First convert to decimal:
Now convert this to base 7:
Now convert this to base 7:
Harry worked 7 hours last week. This is 1/3 as many as Aidan worked. How many hours did Aidan work?
2 answers:
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I.
Given a function f with domain D and range R, and the inverse function
,
then the domain of is the range of f, and the range of is the domain of f.
II. We are given the function
,
the domain of f, is the set of all x for which
makes sense, so x is any x for which x-5
0, that is x≥5.
the range is the set of all values that f can take. Since f is a radical function, it never produces negative values, in fact in can produce any value ≥0
Thus the Domain of f is [5, ∞) and the Range is [0, ∞)
then , the Domain of is [0, ∞) and the Range of is [5, ∞)
III.
Consider
to find the inverse function
,
1. write f(x) as y:
2. write x in terms of y:
take the square of both sides
add 5 to both sides
3. substitute y with x, and x with
:
These steps can be applied any time we want to find the inverse function.
IV. Answer:
, x≥0
y≥0, where y are all the values that
can take
Remark: the closest choice is B
Given a function f with domain D and range R, and the inverse function
then the domain of
II. We are given the function
the domain of f, is the set of all x for which
the range is the set of all values that f can take. Since f is a radical function, it never produces negative values, in fact in can produce any value ≥0
Thus the Domain of f is [5, ∞) and the Range is [0, ∞)
then , the Domain of
III.
Consider
to find the inverse function
1. write f(x) as y:
2. write x in terms of y:
take the square of both sides
add 5 to both sides
3. substitute y with x, and x with
These steps can be applied any time we want to find the inverse function.
IV. Answer:
y≥0, where y are all the values that
Remark: the closest choice is B
Answer: the system has no solution.
Step-by-step explanation:
\displaystyle\\
\left \{ {{x^2y=16\ \ \ \ \ (1)} \atop {x^2+4y+16=0\ \ \ \ \ (2)}} \right. .\\
Multiply\ both\ sides\ of\ the\ equation\ (2)\ by\ y\ (y\neq 0):\\
x^2y+4y^2+16y=0\\
We\ substitute\ equation\ (1)\ into\ equation\ (2):\\
16+4y^2+16y=0\\
4y^2+16y+16=0\\
4*(y^2+4y+4)=0\\
4*(y^2+2*y*2+2^2)=0\\
4*(y+2)^2=0\\
Divide\ both\ sides\ of\ the \ equation\ by\ 4:\\
(y+2)^2=0\\
(y+2)*(y+2)=0\\
So,\ y+2=0\\
y=-2.\\