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Probably....dirt is just really really fine sand.......the dirt particles would have to increase the surface area of the water exposed to air.....Like before....it would be more like adding water TO dirt    rather than adding   dirt TO water....

thank you for your response do you think dirt would result in the same results as sand 

....something that will increase the surface area       Air bubbling through it......      A giant sponge.......a pile of sand      a pile of rocks.....   More like you would be adding the water TO the substance.....

If your water is in a cone...adding anything that raises the water level will increase evaporation....larger surface area....

    Lots of options for your vague question....

yes but like a substance 

Heat.    Decreased pressure.

Or something that will increase the surface area.....

V =250 (10^(t/93.27) )

or

V = 250 (1.02499^t) 

Im still confused, What is the number that is missing from the equation?

\(a^t = e^{\ln(a)t}\)

are there missing decimal points?

\((f.g)(x) = f(x) g(x)\\ \text{so just multiply the two polynomials together}\)

You don't need help.  You're just lazy.

Make a table of values for f(x), and one for g(x)

compare them

If you need to zoom in on an area where they are close but not exactly equal and use closer values of x to find equality, do so

\(f(x) = 4x^2 + 6x = 2x(2x+3)\\ \text{If this problem makes any sense at least one of these terms will be a factor of }g(x)\\ g(x) = 2x^2 + 13x + 15 = (2x+3)(x+5)\\ (f/g)(x) = \dfrac{f(x)}{g(x)} = \dfrac{2x(2x+3)}{(x+5)(2x+3)} = \dfrac{2x}{x+5}\\ \text{HOWEVER}\\ \text{There is a values of }x \text{ for which this is invalid}\\ \text{I leave it to you to identify these}\)

Set the two functions equal to each other (this is what the questions asks) ...and then solve for 'x'

     Then, plug this value of 'x' back in to one of the equations to find the 'y' value.....

Here is the start.....

x^2-6x+8 = x -2    subtract x from both sides of the equation

x^2 -7x +8 = -2     add 2 to both sides

x^2-7x +10 = 0     Now.....factor ....or use quadratic formula to find the x value(s)

(x-5)(x-2) = 0           You SHOULD be able to find the x values and the coreresponding 'y' values from here....at LEAST ..try ! 

OR:   see:

https://web2.0calc.com/questions/algebra-2_145

Average change =   (change in x) / (change in value af f(x) over same 'x' values)

Ave chg = (9-5)/ (14- -2) =  ???

who is Rohit?

Assuming you mean Jack...

\(T = \text{time spent driving first 785 miles}\\ s_1 = \dfrac{785}{T}\\ s_1+12 = \dfrac{785}{T-10}\)

\(\text{Substitute the first equation into the second and solve for } T\\ \text{Find }s_1 \text{ using the first equation}\\ s_2=s_1+12\)

Rate of change on [7, 9]  =     [ 3878 - 1852] / [9 - 7 ] =  2026 / 2  =  1013

Rate of change on [4, 6 ] =  [ 1178 - 358] / [ 6 - 4 ]  =  820 / 2  =   410

So....the rate of change is   1013 - 410  =  603 greater

\(6701^{83} = \sum \limits_{k=0}^{83} ~\dbinom{83}{k} (6000)^{k}(701)^{83-k}\\ \text{It should be clear that all terms but 1, }k=0 \text{ are divisible by }1000\\ \text{we are then left with }\\ (701)^{83} = \sum \limits_{k=0}^{83}\dbinom{83}{k}(700)^k\\ \text{here there are two terms that are not divisible by }1000,~k=0,1\\ (701)^{83}\pmod{1000} = \\ 1+83(700) \pmod{1000} = \\ 1 + (80+3)(700) \pmod{1000}= \\ 1+ 56000+ 2100 \pmod{1000} = \\ 2101 \pmod{1000} = 101\)

(f - g)(x)  = f(x) - g(x)  =

[ 3x^5 + 6x^2 - 5 ] - [ 2x^4 + 7x^2 - 1x + 16 ]  =

3x^5 - 2x^4 - x^2 + x - 21

f(f(x))  =          1                             1                                 x  -  3

               ___________   =       ____________   =       _________

                [1 / {x - 3)] -3               [ 1 - 3(x - 3)]                 10 - 3x

                                                  ___________

                                                      (x -  3)

The numerator has no effect on the domain

The domain will be all x's  except those where 10 - 3x = 0

So....the domain includes all x's   except  x = (10/3)

See the graph here : https://www.desmos.com/calculator/yovhfn2ymb

Thank you!! 

2sin^2 x = 1       divide both sides by 2

sin^2 x  =  1/2      take both roots

sin x = 1 /√2        or   sin x = -1 / √2

sin x = 1/√2   at   pi/4  and 3pi/4      on   [0, 2pi ]

sinx = -1 / √2  at  5pi/4 and 7pi/4    on [ 0, 2pi ]

So.....it appears that this happens  every 90°   after pi/4  or....expressed another way....at

pi/4 +  pi/2 * n      where n is an integer

Thank you! 

We are using this property, GM

a^(m/n)  =     ⁿ√[a^m ]

Also remember that    a^m / a ⁿ  =  a ^{(m - n )}

We just need to be careful about the term that the exponent is applied to

a)  4x^1/2 =  4 * x^1/2  =  4 * ²√ [ x¹ ]   =  4 √x

Note that the exponent doesn't apply to the  4   !!!!

b)  (4x)^1/2   .....note that the exponent applies to  (4x)....this is different from the first one...=

√[4x ]  =  2√x

c)   Note  x ^-a  =  1 / x^a

x^-2/5  =   1/ x^2/5  =  1 /  ⁵√[ x² ]

d)  Note that   (8x³)^1/3  = 8^1/3 * (x³)^1/3    =  ³√8  * x¹   =  2x¹

So....we have   

2x^2/3        

____   =        x ^{(2/3 - 1)}  = x^-1/3  =  1 / x^1/3 =    1 / ³√x

  2x¹

e)   a^1/3

     ____               can you finish this one, GM  ?????

        a

This question is asking for where is the cos of the angle equao to    - sqr3 / 2

This occurs at   5 pi  /6     and   7pi / 6