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3 years ago

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Arielle wants to make a circular garden. She has enough fencing for a garden circumference of 15 feet. There is a 7-foot gap bet

ween her house and the shed where she would like to put the garden. Does she have enough room for the garden? (Hint: Compare the diameter of the garden to the 7-foot gap.)

1 answer:

lesantik [10]3 years ago

3 0

Answer:

yes

Step-by-step explanation:

We need to check if the circle with a circumference of 15 feet is smaller then 7 feet. If so it possible.

Formule for circular circumference = pi * diameter, which should be 15.

So pi * d =15

There fore d = 15/pi

d = 4.77464829275686 feet, which is smaller then 7 feet, so it will fit easilly.

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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:

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

$y_{max} = 84.375\,m$

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Find the midpoint of the segment between (6,7) and (6,-5)

Norma-Jean [14]

Answer:

(6,1)

Step-by-step explanation:

midpoint = ((x2+x1),(y2+y1)/2)

= (6+6)/2, (-5+7)/2

= 12/2, 2/2

= 6,1

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Answer:

(a) 3 ft, 4 ft, 5 ft

Step-by-step explanation:

The triangle inequality requires the sum of the two short sides exceed the long side.

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a) 3 ft + 4 ft = 7 ft > 5 ft . . . . . triangle is possible

b) 4 yd + 1 2/3 yd = 5 2/3 yd < 10 yd . . . . not a possible triangle

c) 3 ft + 3 ft = 6 ft < 7 ft . . . . not a possible triangle

d) 4 in + 4 in = 8 in . . . . not a possible triangle (not greater than 8 in)

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The first set of side lengths can form a triangle: 3 ft, 4 ft, 5 ft.

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Answer:

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Step-by-step explanation:

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side = 1/2 *3*6 = 9 sq. in.

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Answer is A

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