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3 years ago

A manufacturer of rectangular tarpaulins uses the

Mathematics

1 answer:

brilliants [131]3 years ago

5 0

Answer:

f(x) = x + 7 and g(x) = x + 3

Step-by-step explanation:

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On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

The circumference of a circular garden is 147.58 feet What is the diameter of the garden? Use 3.14 for and do not round your ans

Stella [2.4K]

Answer:

The diameter is 47

Step-by-step explanation:

Given

$C = 147.58$ --- Circumference

$\pi = 3.14$

Required

Determine the diameter (d)

The circumference of a circle is:

$C = \pi d$

Make d the subject

$d = \frac{C}{\pi}$

$d = \frac{147.58}{3.14}$

$d = 47$

7 0

2 years ago

4. Find x if PQ = RS,<br> PQ = 9x - 7, and RS = 29.

Gnoma [55]

Answer:

x=4

Step-by-step explanation:

Because we know that PQ = RS, we can use the transitive property to replace PQ in the first equation with 29:

9x-7=29

1) Add 7 to both sides:

9x=36

2) divide by 9 on both sides:

x=4

4 0

3 years ago

Find the surface area of the box with dimensions 125mm×64mm×84mm to the nearest mm?

Contact [7]

Answer:

Step-by-step explanation:

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5 0

2 years ago

The area of a parallelogram is 420cm^2 and the height is 35cm. Find the corresponding base.

bearhunter [10]

The answer is 12cm because area (420) divided bye height (35) gives you the corresponding side 12cm.

4 0

2 years ago