Prove that 6(n+6)-(2n+3) is odd number for all n£N

Mathematics

1 answer:

Mice21 [21]2 years ago

4 0

Answer:

$\begin{aligned}6(n+6)-(2n+3) & = 6n+36-2n-3\\& = 4n+33\\& = 4n + 32 + 1\\& = 2(2n+16) + 1\end{aligned}$

Any number multiplied by 2 is even, therefore 2(2n + 16) is even.

If 2(2n + 16) is even, then adding 1 to this makes it odd.

Thus proving that the given expression is an odd number for all $n \in \mathbb{N}$

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The function h(x) = x2 + 14x + 41 represents a parabola.

defon

Part A:

$h(x)= x^{2} +14x+41$

The first step of completing the square is writing the expression $x^{2} +14x$ as $(x+7)^{2}$ which expands to $x^{2} +14x+49$ .

We have the first two terms exactly the same with the function we start with: $x^{2}$ and $14x$ but we need to add/subtract from the last term, 49, to obtain 41.

So the second step is to subtract -8 from the expression $x^{2} +14x+49$

The function in completing the square form is
$h(x)= (x+7)^{2}-8$

Part B:

The vertex is obtained by equating the expression in the bracket from part A to zero

$x+7=0$
$x=-7$

It means the curve has a turning point at x = -7

This vertex is a minimum since the function will make a U-shape.
A quadratic function $a x^{2} +bx+c$ can either make U-shape or ∩-shape depends on the value of the constant $a$ that goes with $x^{2}$ . When $a$ is (+), the curve is U-shape. When $a$ (-), the curve is ∩-shape

Part C:

The symmetry line of the curve will pass through the vertex, hence the symmetry line is $x=-7$

This function is shown in the diagram below