Answer:
The required expression is .
The value of the expression when y=20 is 2.
Step-by-step explanation:
Consider the provided phase.
3 less than the quotient of a number y and 4
The quotient of a number y and 4 can be written as:
Now 3 less than the quotient of a number y and 4 can be written as:
Hence, the required expression is .
Now evaluate when y=20.
Substitute y=20 in above expression.
Hence, the value of the expression when y=20 is 2.
Answer:
d = -1/3, 0
Step-by-step explanation:
Subtract the constant on the left, take the square root, and solve from there.
(6d +1)^2 + 12 = 13 . . . . given
(6d +1)^2 = 1 . . . . . . . . . .subtract 12
6d +1 = ±√1 . . . . . . . . . . take the square root
6d = -1 ±1 . . . . . . . . . . . .subtract 1
d = (-1 ±1)/6 . . . . . . . . . . divide by 6
d = -1/3, 0
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Using a graphing calculator, it is often convenient to write the function so the solutions are at x-intercepts. Here, we can do that by subtracting 13 from both sides:
f(x) = (6x+1)^ +12 -13
We want to solve this for f(x)=0. The solutions are -1/3 and 0, as above.
