Answer:
False
Step-by-step explanation:
Consider the equations with the same number of equations and variables as shown below,
<u>Case 1</u>
This equation has no solution because it is not possible to have two numbers that give a sum of 0 and 1 simultaneously.
<u>Case 2</u>
This equation has infinitely many possible solutions.
Therefore it is FALSE to say a linear system with the same number of equations and variables, must have a unique solution.
6/(x-8)=2/(x+6)
times both sides by (x-8)(x+6)
6(x+6)=2(x-8)
distribute
6x+36=2x-16
minus 2x both sides
4x+36=-16
miinus 36 both sides
4x=-52
divide both sides by 4
x=-13
Answer:
<h2>k > 4.4</h2>
Step-by-step explanation:
Step1: Define an odd integer.
Define the first odd integer as (2n + 1), for n = 0,1,2, ...,
Note that n is an integer that takes values 0,1,2, and so no.
Step 2: Create four consecutive odd integers.
Multiplying n by 2 guarantees that 2n will be zero or an even number.
Therefore (2n + 1) is guaranteed to be an odd number.
By adding 2 to the odd integer (2n+1), the next number (2n+3) will also be an odd integer.
Let the four consecutive odd integers be
2n+1, 2n +3, 2n +5, 2n +7
Step 3: require that the four consecutive integers sum to 160.
Because the sum of the four consecutive odd integers is 160, therefore
2n+1 + 2n+3 + 2n+5 + 2n +7 = 160
8n + 16 = 160
8n = 144
n = 18
Because 2n = 36, the four consecutive odd integers are 37, 39, 41, 43.
Answer: 37,39,41,43
20/100 * X = 8
X = 8*100/20
X=40
