Answer:
I think it would be 3y+3y-12 because 6y=3y+3y
Step-by-step explanation:
In exponential form: 10^2=3(x+5)
If we were to solve, x=85/3
To do this, you got to square 256.
The square root of 256 is 16.
Therefore, there are 16 small squares on each edge of the mosaic.
Kinda proof:
o o o o O
o o o o O
o o o o O
o o o o O
o o o o O
25 squares. Square root is 5. 5 along each edge. My work shares same concept.
Extremely unnecessary proof:
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
o o o o o o o o o o o o o o o O
There are 256 squares, and you can count 16 on each edge. this shows 16 times 16, or 16 squared, which is 256.
Answer: a) An = An-1 + An-2
b) 55ways
Step-by-step explanation:
a) a nickel is 5 cents and a dime is 10cent so a multiple of 5 cents is the possible way to pay the tolls in both choices.
Let An represents the number of possible ways the driver can pay a toll of 5n cents, so that
An = 5n cents
Case 1: Using a nickel for payment which is 5 cents, the number of ways given as;
An-1 = 5( n-1)
Case 2: using a dime which is two 5 cents, the number of ways is given as;
An-2 = 5(n-2)
Summing up the number of ways, we have
An = An-1 + An-2
From the relation,
If n= 0, Ao= 1
n =1, A1= 1
b) 45 cents paid in multiples of 5cents will give us 9 ways(A9)
From the relation, we have that
Ao = 1
A1 = 1
An =An-1 + An-2
Ao = 1
A1 = 1
A2 = A1+Ao = 1+1= 2
A3 = A2 + A1 = 3
A4 = A3+A2=5
A5=A4+A3=8
A6=A5+A4=13
A7 =A6+A5 = 21
A8= A7+A6= 34
A9= A8+A7= 55
So there are 55ways to pay 45cents.
Hello user
To solve for V we simplify both sides of the equation then isolate the variable to get v <span>≥ 2
Therefor the answer is: </span>v ≥ 2
<span>
I hope this helped
-Chris</span>