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Mila [183]
4 years ago
15

How do I quickly calculate cube roots? (Lord it's been forever since I last been on here.)

Mathematics
1 answer:
Montano1993 [528]4 years ago
4 0

Using a calculator is the quickest method.

Memorizing or just knowing what the first few are.

Then there's prime factorization method. Prime factor and then find 3 groups.

Example:

Square root of 1,728

864 * 2

432 * 2 * 2

216 * 2 * 2 * 2

108 * 2 * 2 * 2 * 2

54 * 2 * 2 * 2 * 2 * 2

27 * 2 * 2 * 2 * 2 * 2 * 2

9 * 3 * 2 * 2 * 2 * 2 * 2 * 2

3 * 3 * 3 * 2 * 2 * 2 * 2 * 2 * 2

Make 3 groups of the same of primes...

(3 * 2 * 2) * (3 * 2 * 2) * (3 * 2 * 2)  

12 * 12 * 12 = 1,728

12³ = 1,728








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What is 0.14x57.05 and what are the steps
liq [111]

Answer:

7.987

Step-by-step explanation:

  1 4

×     5 7 0 5

+         7 0

+ 0 0 0 0 0  

+     9 8    

+   7 0      

=  7 9 8 7 0

Rewrite the product with 4 total decimal places.

Answer = 7.9870

Therefore:

0.14 × 57.05 = 7.9870

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What is the measure of angle z in this figure? Enter your answer in the box. z = ° Two intersection lines. All four angles forme
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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
Verdich [7]

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

36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45

The rock is aloft for 45 seconds.

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When we round this to the nearest whole number:
A =100 \pi = 314.159=314ft^{2}
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