x=a, 3^x=b, and, 5^x=c.#

#:. 4^x+9^x+25^x=a^2+b^2+c^2..............(1).#

Also, #6^x={(2)(3)}^x=2^x3^x=ab, &, ||ly, 10^x=ca, 15^x=bc...(2)#

Hence, using #(1), & (2),#the given eqn. becomes,

#a^2+b^2+c^2-ab-bc-ca=0, or,#

#1/2{(a-b)^2+(b-c)^2+(c-a)^2}=0.#

#rArr a=b=c, or, 2^x=3^x=5^x.#

#2^x=3^x rArr (2/3)^x=1=(2/3)^0#

#x=0.#

#3^x=5^x" also "rArr x=0.#

#:. x=0# is the Soln.

Enjoy Maths.!

Answer:

#z# lies on the #X-# Axis or the Real Axis.

Explanation:

#|(z-3i)/(z+3i)|=1 rArr |z-3i|=|z+3i|#

#:. |x+yi-3i|=|x+yi+3i|,......[because, z=x+yi]#

#:. |x+i(y-3)|=|x+i(y+3)|#

#:. |x+i(y-3)|^2=|x+i(y+3)|^2#

#:. x^2+(y-3)^2=x^2+(y+3)^2,....[because, |a+bi|^2=a^2+b^2]#

#:. -6y=6y, or, y=0.#

This meas that, under the given condition,

#AA z=x+iy in RR^2, y=0.#

Hence, all such #z# lies on the #X-# Axis, or, the Real Axis in

the Argand's Diagram.

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