All Answers 357331 Answers

Let f(x) = In(3x)+2.5 

what is the average year rate of change of f(x) from 2 to 7 

round your answer to the nearest hundredth.

Let f(x) = x^2 -4

what is the average rate of change f(x) from -5 to -2

C. The distance plane A traveled is 1 h greater than the distance plane B traveled in 1 h

Total mechanical energy:  E = (1/2)Mv² + (1/2)kx²  where M is mass, v is velocity

At 18cm, assuming we start with zero velocity,  E = (1/2)*1.1*10³*(0.18)^2  Joules

(a) Let xa be the value of x when the PE is 15% of E:   (1/2)k*xa² = 0.15*(1/2)k*(0.18)^2  or xa = sqrt(0.15)*0.18 m

(b) Let va be the velocity at this point.   (1/2)*4*va^2 + (1/2)*1.1*10³*xa² = (1/2)*1.1*10³*(0.18)^2   solve for va (in m/s)

I'll leave you to do the number crunching.

to find the 5th term of the expression t=4xn you take the number 5 and replaces it with n. so you get t = 4x5. so you multiply it and find out what t is.

Your welcome. 

I also got this question on prodigy. Super cool game btw.

One of these triangles is the image of the other triangle after dilation.

What are the coordinates of the center of dilation?

What is the scale factor?

Let t be a root of

$f(x) = x^3 - x + 2$.

Evaluate

$t^6 - t^2 + 4t$.

1.

$\begin{array}{|rcll|} \hline f(t) = t^3 - t + 2 &=& 0 \\ t^3 - t + 2 &=& 0 \\ \mathbf{t^3} &\mathbf{=}&\mathbf{t-2} \\ \hline \end{array}$

2.

$\begin{array}{|rcll|} \hline t^6 - t^2 + 4t &=& (t^3)^2 - t^2 + 4t \quad | \quad \mathbf{t^3=t-2} \\ &=& (t-2)^2 - t^2 + 4t \\ &=& t^2-4t + 4 - t^2 + 4t \\ \mathbf{t^6 - t^2 + 4t} &\mathbf{=}&\mathbf{4} \\ \hline \end{array}$

From smaller to larger, the scale factor is 4. Just take the ratio of the sides, (which you can find with the Pythagorean Theorem) and that fraction will give you the scale factor. I don't know the official way to do it, but go to a graphing website, plot these points, then draw lines connecting all similar points, A to A', B to B' and so on. Where all the lines intersect is the dilation point. So, the dilation point is (7,-1).

For function f, construct and simplify the difference quotient

f(x) = (x-4)/(x+3)

$\overline{\Delta f} = \dfrac{f(7)-f(2)}{7-2}$

Let $\mathbf{A}$ be a matrix, and let $\mathbf{x}$ and $\mathbf{y}$ be linearly independent vectors such that
$\mathbf{A} \mathbf{x} = \mathbf{y}, \mathbf{A} \mathbf{y} = \mathbf{x} + 2\mathbf{y}$
Then we have that
$\mathbf{A}^{-1} \mathbf{x} = a \mathbf{x} + b\mathbf{y}$
or some scalars a and b.

Find the ordered pair (a,b).

$\begin{array}{|rcll|} \hline Ax &=& y \quad | \quad \cdot A^{-1} \\ A^{-1}Ax &=& A^{-1}y \quad | \quad A^{-1}A = I \\ x &=& A^{-1}y \\ \mathbf{A^{-1}y} &\mathbf{=}&\mathbf{x} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline Ay &=& x + 2y \quad | \quad \cdot A^{-1} \\ A^{-1}Ay &=& A^{-1}x + 2A^{-1}y \quad | \quad A^{-1}A = I \\ y &=& A^{-1}x + 2A^{-1}y \quad | \quad \mathbf{A^{-1}y=x} \\ y &=& A^{-1}x + 2x \\ \mathbf{A^{-1}x} &\mathbf{=}&\mathbf{-2x+y } \\ \hline \end{array}$

$\mathbf{(a,b) = (-2,1)}$

$f(x)=((x+3)^2)^{50}-4x+3=(x+3)^{100}-4x+3$

$(r_i+3)^{100}-4r_i+3=0 (i=1, 2, ...,100)$

$(r_i+3)^{100}=4r_i-3$

Therefore, $(r_1+3)^{100}+(r_2+3)^{100}+...+(r_{100}+3)^{100}=(4r_1-3)+(4r_2-3)+...+(4r_{100}-3)$

$=4(r_1+r_2+...+r_{100})-300$

$=4*(-300)-300=-1500$

(By Vieta's formulas, $-(r_1+r_2+...+r_{100})$ equals the coefficient of $x^{99}$ of $f(x)$, which is $nCr(100, 1)*3^1=300$ by the binomial theorem.

Therefore, $r_1+r_2+...+r_{100}=-300$.)

b. -7

Compute

$\mathbf{\large{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}}}$

$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n^3-n} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n^2-1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n-1)(n+1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{(n-1)n(n+1)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)}\cdot \dfrac{1}{(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}} \\\\ &=& \sum \limits_{n = 1}^{99} \left(\dfrac{1}{n} - \dfrac{1}{n+1} \right) \cdot \dfrac{1}{(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+2)} - \dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+2)} = \dfrac{1}{2}\left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{(n+1)(n+2)}= \dfrac{1}{n+1} - \dfrac{1}{n+2} } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right)\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{2(n+2)} - \dfrac{1}{n+1} + \dfrac{1}{n+2}\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{n+2} \left(1 - \dfrac{1}{2} \right) \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \sum \limits_{n = 1}^{99} \dfrac{1}{n+1} + \sum \limits_{n = 1}^{99} \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n} - \sum \limits_{n = 1}^{99} \dfrac{1}{n+1} + \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n+2} \\\\ && \boxed{\dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 3}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} } \\ && \boxed{\sum \limits_{n = 1}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 2}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 1}^{98} \dfrac{1}{n+2} \\ = \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} } \\ && \boxed{ \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n+2} \\ = \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} } \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \left( \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} \right) \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \dfrac12 - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \dfrac{1}{100} \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}\\ && +\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}+ \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\\\ &\mathbf{=}& \dfrac{5049}{20200} \\\\ &\mathbf{=}& \mathbf{0.24995049505} \\ \hline \end{array}$

1. y=x, (3, -6)

2. 180 degrees counterclockwise, (6, -3)

3. 7 units to the right, 7 units down, (1, -4)

Sorry, I misunderstood the question.

If I remember the story right, then you could do: The inner evils, The unseen demons, More than just a box. I'm not very good at the imagination part of writing, this is just what I came up with off the top of my head

This is good, but you used the formula for a cone, not a cylinder. 

Cylinder - Sphere = Remaining volume.

Cylinder = pi r^2 * h = pi 900 * 100 = 90,000 pi

Sphere = (pi r^3 * 4)/3 = (pi 1728 * 4)/3 = 2304pi

Remaining volume = 87696 pi = 275,365.44 cm^3

If it's gold, then the mineral's structure is not cubic

Never worked with contrapositives before, but I think this is right

First, note that by Vieta's formulas, r^2 + s^2 will be a polynomial in m. Since m is real, we know r^2 + s^2 will be real as well. So, the problem is simply asking us to find the minimum value of r^2 + s^2.

By Vieta's formulas, we have

$r+s=2m \quad\text{and}\quad rs=m^2+2m+3$
Therefore, 

$\begin{align*} r^2+s^2 &= (r+s)^2 - 2(rs) \\ &= (2m)^2 - 2(m^2+2m+3) \\ &= 2m^2-4m-6 \\ &= 2(m^2-2m-3). \end{align*}$
Completing the square gives

$r^2+s^2 = 2\left((m-1)^2-4\right).$
This is minimized when m=1, and the minimum value is $\boxed{-8}$.

I am editing this because Lagtho correctly pointed out I used a cone instead of a cylinder  !!

cone volume - sphere volume = water volume

CYLINDER volume - sphere volume = water volume

pi r^2 h  - 4/3 pi r^3 = water vol     (this assumes the water can squeeze past the sphere to fill the tip of the cone)

pi 30^2  100   - 4/3 pi 12^3 = water vol

= 275365  cm^3