Waiting online for a high score reward good and then add a few high school math problems.

1.

2m=a+b

2n=b+c

b^2=ac

a/m+c/n=(an+cm)/mn=[a(b+c)/2+c(a+b)/2]/mn

=2(ab+ac+ac+bc)/(a+b)(b+c)

=2(ab+2b^2+bc)/(ab+2b^2+bc)

=2

2. (a+b-c)(a+b+c)=a

[(a+b)-c][(a+b)+c]=ab

Square difference formula in (a+b)*(a-b)=a squared-b squared

Taking (a+b) as a whole

So [(a+b)^2]-c^2=ab

a^2+b^2+ab=c^2

So a^2+b^2-c^2=-ab

So cosC=(a^2+b^2-c^2)/2ab

= (-ab )/2ab

=-1/2

So, angle C=120 degrees

3.

From the known x=[6*8‰*a*(1+8)^6]/[1+8‰]^6-1],

The 8% at the back means eight thousandths of a degree, and the symbols have to go to the word to copy them, which is troublesome

y=[12* 8%*a*(1+8%)^12]/[(1+8%)^12-1],

So x/y=[(1+8%)^6+1]/2*(1+8%)^6 <1

So x<y

4.

Let the first term be a

S5=a(1+q+q^2+q^3 +q^4)=2

S10=a(1+q+q^2+q^3+q^4)+aq^5(1+q+q^2+q^3+q^4)

=a(1+q^5)(1+q+q^2+q^3+q^4)=6

S10÷S5=1+q^5=3

q^5=2

a16+a17+a18+a19+a20

=aq^15(1+q+q^2+q^3+q^4)

=(q^5)^3×a(1+q+q^2+q^3+q^4)

=2^3×2

=16

5. (Write 0.5 in place of the expression one-half)

y=x^2+ax+1=(x+a/2)^2-a^2/4+1>=0

Axis of symmetry x=-a/2

If -a/2>0.5

a<-1,the domain of definition is on the left side of the axis of symmetry, and it is a subtractive function

So x=0.5, y min= 1.25+0.5a>=0

a>=-0.25,does not match a<-1

0<=-a/2<=0.5

-1<=a<=0,in the domain of definition of the axis of symmetry

x=-a/2,y min=-a^2/4+1>= 0

-2<=a<=2

Then -1<=a<=0

So a minimum=-1