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Sorry the edit keeps messing it up...

The first "answer" I put Mean on top of, so the mean is 4

the 2nd "answer" I put Difference things on top of.

The charts are the difference thingys...

But from there I have absolutely NO IDEA what to do...

can you solve the initial value problem for:

\(\dfrac{dy}{dx}= xy-2x+y-2,\ y(2)= -1\)

\(\begin{array}{|rclcl|} \hline \mathbf{\dfrac{dy}{dx}} &=&\mathbf{ xy-2x+y-2 } \\\\ \dfrac{dy}{dx} &=& x(y-2)+ (y-2) \\\\ \dfrac{dy}{dx} &=& (y-2)(x+1) \\\\ \dfrac{dy}{(y-2)} &=& (x+1)dx \\\\ \large{ \int } \dfrac{1}{(y-2)}\ dy &=&\large{ \int } (x+1)\ dx \\\\ \mathbf{\ln(y-2) + c_1} &=& \mathbf{\dfrac{x^2}{2} +x + c_2 } \\ \hline && \ln(y-2) + c_1 &=& \dfrac{x^2}{2} +x + c_2 \quad | \quad y(2) = -1 \\ && \ln(-1-2) + c_1 &=& \dfrac{2^2}{2} +2 + c_2 \\ && \ln(-3) + c_1 &=& 4 + c_2 \quad | \quad \boxed{\mathbf{c_1 = -\ln(-3)}} \\ && \ln(-3) -\ln(-3) &=& 4 + c_2 \\ && 0 &=& 4 + c_2 \\ && \mathbf{c_2} &=& \mathbf{-4} \\ \hline \mathbf{\ln(y-2) -\ln(-3)} &=& \mathbf{\dfrac{x^2}{2} +x -4 } \\ \ln\left(\dfrac{y-2}{-3}\right) &=& \dfrac{x^2}{2} +x -4 \\ \dfrac{y-2}{-3} &=& e^\left(\dfrac{x^2}{2} +x -4 \right) \\ y-2 &=& -3e^\left(\dfrac{x^2}{2} +x -4 \right) \\ \mathbf{ y(x)} &=& \mathbf{2-3e^\left(\dfrac{x^2}{2} +x -4 \right)} \\ \hline \end{array} \)

450 is the correct answer

Thank you, Melody !

THX SO MUCH

wow how did I not see that. I suck at functions. THX SO MUCH

f(4) = f(-4)

g(f(4)) = g(f(-4)) = 9

-9

but I heighly dought that this is correct. 

ps. srry for my spelling mistakes:)

f(x) is an even function.

f(4) = f(-4)

so what's the answer?

I'm assuming the pyramid has a square base for this

The formula for the volume of a pyramid is 1/3*base area*height

base area=220²=48400m²

^{(if the area of the base is not square, use the side length to find the area of whatever shape it is and substitute it in)}

height=105m

so the volume of the pyramid is 1/3*48400m²*105m=1,694,000m³

Except the pyramid is only 96% solid so the volume of material used is .96*1694000=1,626,240m³

THX SO MUCH!!​​​

\(\text{so we want to minimize $\left||x|-(-x^2-3x-2)\right|= \left||x| + x^2+3x+2\right|$}\)

\(\text{for $x<0$}\\ \left||x|+x^2+3x+2\right| = \left|x^2 +2x+2\right|= (x+1)^2+1\\ \text{The minimum of this length occurs at $x=-1$ and is length 1}\\~\\ \text{for $x\geq 0$}\\ \left||x|+x^2+3x+2\right| = \left|x^2 +4x+2\right| = \left|(x+2)^2-2\right|\\ \text{Since $x>0$ this has a minimum at $x=0$ and is length 2}\\ \text{Thus the smallest possible length is 1 which occurs at $x=-1$}\)

\(\text{we select two consecutive whole numbers and find their remainder after dividing by 4}\\ \text{the following pairs are possible}\\ (0,1),~(1,2),~(2,3)~(3,0)\\ \text{out of these 4 pairs of remainders 2 of them include 0, i.e.}\\ \text{the original pair had a number divisible by 4}\\ p = \dfrac 2 4 = \dfrac 1 2\)

THANKS!!

THX SO MUCH. 

\(\text{If you study this you find that you must have the pattern $\\MDDM$ embedded in the seating arrangement}\\ \text{There are 3 slots in which these 4 guests can be fit}\\ \text{Having fit them the 2 elves sit in the remaining seats}\\ \text{There are 4 ways to arrange $MDDM$}\\ \text{There are 2 ways to arrange the elves once the others are seated}\\ N = 3 \cdot 4 \cdot 2 = 24\)

answered on the thread at the link

\(\text{if $k>0$ and $x>1$ then $x^k>x$}\\ \text{if $k>0$ and $x=1$ then $x^k=1$}\\ \text{so the range of $f(x)$ for $x \in [1,\infty)$ is $f(x) \in [1,\infty)$}\)

THX SO MUCH.

can you plz help me on this question. That would be amazing . 

 https://web2.0calc.com/questions/halp-plz_6

Yes i believe they should be included.  Even though there is a zero count of one of them, each one is still a valid combination.

what is the problem?  You still have not made your meaning clear!

How many 4-digit numbers have the last digit equal to the sum of the first two digits?

The first two can add to 1,2,3,4,5,6,7,8,9

First I am just looking at combinations for the first 2 digits. 

01     to     09          9 ways                This line was wrong, there is only 9       10,20,30,40,50,60,70,80,90

11   to      18          8+7 ways                11,12,13,14,15,16,17,18,    21,31,41,51,61,71,81

22    to      27          6+5 ways                22,23,24,25,26,27,             32,42,52,62,72,

33  to     36            4+3 ways                  33,34,35,36,                       43,53,63,

44  to    45              2+1ways                     44,45,                                    54

 9+15+11+7+3 = 45

45 combinations for the first two digits

10 combinations for the 3rd digit

The last digit is preset

45*10*1 = 450 possible numbers

OK  Mine is fixed now, i just had a stupid error.     We seem to be in agreement that the answer is 450.

Thanks guest and Rom  

g(4) means f(x) = 4, so 2x - 3 = 4 or x = 7/2.

Hence g(4) = 5 - 4*7/2  or g(4) = -9

Thanks Heureka and Rom.

You both reminded me of things I had almost forgotten.  Thanks :)