Answer:
Step-by-step explanation:
You can split the coins into 3 groups, each of them has 3 coins. Weigh group 1 vs group 2, if one is lighter, that group has the fake coin. If both groups weigh the same, then group 3 has the fake coin.
Continue to split the group that has the fake coin into 3 groups, each group has 1 coin. Now apply the same procedure and we can identify the fake coin.
Total of scale usage is 2
b) if you have 
coins then you can apply the same approach and find the fake coin with just n steps. By splitting up to 3 groups each step, after each step you should be able to narrow down your suspected coin by 3 times.
Step 1: you narrow down to group of 
coins
Step 2: you narrow down to group of 
coins
Step 3: you narrow down to group of 
coins
...
Step n: Step 1: you narrow down to group of 
coin
10% of 320 = 32
So 9.8% of 320 should be some number between 31 and 32
Answer:
30
Step-by-step explanation:
x=6
4 by 6 =24
24 + 6 =30
Answer:
-6
Step-by-step explanation:
We know that since Ax + By = 3 passes through (-7, 2), then if we plug -7 in for x and 2 in for y, the equation is satisfied. So, let's do that:
Ax + By = 3
A * (-7) + B * 2 = 3
-7A + 2B = 3
We also know that this line is parallel to x + 3y = -5, which means their slopes are the same. Let's solve for y in the second equation:
x + 3y = -5
3y = -x - 5
y = (-1/3)x - (5/3)
So, the slope of this line is -1/3, which means the slope of Ax + By = 3 is also -1/3. Let's solve for y in the first equation:
Ax + By = 3
By = -Ax + 3
y = (-A/B)x + 3/B
This means that -A/B = -1/3. So, we have a relationship between A and B:
-A/B = -1/3
A/B = 1/3
B = 3A
Plug 3A in for B into the equation we had where -7A + 2B = 3:
-7A + 2B = 3
-7A + 2 * 3A = 3
-7A + 6A = 3
-A = 3
A = -3
Use this to solve for B:
B = 3A
B = 3 * (-3) = -9
So, B = -9 and A = -3. Then B - A is:
B - A = -9 - (-3) = -9 + 3 = -6
The answer is -6.
<em>~ an aesthetics lover</em>
Reduced ratio<span> of any </span>two<span> corresponding sides is called the scale factor of the similar </span>triangles<span>. </span>
