# 1. Find the value of ((7-6.35)÷6.5+9.9))/((1.2÷36+1.2÷0.25-21/16)÷169/24) SOLUTI0N We shall use BODMAS rule to solve this equation where we solve the numerator first ((7-6.35) ÷6.5+9.9))

1.      Find the prize of     ((7-6.35)÷6.5+9.9))/((1.2÷36+1.2÷0.25-21/16)÷169/24)

SOLUTI0N

We shall use BODMAS   rule to clear-up this equation where we clear-up the numerator original ((7-6.35) ÷6.5+9.9)) we begin we the prize in brackets

0.65/6.5+9.9 then we do division

0.1+9.9=1 thus the prize for numerator is 1

We now profits to denominator

(1.2÷36+1.2÷0.25-21/16)÷169/24)   we begin after a while the prizes in brackets,

(0.03333+4.8-21/16)/169/24)

(4.83333-21/16)/(169/24)

(4.8333-1.3125)/(7.0416667)

3.520833/7.041667= 0.5

Thus the discontinuance for denominator is 0.5

To get our decisive defense we distribute the numerator by denominator

=   1/0.5 =2

Thus     ((7-6.35)÷6.5+9.9))/((1.2÷36+1.2÷0.25-21/16)÷169/24)  = 2

2.      Clear-up for x  ( x2+1/x-4 ) - (x2-1/x+)3 =23

SOLUTION

To clear-up the aloft completion we swell all the identical prizes by the result of all denominator then we equate together

( x2+1/x-4 ) - (x2-1/x+)3 =23/1

(x+3)(X-4)(X2+1)/(x-4) – (x+3)(x-4)(x2-1)/(x+3) = 23(x+3)(x-4) relish provisions abrogate out

(x2+1)(x+3) – (x2-1)(x-4) = 23(x-4)(x+3)

X2(x+3)+1(x+3) – (x2(x-4)-1(x-4)) =23(x(x-4)+3(x-4)

X3+3x2+x+3 – (x3-4x2-x+4) =23(x2-4x+3x-12) garnering relish provisions together

X3-x3+3x2+4x2+x+x+3-4 =23(x2-x-12)

7x2+2x-1=23x2-23x-276   we with the two equations and garner relish provisions together

7x2-23x2+2x+23x-1+276 = 0

-16x2+25x+275=0    we clear-up the quadratic equation

Product = -16*275 = -4400 and sum = 25   the two mass whose sum is 25 and result is 4400 are -55 and 80 we exchange them in our equation to clear-up

-16x2+80x-55x+275 = 0

-16x(x-5)-55(x-5) = 0 consequently -16x -55=0,-16x=55 x = -55/16 = -3.4375

Or x-5=0, x=5 thus x = 5 0r -3.4375

Thus   for (x2+1/x-4) – (x2-1/x+3)=23 , x = 5 or -3.4375

3.      Find the prize of √(25(1/log6 5)+49(1/log87))

SOLUTION

We original clear-up 1/log65   and we recognize from logarithm rules that   for, log AB = C then it implies that A C=B thus we use this basic doctrine to evaluate 1/log65

Let log65= x then we observe for x

If log65= x then 6x=5 thus bring-in log10 to twain sides log6x =log5

Thus x = log5/log6 = 0.898244, 1/log65 = 1/0.898244 =1.1132828

We original clear-up 1/log87   and we recognize from logarithm rules that   for, log AB = C then it implies that A C=B thus we use this basic doctrine to evaluate 1/log87

Let log87= x then we observe for x

If log87= x, then 8x=7 thus bring-in log10 to twain sides log8x =log7

Thus x = log7/log8 = 0.935785, 1/log87= 1/0.935785 =1.0686156

√ (25(1/log6 5) +49(1/log87) = √(25*1.1132828 + 49*1.0686156) = √(80.1945324196) = 8.95514 ~ 9