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V125BC [204]
3 years ago
7

If f(x)=x+7 and g(x)= 1 divided by x-13, what is the domain of (f•g)(x)?

Mathematics
1 answer:
andrew11 [14]3 years ago
5 0
<h2>Hello!</h2>

The answer is:

The domain of the function is all the real numbers except the number 13:

Domain: (-∞,13)∪(13,∞)

<h2>Why?</h2>

This is a composite function problem. To solve it, we need to remember how to composite a function. Composing a function consists of evaluating a function into another function.

Composite function is equal to:

f(g(x))=(f\circ} g)(x)

So, the given functions are:

f(x)=x+7\\\\g(x)=\frac{1}{x-13}

Then, composing the functions, we have:

f(g(x))=\frac{1}{x-13}+7\\

Therefore, we must remember that the domain are all those possible inputs where the function can exists, most of the functions can exists along the real numbers with no rectrictions, however, for this case, there is a restriction that must be applied to the resultant composite function.

If we evaluate "x" equal to 13, the denominator will tend to 0, and create an indetermination since there is no result in the real numbers for a real number divided by 0.

So, the domain of the function is all the real numbers except the number 13:

Domain: (-∞,13)∪(13,∞)

Have a nice day!

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Kay [80]

Answer:

1) d) Square

2) Proofs that PWRS is a rhombus are

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Step-by-step explanation:

The given points (x, y) of the parallelogram are;

E(-2, -4), F(0, -1), G(-3, 1), H(-5, -2)

The slope, m, of the segments are found as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

By computation, the slope of segment EF = 1.5

The slope of segment FG = -0.67

The slope of segment GH = 1.5

The slope of segment HE = -0.67

Therefore, EF is parallel to GH and FG is parallel to HE

The length of the sides are;

\sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

By computation, the length of segment EF = 3.61

The length of segment FG = 3.61

The length of segment GH = 3.61

The length of segment HE = 3.61

The diagonals are;

EG and FH

The length of segment EG = 5.099

The length of segment FH = 5.099

Therefore, the diagonals are equal and the parallelogram is a square

2) The given dimensions are;

P(-1, 3), Q(-2, 5), R(0, 4), S(1, 2)

A rhombus has all sides equal

The length of segment PQ = 2.24

The length of segment QR = 2.24

The length of segment RS = 2.24

The length of segment PS = 2.24

The diagonals are;

QS and PR

The length of segment QS = 4.24

The length of segment PR = 1.41

The slope of segment QR = -0.5

The slope of segment PS = -0.5

The slope of segment RS = -2

The slope of segment QP = -2

Therefore, QS≠QR the parallelogram is a rhombus

The correct option ;

Length of QS ≠ PR and

Slope of segment QR and PS is -1/2 and Slope of segment RS and QP is -2.

Where there are acute angles in parallelogram PQRS, then the correct option is d) Length of QR and PS is 2.2 and Length of RS and QP is 2.2

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Answer:

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Step-by-step explanation:

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The degree of the polynomial, 8y give. and the it is a monomial.

<h3>What is a Polynomial?</h3>

A polynomial is an algebraic expression which may be of varying degrees or number of terms.

According to the question;

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Additionally, since there's only one term in the polynomial, we can conclude it is a monomial.

Read more on polynomials;

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