Answer:
{324, 0, -6, 54}
Step-by-step explanation:
Given that F(x) = 3x² − 3 x − 6 If the domain of a the function f(x) is { -10, -1, 1, 5 }.
To get the range, we will find f(x) for all the domains
at x = -10
f(-10) = 3(-10)²-3(-10) - 6
f(-10) = 3(100)-3(-10) - 6
f(-10) = 300+30-6
f(-10) = 330-6
f(-10) = 324
at x = -1
f(-1) = 3(-1)²-3(-1) - 6
f(-1) = 3(1)-3(-1) - 6
f(-1) = 3+3-6
f(-1) = 6-6
f(-1) = 0
at x = 1
f(1) = 3(1)²-3(1) - 6
f(1) = 3(1)-3(1) - 6
f(1) = 3-3-6
f(1) = 0-6
f(1) = -6
at x = 5
f(5) = 3(5)²-3(5) - 6
f(5) = 3(25)-3(5) - 6
f(5) = 75-15-6
f(5) = 60-6
f(5) = 54
Hence the range of the relation are {324, 0, -6, 54}
Answer: 
Step-by-step explanation:
Given , Micheal bought a car for $15,540.
Rate of depreciation = 5% = 0.05 { To convert a percentage into decimal or fraction , we divide it by 100]
The exponential function used to show depreciation in amount or value ina time period of 'x' is given by :-
, where A = Initial value
r= rate of depreciation.
As per given , we have
A= $15,540
r= 0.05
Then , Required function : 
Answer:
I think x should be equal to 0.
Answer:
Step-by-step explanation:
Hi
Step 1.. Explain the question
From the question,our aim is to know the litres of a 49% critic acid solution Georgina should add to 0.5 litres of water in order to make a 42% critic solution acid
Step 2.. Identify the given and missing variables
Here we not given the litres of a 49% critic solution to be added =?
We were given to critic acid solution to be made =42% which is our solute
Step 3 ...Representing the question in an equation format
Let the number of litres to be added be represented by y
Adding 0.5 litres of water will give y+0.5
Then we proceed
42/100 * (y+0.5) =49/100
0.42(y+0.5)=0.49
Expand the bracket
0.42y +0.21=0.49
0.42y=0.49-0.21
0.42y=0.28
Divide both sides by 0.42
y=3
Answer:
The length of the wire is 3.24451394 × 10⁻⁹ cm
Step-by-step explanation:
The volume of the wire = 66 cm³
The diameter of the wire = 1 mile
We convert the volume from cm³ into mile³ to get;
66 cm³ = 1.5834 × 10⁻¹⁴ mile³
The volume = π × D²/4 × l
Where;
D = The diameter = 1 mile
l = The length of the wire
1.5834 × 10⁻¹⁴ = π × 1²/4 × l
The length of the wire = l = 1.5834 × 10⁻¹⁴/(π × 1²/4) = 2.0160475 × 10⁻¹⁴ mile
1 mile = 160,934.4 cm
2.0160475 × 10⁻¹⁴ mile = 2.0160475 × 10⁻¹⁴ × 160,934.4 cm = 3.24451394 × 10⁻⁹ cm
The length of the wire = 3.24451394 × 10⁻⁹ cm
