Let us say that:
a = ones
b = fives
c = twenties
So that the total money is:
1 * a + 5 * b + 20 * c = 229
=> a + 5b + 20c = 229 -->
eqtn 1
We are also given that:
c = a – 5 -->
eqtn 2
a + b + c = 30 -->
eqtn 3
Rewriting eqtn 3 in terms of b:
b = 30 – a – c
Plugging in eqtn 2 into this:
b = 30 – a – (a – 5)
b = 35 – 2a -->
eqtn 4
Plugging in eqtn 2 and 4 into eqtn 1:
a + 5(35 – 2a) + 20(a – 5) = 229
a + 175 – 10a + 20a – 100 = 229
11a = 154
a = 14
So,
b = 35 – 2a = 7
c = a – 5 = 9
Therefore there are 14 ones, 7 fives, and 9 twenties.
You should have put a picture so ppl would know what the question is askin
the second is a negative slope due to the fact the line is going the the left and the ind variable is breakfast dep is performance
Answer:
Step-by-step explanation:
The parent function here is y = log x, where 10 is the base.
The derivative of y = log x is dy/dx = (ln x) / ln 10.
The derivative of y = log (ax+b) is found in that manner, but additional steps are necessary: differentiate the argument ax + b:
The derivative with respect to 10 of log (ax + b) is:
dy/dx = [ 1 / (ax + b) ] / [ ln 10 ] *a, where a is the derivative of (ax + b).
Alternatively, we could express the answer as
dy/dx = [ a / (ax + b) ] / [ ln 10 ]
Answer:
option C: 5x+1
Explaination:
take 5x common and simplify
