Plz or i fail my test this is a application question. a speedboat takes 1 hour longer to go 24 miles up a river than to return. if the boat cruises at 10 miles per hour in still water, what is the rate of the current? do show work.

time upstream=24/8=3 hours

time downstream=24/12=2 hours

Step-by-step explanation:

If R*T=D, and R is speed + or - current, how can you find the rate without the time, and how can you find the time without the rate?

THE KEY WAS WHEN WE WERE TOLD THAT IT TAKES ONE HOUR LONGER TO GO UPSTREAM THAN DOWNSTREAM BECAUSE T=D/R AND WE ARE TOLD THAT:

D/R(Upstream) minus 1 equals D/R (DOWNSTREAM)

Let x=rate of current

distance(d)=rate(r) times time(t) or d=rt or t=d/r

distance upstream=distance downstream=24 mi

rate upstream=10-x

rate downstream=10+x

time upstream=(distance upstream)/(rate upstream)=24/(10-x)

time downstream=(distance downstream)/(rate downstream)=24/(10+x)

Now, we are told that time upstream minus one hour equals time downstream. So our equation to solve is:

24/(10-x)-1=24/(10+x) Multiply both sides by (10-x)(10+x)

24(10+x)(10-x)/(10-x)-1(10-x)(10+x)=24(10-x)/(10+x)/(10+x) clear fractions

24(10+x)-(10-x)(10+x)=24(10-x) clear parens

240+24x-100-10x+10x+x^2=240-24x subtract 240 from and add 24x to both sides

240-240+24x+24x-100+x^2=240-240-24x+24x collect like terms

x^2+48x-100=0 quadratic equation in standard form

This equation can be factored:

(x+50)(x-2)=0

x=-50 mphdiscount the negative value for speed

and

x=2 mphspeed of the current

CK

10-x=10-2= 8 mph speed upstream

10+x=10+2=12 mphspeed downstream

time upstream=24/8=3 hours

time downstream=24/12=2 hours

so 3-2=1takes 1 hour longer upstream

what is the question? i am guessing you are looking for the area.

to find an area, you have to do length x width.

in this case, it would be 12 ft x 4 ft = 48 sq. ft

for this case we have that by definition, the surface area of a cylinder is given by:

where:

r: it is the radius of the cylinder

h: it is the height of the cylinder

we can also rewrite it as:

where:

la: it's the lateral area

ba: area of the base

option a