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If you mean all combinations taken 1 at a time plus those taken 2 at a time plus ... plus those taken 10 at a time, then:

$$\sum_{k=1}^{10}nCr(10,k)=1023$$

(I've assumed "between" is inclusive of 1 and 10).

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How about 24 in 20 classes and 25 in 8 classes?

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sin^-1 is the same as asin.  Press 2nd followed by asin

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x = √[64 - 25] = √39

So  cos  = √39 / 64   and tan  = 5/√39

The  csc, sec and cot will be the reciprocals of the sin, cos and tan, respectively

Since the starting point is (-2, 9), the slope is -3/4, and t increases as x increases,  we have....

x(t) = -2 + 3t

y(t)  = 9 - 4t

Notice that when t = 1, x = 1 and y = 5

And the distance between the ponts is  √[(-2 -1)^2 + (9 - 5)^2] = √[3^2 + 4^2] = √25......which means that some object traveling on this line starting at (-2,9) and moving at this speed (per some time unit) will reach (1, 5)  after one of these time units.....just what we want....!!!

1250 revs  every 5 minutes = 250 revs every minute

x(t)  = 6.5cos[(-500pi)t]  

y(t)  = 6.5sin [(-500pi) t ]

x = e^2t means ln(x) = 2t

y = e^8t means ln(y) = 8t

Therefore ln(y) = 4*ln(x)  or ln(y) = ln(x⁴)  or y = x⁴

Alternatively you could write directly:  y = e^8t = (e^2t)⁴ = x⁴

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x(t) = 7√t   →    √t = x /7   → t = x^2 / 49

y(t) = 5t + 2         and substituting, we have   

y(x)  =  (5/49)x^2 + 2  .

Notice that we must restrict the domain of this function to  x ≥ 0....the reason for this is that x(t) in the original parametric equation isn't defined for  t < 0

The volume is  base area (8² square inches) multiplied by the height (12").

For surface area you have twice the base area (a square of side 8") plus the area of four rectangles each with width 8" and height 12". 

You can do the number crunching.

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You can do this by integrating twice with respect to t:

acceleration: dv/dt = (10/3)t³ - 30t² + 60t

velocity: v = ds/dt = (5/6)t⁴ - 10t³ + 30t² + constant  where s is distance.  If we assume the velocity is zero when t = 0, then we just have:

ds/dt = (5/6)t⁴ - 10t³ + 30t² 

Integrate this:

distance: s = t⁵/6 - (5/2)t⁴ + 10t³ + constant.  If we assume distance is zero when t = 0, then

distance: s = t⁵/6 - (5/2)t⁴ + 10t³

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x(t)  = 5 - t   →    t  = 5 - x  

y(t)  =  2 - 4t    so we have, by substitution...

y(x)  = 2 - 4(5 - x)   →   y = 2 - 20 + 4x    →   y = 4x - 18

150 milligrams is 150*10^-3 grams, so you have:

$$\frac{150\times10^{-3}}{75\times10^8}\rightarrow 2\times10^{-11}$$  grams per cell

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In general:  z^a/z^b = z^a-b so you will be left with:

-z^12-3 - 4z^9-3 - 4z^5-3 → -z⁹ - 4z⁶ -4z²

You could also write this as  -z²(z⁷ + 4z⁴ + 4)

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I'm assuming you mean that when t changes by 1, y changes by 1...if so, we have....

x(t) = 1      notice that this is constant  because we have a verical line

y(t)  = t + 10

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x(t)  = 1.5cos[300pi(t)]

y(t)  = 1.5sin [300pi(t)]

Edit....I made a slight mistake here.... the particle starts at (0, 1.5)....so.....the correct equations are actually

x(t)  = 1.5sin(-300pi/t)  

y(t) = 1.5cos(300pi/t)

   7343₈

+   646₈

----------

  10211₈   or  just 10211 without the subscript

x(t)  = 6.5cos[(-pi/5)t]   

y(t) = 6.5sin[(-pi/5)t]

A verticle asymptote occurs where a value of x makes the denominator equal to zero but not the numerator:

In this case the only value of x that makes the denominator = 0 is x=-1/2 and this value also doesn't make the numerator = 0 therefore the vertical asymptote is at x=-1/2.

What if the function was this however: $$y(x)=\frac{(x^2+7x+10)(2x+1)}{2x+1}$$

If you dont cancel the 2x+1 and substitute in x=-1/2 you will get 0 for both the numerator and the denominator therefore this wouldnt be  a vertical asymptote. The way to go about this situation is to cancel the 2x+1 to get $$y(x) = x^2+7x+10$$ but you Have to mention that x cannot equal -1/2 as this isn't defined for the original function and is a discontinuity and then you can leave it at that.

Saul from Khan Academy Explains this nicely: https://www.khanacademy.org/math/algebra2/rational-expressions/rational-function-graphing/v/finding-asymptotes-example

   14A6₁₂

     -  5B9₁₂

     ----------

       AA9₁₂ or just AA9 without the subscript

  101₂

+  11₂

-------

  1000₂

+1100₂

---------

  10100₂

+11101₂

----------

  110001₂   or just 110001 without the subscript

2.  ... I think this is   x/10= (x+1)/(2x-2) - (1)/(x-1) =

And getting a common denominator on the right, we have

x/10= (x+1)/(2(x-1)) - (2)/(2(x-1))   =  

x/10  = (x -1)/(2(x- 1))     cross multiply

x(2(x-1) = 10(x - 1)    simplify

2x^2 - 2x  = 10x - 10   simplify, again

2x^2 - 12x + 10 = 0      divide everything by 2

x^2 - 6x + 5 = 0     factor

(x - 5) ( x - 1)  = 0      and setting both factors to 0, we have that x = 5 or x = 1

We have to reject the second answer since it makes a denominator in the original problem = 0