Answer:
A.The answer four square root of two plus ten is irrational because the sum of a rational number and irrational number is an irrational number.
Step-by-step explanation:
Given the question as 4√2 + 5√4 the sum is
4√2 + 5*√4
4√2 + 5*2
4√2 + 10
Putting an irrational number in decimal form, it goes on forever without repeating.
4*√2 +10 = 4× 1.41421356237 +10
=15.6568542495----------------which irrational
Answer:
............................
Step-by-step explanation:
Answer:
x = -2
y = -1
(-2, -1)
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
<u>Algebra I</u>
- Solving systems of equations using substitution/elimination
- Solving systems of equations by graphing
Step-by-step explanation:
<u>Step 1: Define systems</u>
y = x + 1
3x + 3y = -9
<u>Step 2: Solve for </u><em><u>x</u></em>
<em>Substitution</em>
- Substitute in <em>y</em>: 3x + 3(x + 1) = -9
- Distribute 3: 3x + 3x + 3 = -9
- Combine like terms: 6x + 3 = -9
- Isolate <em>x</em> term: 6x = -12
- Isolate <em>x</em>: x = -2
<u>Step 3: Solve for </u><em><u>y</u></em>
- Define original equation: y = x + 1
- Substitute in <em>x</em>: y = -2 + 1
- Add: y = -1
<u>Step 4: Graph systems</u>
<em>Check the solution set.</em>
Hello there!
This is a conceptual question about quadratic functions.
Remember that a solution of ANY function is where it intersects the x-axis, so if the quadratic function intersects the x-axis TWO times, this means that there are TWO real solutions.
Here's a list of things to remember that will help you out for quadratic functions...
•if a quadratic function intersects the x-axis twice, it has two real solutions.
•if a quadratic function intersects the x-axis once, it has one real solution and one imaginary solution.
•if a quadratic function intersects the x-axis zero times, it has zero deal solutions and two imaginary solutions.
Please NOTE: If you want to know how many solutions a polynomial function has, look at it's highest exponent. If it is 2, then it has 2 solutions whether they be real or imaginary. If it is 3, then it has 3 solutions.
Also, if one of the factors are the same for a polynomial function, the way it hits the x-axis changes! This is just some extra information to help you in the long run!
I hope this helps!
Best wishes :)
Answer and Step-by-step explanation:
(a) Given that x and y is even, we want to prove that xy is also even.
For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.
(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:
Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.
(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.
