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2 years ago

70 points!!!! Find the horizontal and the vertical components of f.

Mathematics

1 answer:

choli [55]2 years ago

6 0

Answer:

F = (16.74, -2.95)

Explanation:

1) Horizontal component

The horizontal component is related to the magnitude of the force by the cosine ratio:

cos (α) = Fx / | F |

Where α in the angle (-10º), Fx is the horizontal component, and | F | is the magnitude of the force (17).

cos(-10º) = Fx / 17 ⇒ Fx = cos (-10º) × 17 = 16.74

2) Vertical component

The vertical component is related to the magnitude of the force by the sine ratio:

sin (α) = Fy / | F |

Where α in the angle (-10º), Fy is the vertical component, and | F | is the magnitude of the force (17).

sin(-10º) = Fy / 17 ⇒ Fy = sin (-10º) × 17 = -2.95

3) Form (x,y)

F = (16.74, -2.95)

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

3 0

2 years ago

a gardener is putting a wire fence along the edge of his garden to keep animals from eating his plants. if he has 20meters of fe

Alexeev081 [22]

Answer:

The largest rectangular space is $24m^{2}$

Step-by-step explanation:

Given a 20m long wire fence, implies that the perimeter of the fence has been given, because Perimeter can be defined as the length of the boundary of a figure (in this case, a rectangle)

So, from the formula: Perimeter = 2 (L + B) = $20m^{2}$

We have that L + B = $10m^{2}$

The possible values of L and B are:

(L, B) = (9, 1), (8, 2), (7, 3) and (6, 4).

The areas of the rectangle that will be formed using these values are respectively; $9m^{2}$ , $16m^{2}$ , $21m^{2}$ and $24m^{2}$ , the largest area of which is the last which corresponds to the values (6, 4).