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2x8+2-2/2

16+2-2/2

16+2-1

16-1

17

5=20-5x   |+5x-5

5x=15      | :5

  x=3

A Titan IV rocket has put your spacecraft in a circular orbit around Earth at an altitude of 390 km. Calculate the force due to gravitational attraction between the Earth and the spacecraft in N if the mass of the spacecraft is 2050 kg.

Find the x-coordinates of the points where the line y=5x-1 meets the curve y=2x^3+x^2+1. I'm a bit confused here, I don't know if I'm suppose to plug anything in or not.

Given a function in factored form is identified as y= a(x-3)(x+4) and that the coordinates ( 1, -10) is located on the graph; state the equation in vertex form. 

1.1392172e+18

0.6464... (repeating) as a Fraction?

\(\begin{array}{|rcll|} \hline 1\cdot a &=& 0.6464... \text{ (repeating)} \\ 100\cdot a &=& 64.6464... \text{ (repeating)}) \\\\ 100\cdot a - 1\cdot a &=& 64.6464... \text{ (repeating)} - 0.6464... \text{ (repeating)}\\ 99\cdot a &=& 64 \\ a &=& \frac{64}{99} \\ \mathbf{ 0.6464... \text{ (repeating)} } &\mathbf{=}& \mathbf{ \frac{64}{99} }\\ \hline \end{array}\)

can someone please help me find the derivative of this:
\(f(x)=\sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}}\)

\(\begin{array}{|rcllll|} \hline f(x) &=& \sqrt[3]{x}+\frac{2}{{x}^{2016}}+\frac{1}{\sqrt{x^3}} \\ && & \sqrt[3]{x} = x^{\frac13} \\ && & \frac{2}{{x}^{2016}} = 2 \cdot x^{-2016} \\ && & \frac{1}{\sqrt{x^3}} = \frac{ 1 } { x^{\frac32} } = x^{-\frac32} \\ f(x) &=& x^{\frac13} + 2 \cdot x^{-2016} + x^{-\frac32} \\ && & \text{Formula: } \\ && & y = x^n \\ && & y' = n\cdot x^{n-1} \\\\ f'(x) &=& \frac13 \cdot x^{\frac13-1} + 2 \cdot (-2016) \cdot x^{-2016-1} + (-\frac32) x^{-\frac32-1} \\ f'(x) &=& \frac13 \cdot x^{-\frac23 } -4032 \cdot x^{-2017} -\frac32\cdot x^{-\frac52} \\ && & x^{-\frac23 } = \frac{ 1 } { x^{\frac23} } = \frac{1}{\sqrt[3]{x^2}} \\ && & x^{-2017} = \frac{1}{{x}^{2017}} \\ && & x^{-\frac52 } = \frac{ 1 } { x^{\frac52} } = \frac{1}{\sqrt{x^5}}\\ f'(x) &=& \frac13 \cdot \frac{1}{\sqrt[3]{x^2}} -4032 \cdot \frac{1}{{x}^{2017}} -\frac32\cdot \frac{1}{\sqrt{x^5}} \\ \mathbf{f'(x)} & \mathbf{=} & \mathbf{\frac{1}{3 \sqrt[3]{x^2}} - \frac{4032}{{x}^{2017}} - \frac{3}{2 \sqrt{x^5}} }\\ \hline \end{array}\)

Your answer is the same and also correct.

1500 = 2^2 * 3 * 5^3

2625 = 3 * 5^3 * 7

Thank you! 

You can go online and watch this video which explains it very well:

https://www.khanacademy.org/math/pre-algebra/pre-algebra-factors-multiples/pre-algebra-divisibility-tests/v/the-why-of-the-3-divisibility-rule

a) No, clearly the starting points are different for each support: (0,0) is not ((2,1), etc.

b) No, similar to a) answer: (2,1) is not (6,3) etc

c) No, 2 does not equal 4, etc.

d) No, 1 does not equal 2, etc.

e) Yes, this is just the slope of the diagonal line: 1/2 = 2/4

I'm sorry but I am not quite understanding the question. 

If the triangles are not the same size then only e will be ' YES' because the gradient of the roof does not change.

lol

I have worked out the answer so when you post the question as a member I will give it to you. I am tired of posting providing answers for non members and not even knowing if the answer has been seen. Some people are very rude with their lack of response/appreciation.

if you divide 9/11 you get 0.8181818181818182, but if you round it the answer would be 0.8

I'm not sure if this is the right answer, but this is how I attempted to solve it.

So...

Displacement = [(V1 + V2) / 2]*change in T

D = [(119.722+0)/2]*10

D = [(119.722)/2]*10

D = (59.861)*10

D = 598.61

Therefore the distance travelled is 598.61m.

Hope this helps!

If you post this as a member then I'll take a look.  :)

Help please! 

You just know.

Its phychic power

1) Find the surface area of Mars using: SA=4 x pi x r^2, r=radius + 30 meters.

2) Find the surface area of 1 asteroid

3) divide the answer in 1 by the answer in 2 =number of asteroids to cover Mars with water, 30 meters deep.

4) Divide the answer in 3 by 1 billion years to get the average time between impacts.

YEEEEEEEEEEEEEEEEEEE!!!!! First question. Noice