First we try to frame the equation between the sale price and the final cost.
Let us consider, the price to be x and the final cost to be y, since the final cost is 75% of the price, we have the below equation
y = 0.75x
This is of the form of straight line equation y = mx +c where c =0 and m = 0.75
This means the function is linear since it satisfies a straight line equation.
So, we can eliminate options (3) and (4)
Now, the function is not a continuous one because the function is has definite value which is 0.75x and this make the function discrete. Hence option (2) is eliminated
Therefore we are left with option (1) which is the answer
Answer is (1) It is linear because the ratio of the change in the final cost compared to the rate of change in the price tag is constant
Answer:
x = 2
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Distributive Property
Equality Properties
- Multiplication Property of Equality
- Division Property of Equality
- Addition Property of Equality
- Subtraction Property of Equality
<u>Algebra I</u>
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
2(x + -5) + x = x + (-6)
<u>Step 2: Solve for </u><em><u>x</u></em>
- [Distributive Property] Distribute 2: 2x - 10 + x = x - 6
- [Addition] Combine like terms (x): 3x - 10 = x - 6
- [Subtraction Property of Equality] Subtract <em>x</em> on both sides: 2x - 10 = -6
- [Addition Property of Equality] Add 10 on both sides: 2x = 4
- [Division Property of Equality] Divide 2 on both sides: x = 2
Answer:
There are 60 permutations that can be formed from the first five letters of the alphabet.
Step-by-step explanation:
The number of three letter permutations that can be formed from the first five letters of the alphabet is
.


= 60
Hence, there are 60 permutations that can be formed from the first five letters of the alphabet.
10+15=25, so when you add 2+3 it equals 5 and when you times 5 by 5 it equals 25.
Answer:
Linearly Dependent for not all scalars are null.
Step-by-step explanation:
Hi there!
1)When we have vectors like
we call them linearly dependent if we have scalars
as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.
When all scalar coefficients are equal to zero, we can call them linearly independent
2) Now let's examine the Matrix given:

So each column of this Matrix is a vector. So we can write them as:
Or
Now let's rewrite it as a system of equations:

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.
Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.


