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Natasha2012 [34]
1 year ago
11

Find the domain of the following functions and simplify their expressions.

Mathematics
1 answer:
erica [24]1 year ago
7 0

The domain of the function is  (-∞,∞) and the simplified expression is x²-6x-40 .

A function's domain and range are its constituent parts. A function's  range is its potential output, whereas its domain is the set of all possible input values. Range, Domain, and Function. A is the domain and B is the co-domain if a function f: A B exists that maps every element of A to an element in B. 'b', where (a,b) R, provides the representation of an element 'a' under a relation R. The set of images is the function's range.

The given functions are

f(x)=6x+24

g(x)=x²-16

Now we have to find (g-f)(x) .

(g-f)(x)=x²-16-6x-24

or,(g-f)(x)=x²-6x-40

or,(g-f)(x)=x²-10x+4x-40

or,(g-f)(x)=x(x-10)+4(x-10)

or,(g-f)(x)=(x-10)(x+4)

So the domain of the function (g-f)(x) are the values for which the function exists. we can see that the function exists for all values of x.

Domain in set builder notation={x|x∈R}

Domain in interval notation=(-∞,∞)

To learn more about domain and range:

brainly.com/question/28135761

#SPJ9

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Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

\sf\longrightarrow sin^2\theta + cos^2\theta = 1

Imagine a right angled triangle with one of its acute angle as \theta .

  • The side opposite to this angle will be perpendicular .
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\sf\longrightarrow sin\theta =\dfrac{p}{h} \\

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And by Pythagoras theorem ,

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Where the symbols have their usual meaning.

Now , taking LHS ,

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\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\

\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\

\sf\longrightarrow \dfrac{p^2+b^2}{h^2}

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\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\

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Since LHS = RHS ,

Hence Proved !

I hope this helps.

5 0
2 years ago
Answer please and thank you
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Answer:

b

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Step-by-step explanation:

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