Melody

Answered.

\(\overline{FG}=\sqrt{3.5^2+7^2}\\ \overline{FGE}=2\sqrt{3.5^2+7^2}\\ [\overline{FGE}]^2=4(3.5^2+7^2)\\ [\overline{FGE}]^2=245\\ l^2=245 \)

Thanks answering guest.

Sorry asker guest, I did not read your question properly but you can still use the table only instead of using the headings

1,2,3,4,5,6

jus use

1,2,2,4,6,6

It will work well :)

You can count the number of boxes that make it true and put that over 36 because there are 36 squares (in the grid) in total.

Use this table, or one the same to work out all your possibilities.

The top heading row is the number on dice 1

The left heading row it the number on dice 2

You fill out the sums and answer the question/s accordingly.

I do not think your description makes sense.

I take it that X is between A and B

So there is no arc or interval XAB, it would have to be AXB or BXA I suppose.

Can you write your problem so that it makes sense please?

\((2 \times 3^x)^3 +(9^x - 3)^3 = (9^x + 2 \times 3^x - 3)^3\\ let\;\; y=3^x\\ (2y)^3 +(y^2 - 3)^3 = (y^2 + 2y - 3)^3\\ LHS=y^6-9y^4+8y^3+27y-27\\ RHS=y^6+6y^5+3y^4-28y^3-9y^2+54y-27\\ so\\ -9y^4+8y^3+27y=6y^5+3y^4-28y^3-9y^2+54y\\ \color{red}y=0\;\;\text{could lead to one solution}\qquad \color{black}or\\ -9y^3+8y^2+27=6y^4+3y^3-28y^2-9y+54\\ 6y^4+12y^3-36y^2-36y+54=0\\ y^4+2y^3-6y^2-6y+9=0 \)

\(consider\\ f(y)=y^4+2y^3-6y^2-6y+9\\ \text{Any integer roots will be factors of 9 so they could be}\pm1,\;\;\pm3,\;\;\pm9\\ f(1)=0\;\;\text{therefore it is divisible by }(y-1)\\ f(-3)=0\;\;\text{therefore it is divisible by }(y+3)\\ \text{by division I found another factor to be}\;\; (y^2-3)\\ so\\ f(y)=(y-1)(y+3)(y^2-3)\\ (y-1)(y+3)(y-\sqrt3)(y+\sqrt3)=0\\ \text{So I have}\\ y=0,1,-3,\sqrt3, -\sqrt3\\~\\ y=3^x\qquad x=log_3y\qquad \text{y has to be greater than 0 so}\\ y=1,\sqrt3\\ x=0\;\;or\;\;0.5 \)

Latex:

(2 \times 3^x)^3 +(9^x - 3)^3 = (9^x + 2 \times 3^x - 3)^3\\
let\;\; y=3^x\\
(2y)^3 +(y^2 - 3)^3 = (y^2 + 2y - 3)^3\\
LHS=y^6-9y^4+8y^3+27y-27\\
RHS=y^6+6y^5+3y^4-28y^3-9y^2+54y-27\\
so\\  -9y^4+8y^3+27y=6y^5+3y^4-28y^3-9y^2+54y\\
\color{red}y=0\;\;\text{could lead to one solution}\qquad \color{black}or\\
-9y^3+8y^2+27=6y^4+3y^3-28y^2-9y+54\\
6y^4+12y^3-36y^2-36y+54=0\\
y^4+2y^3-6y^2-6y+9=0

consider\\
f(y)=y^4+2y^3-6y^2-6y+9\\
\text{Any integer roots will be factors of 9 so they could be}\pm1,\;\;\pm3,\;\;\pm9\\
f(1)=0\;\;\text{therefore it is divisible by }(y-1)\\
f(-3)=0\;\;\text{therefore it is divisible by }(y+3)\\
\text{by division I found another factor to be}\;\; (y^2-3)\\
so\\
f(y)=(y-1)(y+3)(y^2-3)\\
(y-1)(y+3)(y-\sqrt3)(y+\sqrt3)=0\\
\text{So I have}\\
y=0,1,-3,\sqrt3, -\sqrt3\\~\\
y=3^x\qquad x=log_3y\qquad \text{y has to be greater than 0  so}\\
y=1,\sqrt3\\
x=0\;\;or\;\;0.5

Find the coefficient of $y^4$ in the expansion of $(2y - 5)^7.$

\(\begin{pmatrix} 7\\ 4 \end{pmatrix} (2y)^4(-5)^3\)

you can finish it

PLUS     0     

-5*-2 = +10 

That is the largest possible value.