Answer:
b+6
Problem:
If the average of b and c is 8, and d=3b-4, what is the average of c and d in terms of b?
Step-by-step explanation:
We are given (b+c)/2=8 and d=3b-4.
We are asked to find (c+d)/2 in terms of variable, b.
We need to first solve (b+c)/2=8 for c.
Multiply both sides by 2: b+c=16.
Subtract b on both sides: c=16-b
Now let's plug in c=16-b and d=3b-4 into (c+d)/2:
([16-b]+[3b-4])/2
Combine like terms:
(12+2b)/2
Divide top and bottom by 2:
(6+1b)/1
Multiplicative identity property applied:
(6+b)/1
Anything divided by 1 is that anything:
(6+b)
6+b
b+6
The answer is 51.3
Explanation:
You’re finding angle Y, so write the SOH CAH TOA. the two sides your are given are the adjacent side to Y and the opposite side so you use tangent. You then get the equation tanY=10/8 Since you’re finding an angle you use inverse tangent so you get y=tan^-1(10/8) and then put that into the calculator!
Let p = points. 0 -6(3) + 4 - 2 = p. Simplify multiplication to get -18 + 4 - 2. -18 + 4 = -14 - 2 = -16.
Likely best describes the likelihood of the probability 2/3. It is not completely certain, nor is it unlikely, and it definitely cannot be impossible.
Answer:
3[(u × t) - v]
Step-by-step explanation:
I'm not sure if this is correct, but this is the only equation that I can come up with.
