Music Read aloud 1) Sound effects

Mathematics

2 answers:

zvonat [6]3 years ago

4 0

Answer:

Step-by-step explanation:

What am I supposed to do? If you don’t mind me asking. Like I’m not trying to waste your points at all just I’m confused I can def help tho if you tell me what I need to do

Marysya12 [62]3 years ago

4 0

Answer:

Huh-?

Step-by-step explanation:

You might be interested in

1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff

Fantom [35]

Answer:

a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

c) True

d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

c) This is true, if f were surjective, then for every element of B there should exist an element a in A such that f(a) = b. This means that f−¹({b}) has positive cardinal for each element b from B. since f⁻¹(b) ∩ f⁻¹(b') = ∅ for different elements b and b' (because an element of A cant return two different values with f). Therefore, each element of B can be assigned to a subset of A (f⁻¹(b)), with cardinal at least 1, this means that |B| ≤ |A|, and as a consequence, B is finite.

b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

4 0

4 years ago

Write a polynomial Of least degree With roots zero and -3

GuDViN [60]

Answer:

X^2+3x

Step-by-step explanation:

in order to get these roots, we would set up the equation to be (x)(x+3). We can then multiply it to get x^2+3x

3 0

3 years ago

While constructing an equilateral triangle or a regular hexagon inscribed in a circle, you may have noticed that several smaller

Sauron [17]

Answer:

Using the transitive property of geometry, we proved that △PQR is an equilateral triangle.

Step-by-step explanation:

The diagram of the given scenario is in the attachment.

Here, PQ is the radius of circle Q

Also, PQ is the radius of circle P

Hence, circle-P and circle-Q have same radii.

Now, PQ=RQ as both are radii of circle Q

And PQ=PR, as both are radii of circle P

Hence as per transitive property, which states that, if any two angles, lines, or shapes are congruent to a third angle, line, or shape respectively, then the first two angles, lines, or shapes are also congruent to the third angle, line, or shape

PR=QR

Now, PQ = QR = PR

Hence, ∆PQR is equilateral.

For more explanation, refer the following link:

brainly.com/question/27884420

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