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Vikentia [17]
3 years ago
11

Can someone please help me?

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

Answer:

a

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=f%20%3D%28x%29%20%3D%20%5Cfrac%7Bx%7D%7B4%7D%20" id="TexFormula1" title="f =(x) = \frac{x}{4}
Inga [223]

Answer:

g(x) = \frac{x+3}{x-3}

Step-by-step explanation:

From the picture attached,

Given function is,

f(x) = x + 1

We have to find the value of g(x) if the composite function has been given as,

f[g(x)] = g(x) + 1 = \frac{2x}{(x-3)}

g(x) = \frac{2x}{(x-3)}-1

      = \frac{2x-(x-3)}{(x-3)}

      = \frac{(x+3)}{(x-3)}

Therefore, g(x) = \frac{x+3}{x-3} will be the answer.

5 0
2 years ago
What is the following sum? 2(^3√16x^3y)+4(^3√54x^6y^5)
victus00 [196]
Option A: 4x(^3√2y)+12x^2y(^3√2y^2)
6 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
There is two questions
Mkey [24]

Answer:

1 is the 4th answer

2 is 1.875

Step-by-step explanation:

1/2*3 3/4=1.875

4 0
2 years ago
Read 2 more answers
Two friends leave their houses at the same time and run across town towards each other. The first friend runs at a speed of 8 mi
GalinKa [24]
2 and a half hours estimate
5 0
3 years ago
Read 2 more answers
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