5 + 6 − 1 2 − 1 1 -11y 5+6y−12−11y − 7 + 6 − 1 1
− 7 + 6 − 1 1 −7+6y−11y − 7 − 5
− 7 − 5 −7−5y − 5 − 7
Solution: −5y−7
Answer:
Step-by-step explanation:
Not exactly sure what your question is - I am assuming that it is something like:
Show/prove that for any integer x, x^2 - x is even.
Suppose that x is an even integer. The product of an even integer and any other integer is always even (x = 2n, so x * y = 2 n * y which is even. Therefore x^2 is even. An even minus an even is even. (The definition of an even number is that it is divisible by 2 or has a factor of 2. So the difference of even numbers could be written as 2*( the difference of the two numbers divided by 2); therefore the difference is even)
Suppose that x is an odd integer. The product of 2 odd numbers is odd - each odd number can be written as the sum of an even number and 1; multiplying the even parts with each other and 1 will produce even; multiplying the 1's will produce 1, so the product can be written as the sum of an even number and 1 - which is an odd number. The difference between two odd numbers is even - the difference between the even parts is even (argument above), the difference between 1 and 1 is zero, so the result of the difference is even.
x^2 is therefore even if x is even and odd if x is odd; The difference x^2 - x is even by the arguments above.
Answer:
Step-by-step explanation:
Step 1: 6(x^2-1) [(6x-1)/(6(x+1)]
Step 2: [6(X+1) (X-1) (6X-1)] / 6(X+1)
Explanation: Factor out the (x^2-1)
Step 3: (6X-1) (X-1)
Explanation: Cancel out common terms on the top and the bottom of the fraction.
Step 4: 6X^2 - 6X-X+1
Explanation: Multiply out and Simplify
Answer: 6X^2 - 7x+1
Answer:
Step-by-step explanation:
The number is already defined as ' h '. "Twice" means times by 2. So, 13 and twice 'h' are being added together. Here 13 and h are to be added, and the answer is doubled.
