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liubo4ka [24]
3 years ago
15

PLEASE HELP THANK YOU!!!!

Mathematics
1 answer:
insens350 [35]3 years ago
3 0
//// For G(x) = -1/4x - 1////
G(-5) = -1/4(-5) - 1 = 0.25;
G(-2) = -1/4(-2) - 1 = -0.25;
G(-1) = -1/4(-1) - 1 = -0.75;
///////////////////////////////////

////For G(x) = (x-1)^2 - 3////
G(-1) = ((-1) -1)^2 - 3 = 1;
G(-2) = ((-2) - 1)^2 - 3 = 6;
G(-5) = ((-5) - 1)^2 - 3 = 33;
//////////////////////////////////////////

////For G(x) = 1/4x + 2////////
G(-5) = 1/4(-5) + 2 = 0.75;
G(-2) = 1/4(-2) + 2 = 1.5;
G(-1) = 1/4(-1) + 2 = 1.75;

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3 years ago
A) Write the sequence of natural numbers which are multiplied by 3 ?
Volgvan

Answer:

a) 3, 6, 9, 12, 15,...,3\cdot n, b) 4, 7, 10, 13, 16,...,3\cdot n +1, c) Both sequences are arithmetic.

Step-by-step explanation:

a) The sequence of natural numbers which are multiplied by 3 are represented by the function f(n) = 3\cdot n, n\in \mathbb{N}. Let see the first five elements of the sequence: 3, 6, 9, 12, 15,...

b) The sequence of natural numbers which are multiplied by 3 and added to 1 is represented by the function f(n) = 3\cdot n + 1, n\in \mathbb{N}. Let see the first five elements of the sequence: 4, 7, 10, 13, 16,...

c) Both sequences since differences between consecutive elements is constant. Let prove this statement:

(i) f(n) = 3\cdot n

\Delta f = f(n+1) -f(n)

\Delta f = 3\cdot (n+1) -3\cdot n

\Delta f = 3

(ii) f(n) = 3\cdot n +1

\Delta f = f(n+1)-f(n)

\Delta f = [3\cdot (n+1)+1]-(3\cdot n+1)

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8 0
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What is the value of i 97 – i?<br> –i<br> 0<br> –2i<br> i Superscript 96
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<em>The question is not clearly readable, but I'm assuming the expression which makes more sense, so you can have a clue</em>

Answer:

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Step-by-step explanation:

<u>Powers of The Imaginary Unit</u>

The imaginary unit i is defined as

i=\sqrt{-1}

The first powers of i are

i^0=1

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i^3=i.i^2=-i

i^4=i^2.i^2=1

And so on the cycle repeats every four numbers. To find the value of i^{96} we can find the remainder of 96/4=0. So i^{96}=i^0=1

The given expression is

i^{97}-i

Factoring

i(i^{96}-1)

Since i^{96}=1

=i(1-1)=0

The required value is 0

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