Domain of - 5 : x ≥ 25 .
Range of √x - 5 : y ≥ -5 .
Given,
f(x) = √x - 5
Here,
To find the domain,
√x - 5 ≥ 0
√x ≥ 5
Squaring both sides ,
x ≥ 25 .
Thus domain of √x - 5 is x ≥ 25 .
Range of √x - 5,
To define the range of the function the value of x can not be negative as it will make the function complex .
So,
Range will be ,
y ≥ -5 .
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line y= 2x - 3
The required point for the given line is (0, -3). The correct option is (D).
A linear equation in two variable has the general form as y = ax + by + c, where a, b and c are integers and a, b ≠ 0.
It can be represented as a straight line on a graph.
The equation of the given line is y = 2x - 3.
In order to find the point lying on it, consider each of the options one by one as follows,
(a) (2 , 3)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 7
RHS = 2x - 3
= 2 × 2 - 3
= 1
Since, LHS ≠ RHS, the given point is not the solution.
(b) (-2, -1)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 7
RHS = 2x - 3
= 2 × -1 - 3
= -6
Since, LHS ≠ RHS, the given point is not the solution.
(c) (4, 1)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 4
RHS = 2x - 3
= 2 × 1 - 3
= -1
Since, LHS ≠ RHS, the given point is not the solution.
(d) (0, -3)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= -3
RHS = 2x - 3
= 2 × 0 - 3
= -3
Since, LHS = RHS, the given point is the solution.
Hence, the point (0, -3) is the solution of the given equation.
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sell a minimum of $8300 worth of chairs and
tables each day. Write an inequality that could
represent the possible values for the number of
tables sold, t, and the number of chairs sold, c,
that would satisfy the constraint.
Answer:
13 3/4
Step-by-step explanation:
Answer:
x = − 19
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Mon | 0.3
Tues | 15%
Wed | 1/6
Thurs | 0.2
Fri | 1/8
ON WHICH DAY DID YOU SPEND THE LEAST AMOUNT OF TIME ON THE RUNNING MACHINE?
Answer: On Friday, he spend the least amount of time on the running machine.
Step-by-step explanation:
Since we have given that
Monday 0.3 =
Tuesday = 15%
Wednesday =
Thursday =
Friday =
Hence, we can compare all the days, and we get that
On Friday, he spend the least amount of time on the running machine.