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3 years ago

An airport has two long term parking lots. the cost to park,y, in each lot for x days is shown in the tables

Mathematics

1 answer:

Licemer1 [7]3 years ago

8 0

Answer:

y = 4x + 12 will be the other equation.

Step-by-step explanation:

Data given in the tables show a linear relation (has a common data).

To get the linear relation, we will choose the two points from table (1) .

Let the points are (1, 16) and (2, 20).

Slope of the line 'm' = $\frac{\triangle{y}}{\triangle{x}}$

m = $\frac{20-16}{2-1}$

m = $\frac{4}{1}$

m = 4

Equation of the line passing through (1, 16) having slope = 4

y - 16 = 4(x - 1)

y = 4(x - 1) + 16

y = 4x - 4 + 16

y = 4x + 12

Now we take second set of data,

We choose two points (1, 6) and (2, 12).

Slope 'm' = $\frac{\triangle{y}}{\triangle{x}}$

m = $\frac{12-6}{2-1}$

m = 6

Equation of the line passing through (1, 6) having slope = 6

y - 6 = 6(x - 1)

y = 6x - 6 + 6

y = 6x

Therefore, other equation of the system of equations will be,

y = 4x + 12

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3 years ago

Question (c)! How do I know that t^5-10t^3+5t=0?<br> Thanks!

astra-53 [7]

(a) By DeMoivre's theorem, we have

$(\cos\theta+i\sin\theta)^5=\cos5\theta+i\sin5\theta$

On the LHS, expanding yields

$\cos^5\theta+5i\cos^4\theta\sin\theta-10\cos^3\theta\sin^2\theta-10i\cos^2\theta\sin^3\theta+5\cos\theta\sin^4\theta+i\sin^4\theta$

Matching up real and imaginary parts, we have for (i) and (ii),

$\cos5\theta=\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta$
$\sin5\theta=5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta$

(b) By the definition of the tangent function,

$\tan5\theta=\dfrac{\sin5\theta}{\cos5\theta}$
$=\dfrac{5\cos^4\theta\sin\theta-10\cos^2\theta\sin^3\theta+\sin^5\theta}{\cos^5\theta-10\cos^3\theta\sin^2\theta+5\cos\theta\sin^4\theta}$

$=\dfrac{5\tan\theta-10\tan^3\theta+\tan^5\theta}{1-10\tan^2\theta+5\tan^4\theta}$
$=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

(c) Setting $\theta=\dfrac\pi5$ , we have $t=\tan\dfrac\pi5$ and $\tan5\left(\dfrac\pi5\right)=\tan\pi=0$ . So

$0=\dfrac{t^5-10t^3+5t}{5t^4-10t^2+1}$

At the given value of $t$ , the denominator is a non-zero number, so only the numerator can contribute to this reducing to 0.

$0=t^5-10t^3+5t\implies0=t^4-10t^2+5$

Remember, this is saying that

$0=\tan^4\dfrac\pi5-10\tan^2\dfrac\pi5+5$

If we replace $\tan^2\dfrac\pi5$ with a variable $x$ , then the above means $\tan^2\dfrac\pi5$ is a root to the quadratic equation,

$x^2-10x+5=0$

Also, if $\theta=\dfrac{2\pi}5$ , then $t=\tan\dfrac{2\pi}5$ and $\tan5\left(\dfrac{2\pi}5\right)=\tan2\pi=0$ . So by a similar argument as above, we deduce that $\tan^2\dfrac{2\pi}5$ is also a root to the quadratic equation above.

(d) We know both roots to the quadratic above. The fundamental theorem of algebra lets us write

$x^2-10x+5=\left(x-\tan^2\dfrac\pi5\right)\left(x-\tan^2\dfrac{2\pi}5\right)$

Expand the RHS and match up terms of the same power. In particular, the constant terms satisfy

$5=\tan^2\dfrac\pi5\tan^2\dfrac{2\pi}5\implies\tan\dfrac\pi5\tan\dfrac{2\pi}5=\pm\sqrt5$

But $\tanx>0$ for all $0$ , as is the case for $x=\dfrac\pi5$ and $x=\dfrac{2\pi}5$ , so we choose the positive root.

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