Iam not sure for the awnser but maybe its 3
Answer:
Gh 2
Step-by-step explanation:
2
bearing in mind that standard form for a linear equation means
• all coefficients must be integers, no fractions
• only the constant on the right-hand-side
• all variables on the left-hand-side, sorted
• "x" must not have a negative coefficient

![\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{8}{9}}[x-\stackrel{x_1}{(-4)}]\implies y+1=\cfrac{8}{9}(x+4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{9}}{9(y+1)=9\left( \cfrac{8}{9}(x+4) \right)}\implies 9y+9=8(x+4)\implies 9y+9=8x+32 \\\\\\ 9y=8x+23\implies -8x+9y=23\implies 8x-9y=-23](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20%5Ctextit%7Bpoint-slope%20form%7D%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y-y_1%3Dm%28x-x_1%29%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%5Cimplies%20y-%5Cstackrel%7By_1%7D%7B%28-1%29%7D%3D%5Cstackrel%7Bm%7D%7B%5Ccfrac%7B8%7D%7B9%7D%7D%5Bx-%5Cstackrel%7Bx_1%7D%7B%28-4%29%7D%5D%5Cimplies%20y%2B1%3D%5Ccfrac%7B8%7D%7B9%7D%28x%2B4%29%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B9%7D%7D%7B9%28y%2B1%29%3D9%5Cleft%28%20%5Ccfrac%7B8%7D%7B9%7D%28x%2B4%29%20%5Cright%29%7D%5Cimplies%209y%2B9%3D8%28x%2B4%29%5Cimplies%209y%2B9%3D8x%2B32%20%5C%5C%5C%5C%5C%5C%209y%3D8x%2B23%5Cimplies%20-8x%2B9y%3D23%5Cimplies%208x-9y%3D-23)
Answer:
x=13
Step-by-step explanation:
9^2 * 27^3 = 3^x
We need to get each term with a base of 3
9^2 = (3^2) ^2
We know that a^b^c = a^(b*c)
(3^2) ^2 = 3^(2+2) = 3^4
27^3 = (3^3) ^3 = 3^(3*3) = 3^9
Replacing these in the original equation
3^4 * 3^9 = 3^x
We know that a^b *a^c = a^(b+c)
3^4 * 3^9 =3^(4+9) = 3^13 = 3^x
The bases are the same, so the exponents must be the same
x=13