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german
3 years ago
12

20 + 27x + 9x^2 = (?) (3x + 5)

Mathematics
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

(3x + 4)

Step-by-step explanation:

9x²+27x+20 can be factored out to (3x + 4)(3x + 5)

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Help me please I have 10 minutes to do this
Olin [163]

Answer:

Distributive

Step-by-step explanation:

The 4y is distributed to the terms in parentheses.

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3 years ago
Which of the following could be the ratio between the lengths of the two legs
skad [1K]

9514 1404 393

Answer:

  B, E

Step-by-step explanation:

The ratios of side lengths in a 30-60-90 triangle are ...

  1 : √3 : 2

The two legs are the shorter sides, so any ratio that reduces to 1 : √3 is an appropriate choice:

  B.  1 : √3

  E.  2√3 : 6

3 0
2 years ago
HELP ME ASAP PLEASE IM REALLY STRESSED :c
sveta [45]

Answer:

0.3 miles.

Step-by-step explanation:

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5 0
3 years ago
Read 2 more answers
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
How do you do it<br> .Thank you
GrogVix [38]
An improper fraction is when the numerator is greater. For example, 1 7/8 = 15/8
5 0
3 years ago
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