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ss7ja [257]
3 years ago
6

Prove that there is a positive integer that equals the sum of the positive integers not exceeding it. Is your proof constructive

or nonconstructive?
Mathematics
1 answer:
Elena L [17]3 years ago
8 0

Answer:

Constructive Proof

Step-by-step explanation:

Let x be a positive integer

x must be equal to sum of all positive integers exceeding it

i.e.

x = x + (x - 1) + ( x - 2) + ......... + 2 + 1

Equivalently,

x = ∑i (where i = 1 to x)

The property finite sum;

∑i (i = 1 to x) = x(x + 1)/2

So,

x = x(x + 1)/2 ------- Multiply both sides by 2

2 * x = 2 * x(x + 1)/2

2x = x(x + 1)

2x = x² + x ------- subtract 2x from both sides

2x - 2x = x² + x - 2x

0 = x² + x - 2x ----- Rearrange

x² + x - 2x = 0

x² - x = 0 ------ Factorise

x(x - 1) = 0

So,

x = 0 or x - 1 = 0

x = 0 or x = 1 + 0

x = 0 or x = 1

But x ≠ 0

So, x = 1

The statement is only true for x = 1

This makes sense because 1 is the only positive integer not exceeding 1

1 = 1

It is a Constructive Proof

A proof is constructive when we find an element for which the statement is true.

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2, 6, 18, 54. 162,...
grigory [225]

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Answer:

  \text{c. }f(n)=3f(n-1);f(1)=2

Step-by-step explanation:

The first term is 2, so part of the recursive definition is ...

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The common ratio is 6/2 = 3, so each term is 3 times the previous one. That part of the recursive definition is ...

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These two parts of the definition match choice C.

3 0
3 years ago
Simplify the expression below, I really just need the steps I have the answer (also can someone tell me how to edit points on th
Sloan [31]

Answer:  3x^2y\sqrt[3]{y}\\\\

Work Shown:

\sqrt[3]{27x^{6}y^{4}}\\\\\sqrt[3]{3^3x^{3+3}y^{3+1}}\\\\\sqrt[3]{3^3x^{3}*x^{3}*y^{3}*y^{1}}\\\\\sqrt[3]{3^3x^{2*3}*y^{3}*y}\\\\\sqrt[3]{\left(3x^2y\right)^3*y}\\\\\sqrt[3]{\left(3x^2y\right)^3}*\sqrt[3]{y}\\\\3x^2y\sqrt[3]{y}\\\\

Explanation:

As the steps above show, the goal is to factor the expression under the root in terms of pulling out cubed terms. That way when we apply the cube root to them, the exponents cancel. We cannot factor the y term completely, so we have a bit of leftovers.

4 0
3 years ago
Given that f(x)=6x+2 and g(x)=2x+4/5 solve for g(f(1))<br><br> A.1<br> B.3<br> C.4<br> D.8
GrogVix [38]

Answer:

The value of g(f(1)) is 84/5

Step-by-step explanation:

To find the answer to a composite function, start with the function on the inside, which is f(1). So, we input 1 into the f(x) equation and evaluate.

f(x) = 6x + 2

f(1) = 6(1) + 2

f(1) = 6 + 2

f(1) = 8

Now that we have the answer of 8, we can input that in for x in the outside function, which is g(x).

g(x) = 2x + 4/5

g(8) = 2(8) + 4/5

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6 0
3 years ago
20x+6=y<br><br> 19x+7=y<br><br> answer please
morpeh [17]

Step-by-step explanation:

what we trying to find the value of x and y

20x+6=y

x intercept,y=0

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×=-0.3

y intercept,x=0

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(-0.3,0)

19x+7=y

y intercept,x =0

19(0)+7=y

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7=y

x intercept,y=0

19x+7=0-7

19x=-7

19x/19=-7/19

x=-0.37

4 0
2 years ago
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