Explain why the initial value of any function of the form f (x) = a(b') is equal to a.

Mathematics

1 answer:

tankabanditka [31]2 years ago

3 0

Answer:

THE ANSWER ON E2020

Step-by-step explanation: When you substitute 0 for the exponent x, the expression simplifies to a times 1, which is just a. This is because any number to the 0 power equals 1. Since the initial value is the value of the function for an input of 0, the initial value for any function of this form will always be the value of a.

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2 years ago

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3 years ago

If PQRS is a rhombus, which statements must be true? Check all that apply.

bezimeni [28]

These statements are true:

$\bold{First.} \ \overline{PS} \ is \ parallel \ to \ \overline{QR}$

This is true because of the definition of a rhombus which states that opposite sides are parallel.

$Then \ \overline{PS} \ is \ opposite \ to \ \overline{QR}$

$\bold{Second.} \ \overline{PR} \ is \ perpendicular \ to \ \overline{QS}$

This is true because the diagonals of a rhombus bisect each other at right angles.

$\bold{Third.} \ \overline{PQ}=\overline{RS}$

This is true because the four sides of a rhombus are all equals.

$\bold{Fourth.} \ \angle PQR \ is \ supplementary \ to \ \angle QPS$

This is also true. Two angles are supplementary when they add up to 180 degrees. Angles in rhombus are equal two to two. Let's name:

$\angle PQR = \alpha \ and \ \angle QPS = \beta$

Then it is true that:

$\angle PQR=\angle PSR \ and \ \angle QPS=\angle QRS$

Besides, a rhombus is a quadrilateral and it is true that the interior angles of a quadrilateral add up to 360 degrees, thus:

$2\alpha+2\beta=360 \\ \\ \therefore \boxed{\alpha+\beta = 180^\circ}$

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