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andrey2020 [161]
2 years ago
5

What is the difference between a relation and function? Classify each of the following as a function, or not a function. State t

he domain and range
{(1,7),(1,14),(1,21)}

The relation is/isn't a function
Domain _ _ _
Range _ _ _
Leave Unused Fields Blank
Mathematics
1 answer:
Levart [38]2 years ago
5 0

Answer:

A relation is a subset of cartesian product of two non empty sets whereas A function is a type of relation in which every element of first set has one and only image in the second set.

In a relation an element of the first set can have many images in the second set whereas in a function the first element can have only one image in the second set.

The given relation is not a function as the element 1 is related to 3 different elements in the second set.

Domain={1}

Range={7,14,21}

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What is the factorization of the polynomial below?
katrin [286]

Answer:

A. (2x + 5)(2x - 5)

Step-by-step explanation:

Factor out the polynomial given.

(4x² - 25) = (2x - 5)(2x + 5)

Check: Use the FOIL method.

(2x)(2x) = 4x²

(2x)(5) = 10x

(2x)(-5) = -10x

(-5)(5) = -25

Simplify. Combine like terms: 4x² + 10x - 10x - 25 = 4x² - 25

A. (2x + 5)(2x - 5) is your answer

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5 0
3 years ago
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Can someone please explain to me how to solve this?
lyudmila [28]
I hope this helps you



x= -2


h (-2)= 3. (-2)+1


h (-2)=-6+1


h (-2)= -5
4 0
3 years ago
Use the properties of logarithms to expand the following expression as much as possible. Simplify any numerical expressions that
sweet [91]

Answer:

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

Step-by-step explanation:

Let be f(x,y) = \log_{4}(2\cdot x^{2}-20\cdot x +12\cdot y), this expression is simplified by algebraic and trascendental means. As first step, the second order polynomial is simplified. Its roots are determined by the Quadratic Formula, that is to say:

r_{1,2} = \frac{20\pm \sqrt{(-20)^{2}-4\cdot (2)\cdot (12\cdot y)}}{2\cdot (2)}

r_{1,2} = 5\pm \sqrt{25-6\cdot y}

The polynomial in factorized form is:

(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})

The function can be rewritten and simplified as follows:

f(x,y) = \log_{4} [(x-5-\sqrt{25-6\cdot y})\cdot (x-5+\sqrt{25-6\cdot y})]

f(x,y) = \log_{4} (x-5-\sqrt{25-6\cdot y})+\log_{4} (x-5+\sqrt{25-6\cdot y})

3 0
3 years ago
Solve the equation with a variable term on both sides: 5.1y+21.3=-0.3y-24.6.
ivann1987 [24]

To solve this equation, we will need to isolate y, the variable. These are the steps that follow:


1. Add 0.3y to each side.

2. Subtract 21.3 from each side.

3. Divide both sides by 5.4y.

Try and figure it out yourself! If you have trouble...


The answer is y = -8.5

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ANSWER IT FOR BRIANLESTT AND POINTS
kvasek [131]

Answer:

Top right square. It's mirroring the image on the other side of the y axis. So Quadrant 2

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