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Greeley [361]
3 years ago
8

Michael puts 3 beads next to each other on a pice of string to make a bracelet. Each bead has a width of 5/6 inch. Which express

ion and fraction both represent the total width of the three beads?
Mathematics
1 answer:
pentagon [3]3 years ago
8 0

Answer:

3 (5/6)

 5/2

Step-by-step explanation:

Hi, to answer this question we have to write an expression with the information given.

The total width of the three beads  must be equal to the number of beads (3) multiplied by the length of each bead (5/6)

Mathematically speaking:

3 (5/6)

Solving:

5/2

Feel free to ask for more if needed or if you did not understand something.

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The area of the given figure is 576 cm sq.

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Step-by-step explanation:

The shape consists of one semicircle and one square which is cut by a semicircle.

<u>To find Area:</u>

Find the area of semicircle= (πr^2)/2

where r is the radius= diameter/2= 24/2= 12 cm

Area of semicircle= π(12)(12)/2= 144π/2= 72π cm sq.

Find the area of square= a^2

where a is the side of square, a=24 cm.

Area of the square= 24*24= 576 cm sq.

To find the total area of the given figure,

Area of the figure= Area of semicircle+(area of square- area of semicircle)

                             = 72π+576-72π =576 cm sq.

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Perimeter of semicircle= (2πr)/2

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A rock thrown vertically upward from the surface of the moon at a velocity of 36​m/sec reaches a height of s = 36t - 0.8 t^2 met
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Answer:

a. The rock's velocity is v(t)=36-1.6t \:{(m/s)}  and the acceleration is a(t)=-1.6  \:{(m/s^2)}

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

  • Velocity is defined as the rate of change of position or the rate of displacement. v(t)=\frac{ds}{dt}
  • Acceleration is defined as the rate of change of velocity. a(t)=\frac{dv}{dt}

a.

The rock's velocity is the derivative of the height function s(t) = 36t - 0.8 t^2

v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t

The rock's acceleration is the derivative of the velocity function v(t)=36-1.6t

a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6

b. The rock will reach its highest point when the velocity becomes zero.

v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = \frac{405}{2} =202.5 \:m

To find the time it reach half its maximum height, we need to solve

36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0

For a quadratic equation of the form ax^2+bx+c=0 the solutions are

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

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The rock is aloft for 45 seconds.

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