## Answers by Michael F.

m=man's age s1,s2 son's ages m=3s1+3s2 m+5=2(s1+5)+2(s2+5) Now let s=s1+s2 and have m=3s m+5=2s+20 3s+5=2s+20 s=15,m=45 Please forgive the nonsense comment above. y+5=y+10 has no sense

Let the amount of added water be W gallons. After it is added we have 1/(1+W)=.95 You may take it from here.

(1/(a+h) - 1/a)/h add the two fractions using the common denominator a(a+h) a/(a(a+h))-(a+h)/(a(a+h))=(a-(a+h))/(a(a+h))=-h/(a(a+h)) Now divide it by h to get -1/(a(a+h))

If x=logbN and y=logb2 N we have bx  = (b2 )y=b2y or that y=x/2 or x=2y  Also logcx/logcy=logyx Now log38÷( log916×log410 )=log38/(2log316×log410)=log168/2log410=2log48/(2log410)=log108=3log102

The ancient Greeks thought this ratio gave a most pleasing appearance to the structure. It is also called the golden mean. The reason for this is that it is the mean proportional between 1 and 1+itself, that is the positive solution to the equation 1/g=g/(1+g).

Assuming that f(x)=exp(x2) the equation f''g'=(fg)'=f'g+fg'  now reads 2xexp(x2)g'=2xexp(x2)g+exp(x2)g' (2x-1)exp(x2)g'=2xexp(x2)g since exp(x2) is never 0  (2x-1)g'=2xg g'/g=2x/(2x-1)=(2x-1+1)/(2x-1)=1+1/(2x-1)  Integrating we have logg(x)=x&#...

16 cos(2θ) = 16 cos2(θ) − 9 cos(2θ)=cos2θ-sin2θ=cos2θ-9/16 so 9/16=sin2θ sinθ=±3/4 θ=arcsin(±3/4)

The formula to use is sinA+sinB=2sin((A+B)/2)cos((A-B)/2) sin(x) + sin(3x) = 4 sin(x) cos^2(x) 2sin((x+3x)/2)cos((3x-x)/2)=2sin2xcosx=4sinxcosxcosx=4sinxcos2x

This is a form you should learn Let a be the larger number, b the smaller a+b=13 a-b=1 add the equations 2a=14 so a=7 subtract the equations 2b=12 so b=6

Let p = original price 39=p-.25p=.75p p=39/.75=52

1 gal/min = 4 qts/min = 4/60 qts/sec = 1/15 qts/sec  so  15 sec/qt

sin(x-π)=sinxcosπ-coxsinπ=-sinx siπ(x - π) + 2 = 1 now reads -sinx+2=1 or sinx=1 so in [0,2π] the only answer is x=π/2

Let the first term be a, the common difference d. the ninth term is a+8d=0, which makes a=-8d The ratio of the 29th term to the 19th term is (a+28d)/(a+18d)=20d/10d=2

A+B+C=180 so that B+C=180-A A-B-C=A-(B+C)=A-(180-A)=2A-180 tan(x-y)=(tanx-tany)/(1+tanxtany) tan(2A-180)=(tan2A-tan(180))/(1_+tan2Atan(180))=tan2A/1-tan2A) since tan(180)=-1

− 4 < − 2 x − 14 < 3 This is really two inequalities. -4<-2x-14   and   -2x-14<3 -4<-2x-14 multiply by -1 to get 4>2x+14 subtract 14 from each side to get -10>2x or 2x<-10 or x<-5 − 2 x − 14 < 3add 14 to each side to get -2x<17...

I assume that you want the value of sinθ, given the value of cos2θ. The relevant half-angle formula is sinθ=√((1-cos2θ)/2) sinθ=√((1+24/25)/2)=7/(5√2)

The terms of a geometric sequence may be written as a, ar, ar2,...,arn-1 for thre first n terms 8 is the first term, 8r5 is the sixth term 8r5=243/128=35/27 divide by 8=23 r5=35/210 and so r=3/22=3/4

sinθ=2/5.  cos2θ=1-sin2θ gives cos2θ=1-4/25=21/25 cosθ<0 gives cosθ=-(√21)/5 sin2θ=2sinθcosθ=2×(2/5)×(-(√21)/5)=-4(√21)/25 cos2θ=cos2θ-sin2θ=21/25-4/25=17/25

or you could divide the numbers and see whether the answer is <=>1, as first number 4 3/8=35/8,  second number 34/7 35/8÷34/7=35/8×7/34=245/272 <1 so 34/7 is larger.

2x−y=1 −6x+3y=2 multiply the first equation by -3 to-3 get -6x+3y=-3 the two equations are a set of two parallel lines −6x+3y=2 and -6x+3y=-3 No solution.