Answer:
[tex]3^{11}[/tex]
Step-by-step explanation:
It is given in the question that:
[tex]6x-2y=11[/tex]
Dividing both hand sides by 2, we get:
[tex]3x-y=5.5[/tex]
We have to find the value of [tex]\frac{9^{3x} }{9^{y} }[/tex].
From laws of exponents we know:
[tex]\frac{a^{m} }{a^{n}}=a^{m-n}[/tex]
So we can write :
[tex]\frac{9^{3x} }{9^{y} }=9^{3x-y}[/tex]
[tex]=9^{5.5}=3^{11}[/tex]
Answer:option C is the correct answer
Step-by-step explanation:
The given equation is
6x - 2y = 11
Let us make x the subject of the formula. The first step is to add 2y to both the left hand side and the right hand side of the equation. It becomes
6x - 2y + 2y = 11 + 2y
6x = 11 + 2y
Dividing both the left hand side and the right hand side of the equation by 6. It becomes
6x/6 = (11 + 2y)/6
x = (11 + 2y)/6
We would substitute x = (11 + 2y)/6 into (9^3x)/(9^y). It becomes
[9^3(11 + 2y)/6 ] / (9^y)
= [9^(11 + 2y)/2 ] / (9^y)
= [9^(5.5 + y)/] / (9^y)
We would apply the law of indices
a^b/a^c = a^(b - c)
Therefore
[9^(5.5 + y)/] / (9^y) = 9^(5.5 + y - y)
= 9^(5.5)
= 3^2(5.5)
= 3^11