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Bogdan [553]
3 years ago
11

If f(x) = 6x +5 and g(x) = 2x – 7 find (fºg)(x)

Mathematics
1 answer:
adell [148]3 years ago
7 0

Answer:

f(g(x)) = 12x - 37

Step-by-step explanation:

Step 1: Define

f(x) = 6x + 5

g(x) = 2x - 7

Step 2: Find f(g(x))

f(g(x)) = 6(2x - 7) + 5

f(g(x)) = 12x - 42 + 5

f(g(x)) = 12x - 37

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Step-by-step explanation:3,360 is your answer

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3 years ago
Read 2 more answers
(1) Let {v1,v2,v3} be a set of vectors in Rn . If u is Span {v1,v2,v3}, show that 3u is in Span {v1,v2,v3}.
Evgesh-ka [11]

Answer:

(1)

Multiplying by 3 both sides of the equality you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

3u  is in the Span of the vectors \{v_1,v_2,v_3\}.

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

Step-by-step explanation:

(1)

As the hint indicates, you know that

u = c_1 v_1 + c_2v_2+c_3v_3

Then, if you multiply both sides of the equality by 3, you get that

3u = 3c_1v_1+3c_2v_2+3c_3v_3

And that's it. 3u  is in the Span of the vectors \{v_1,v_2,v_3\}

(2)

That's not true, consider the following counter example.

v_1 = (0,0,0,1)\\v_2 = (0,0,1,0)\\v_3 = (0,1,0,0)\\v_4 = (1,0,0,0)\\u = (0,1,1,1)

u is a linear combination of v_1,v_2,v_3 but is NOT a linear combination of v_1,v_2,v_3,v_4.

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3 years ago
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about 50 students maybe a few less

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2 years ago
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_nC_k=\dfrac{n!}{k!(n-k)!}\\\\\\_{34}C_{34}=\dfrac{34!}{34!0!}=1

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2 years ago
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irina1246 [14]

Answer:

(a) It is a function.

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c) What is the range?

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