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2 years ago

The total cost of Anfä's trip to the dentist was $628.35. She paid a flat fee of $89.95 which included the checkup:

Mathematics

1 answer:

daser333 [38]2 years ago

5 0

Answer:

The cost of each cavity filling was $ 134.60.

Step-by-step explanation:

Given that the total cost of Anfa's trip to the dentist was $ 628.35, and she paid a flat fee of $ 89.95 which included the checkup: cleaning and then had 4 cavities filled, each of which cost the same amount, to determine which shows the correct equation and value of x, the cost of each cavity filling, the following calculation must be performed:

(628.35 - 89.95) / 4 = X

538.4 / 4 = X

134.6 = X

Therefore, the cost of each cavity filling was $ 134.60.

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PLEASE HELP FAST!! WILL MARK BRAINLIEST!!!!

Aleonysh [2.5K]

In 4 days Bryn sells 24 tickets (50-26)
So in 1 day he sells 6 tickets --> B is right
The +33 in the equation means, that Eddie sold 33 tickets oon the first day
--> So C.) and D.) are right
When you replace the x in the equation with 1, you'll get the tickets, that eddie sold on the first day. --> y = 5 *(1)+33 = 38
38 tickets - 33tickets = 5 tickets ( the rise)
-->A.) is wrong, because Bryn sold two tickets more as Eddie per day

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3 years ago

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Please Help! Thanks to whoever can answer this!

Illusion [34]

Here, we are missing the slope and the y intercept.

Lets look for the y intercept first.

Where does it cross the y axis? At 3!!

3 is the y intercept.

Now, find the slope by using the slope formula.

$m = \frac{y2 - y1}{x2 - x1}$
(0,3) and (2,1)

m = 1-3/2-0 = -2/2 = -2/2 = -1
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Okay, so we found our slope, now just write the equation.

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3 years ago

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.

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