Answer:
k = 4
Step-by-step explanation:
Given that a varies directly as b then the equation relating them is
a = kb ← k is the constant of variation
To find k use the condition a = 8 when b = 2
k =
=
= 4
Answer:

Step-by-step explanation:
![\sf h(x) = 5x+2\\\\Put \ h(x) = -8\\\\-8 = 5x+2\\\\Subtract \ 2 \ to \ both \ sides\\\\-8-2 = 5x\\\\-10 = 5x\\\\Divide\ both \ sides \ by \ 5\\\\-10 / 5 = x \\\\x = -2 \\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Csf%20h%28x%29%20%3D%205x%2B2%5C%5C%5C%5CPut%20%5C%20h%28x%29%20%3D%20-8%5C%5C%5C%5C-8%20%3D%205x%2B2%5C%5C%5C%5CSubtract%20%5C%202%20%5C%20to%20%5C%20both%20%5C%20sides%5C%5C%5C%5C-8-2%20%3D%205x%5C%5C%5C%5C-10%20%3D%205x%5C%5C%5C%5CDivide%5C%20both%20%5C%20sides%20%5C%20by%20%5C%205%5C%5C%5C%5C-10%20%2F%205%20%3D%20x%20%5C%5C%5C%5Cx%20%3D%20-2%20%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Hope this helped!
<h3>~AH1807</h3>
Start by putting two coins on each side of the scale. If one side is higher, then the light coin must be in that pair (use a second weighing to determine which of that pair is the lightest.)
If the initial weighing shows the two pairs to be even, discard those four and go to the three others. Put two of the coins on the scale - one on each side. If one side stays higher, that coin is your light one. If however they are equal, the final unweighed coin must be the lighter coin.
$115.45 This should be the correct answer. Hope this help need brainlest.