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Aneli [31]
3 years ago
5

34. 6a + 11 if a = 8 59

Mathematics
2 answers:
skelet666 [1.2K]3 years ago
8 0

Answer:

6a+11= 59

Step-by-step explanation:

8x6= 48

48+11= 59

sergiy2304 [10]3 years ago
4 0

Answer:

Evaluate for a = 859

( 6 ) ( 859 ) + 11

( 6 ) ( 859 ) + 11

= 5165

Hope it helps

Please mark me as the brainliest

Thank you

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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
3 years ago
Wich expression represent the sum of 59 and x
KIM [24]
I don't see the answers, but it would most likely be 59 + x.
7 0
3 years ago
Need help on number 2 plz I appreciate it thank you so much
vladimir1956 [14]

Answer:

\angle PQR, \angle SQR, and \angle PQS

Or

angle PQR, angle SQR and angle PQS

Step-by-step explanation:

The three different angles in the diagram are angle PQR, angle SQR and angle PQS.

Another way of writing this is using an angle sign before the alphabets follows. Thus:

\angle PQR, \angle SQR, and \angle PQS

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3 years ago
-7 a rational or irrational number
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Wetting the cloth cover of your canteen on a hot day is a good or bad idea
Lyrx [107]
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