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If you are going to call Heureka out you need justify yourself.

https://web2.0calc.com/questions/physics_95

Please do not post questions more than once. 

The 'net ' forces acting on her are

f = ma

  = 57 (9.81 + 1.25) = 11.06*57 = 630.42 N  Downward

the answer is 235

i just took the test

sadasdasd

its too dry 

How about this website?

\(F_n = mg\), so let's find the mass

We know the force, \(F\), and the acceleration, \(a\), so we can use \(F = ma\)

\(F = ma \rightarrow 1050 = m(0.785) \rightarrow m = 1337.58 kg\)

Now that we know the mass, we can go back to \(F_n\)

\(F_n = mg \rightarrow F_n = (1337.58)(9.8) \rightarrow F_n = 13108.3 N\)

(x + 1)²   =   (x + 1)(x + 1)

Multiply each item in the first set of parenthesees by each item in the second set of parenthesees.

(x + 1)(x + 1)   =   (x)(x) + (x)(1) + (1)(x) + (1)(1)

Simplify.

(x)(x) + (x)(1) + (1)(x) + (1)(1)   =   x² + 2x + 1

So....

(x + 1)²   =   x² + 2x + 1

I'm a lil confused. Are there no bounds to confine the area to? Otherwise, due to the nature of this function, the area under the graph but above the x-axis is \(\infty\).

Ima reformat da question right quick nawmsayn:

Find the positive value of \(n\) such that the equation \(9x^2+nx+1=0\) has exactly one solution in \(x\).

When working with quadratic equations and the question mentions something about a number of solutions, we definitely want to use the discriminant \(b^2-4ac\).

Let's review what the output of the discriminant tells you about the solutions to a quadratic equation.

If \(b^2-4ac > 0\), there are two solutions.

If \(b^2-4ac = 0\), there is one solution.

If \(b^2-4ac < 0\), there are no solutions.

Since we want only one solution, we want \(b^2-4ac = 0\). Where currently \(a = 9, b = n, c = 1\)

\(b^2 - 4ac = 0 \rightarrow n^2 - 4(9)(1) = 0\rightarrow n^2 = 36 \rightarrow n = \pm 6\)

The question asks for the positive value of \(n\), therefore \(n = 6\)

\(f(x)=\sum_{k=0}^\infty \frac{f^k(1)}{k!}(x-1)^k\)

With these problems, I like to change everything to 3D coordinates to make it easier for me to understand.

Currently, the room is a rectangular prism with important vertices of \((0,0,0)\) and \((12,10,8)\).

Now we should find out the gecko's current position. It shouldn't matter what side wall is chosen for the lil guy to be on, because the answer will be the same in this case, so we can choose the side wall closest to the origin, with an x-value of 0. Also depending on what you choose to be the front and back wall, this will affect the coordinates. I have illustrated how I chose such. Definitely not to scale.

In this sense, the Gecko's coordinates are \((0,0+1,8-1) = (0,1,7)\) and the fly's are \((12,10-1,0+1) = (12,9,1)\). With the tricky bit out of the way, we can now just use the 3D distance formula.

\(D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\), or in our case \(D = \sqrt{(x_F-x_G)^2+(y_F-y_G)^2+(z_F-z_G)^2}\)

\(D = \sqrt{(12-0)^2+(9-1)^2+(1-7)^2} = \sqrt{144+64+36} = \sqrt{244} = 2\sqrt{61}\)

Go online to this link and it should answer your question:     http://math2.org/math/expansion/log.htm

The two correct answers for the first question is: (The question is biased toward a No response.) & (The sample is not biased.) 

Hope this helps. 

\(\ \phantom{=\qquad}3x^2+bx+10\\~\\ =\qquad3(x^2+\frac{b}{3}x)+10\\~\\ =\qquad3(x^2+\frac{b}{3}x+(\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3((x+\frac{b}{6})^2-(\frac{b}{6})^2)+10\\~\\ =\qquad3(x+\frac{b}{6})^2-3(\frac{b}{6})^2+10\)

Now it is in the form  a(x + m)² + n   where

\(a=3\qquad \text{and}\qquad m=\frac{b}{6}\qquad \text{and}\qquad n=-3(\frac{b}{6})^2+10\)

In order for  m  to be an integer,  b  must be a multiple of  6

In order for  n  to be an integer,  b  must be a multiple of  6

The largest integer that must be a divisor of  b  in order for both  m  and  n  to be integers is  6

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How many ordered triplets \((a,b,c) \) of rational numbers are there where\( a,b,c\) are the roots of \(x^3 + ax^2 + bx + c = 0\) ?

I assume:

\(\begin{array}{|rcll|} \hline x^3 + ax^2 + bx + c = 0 &=& (x-a)(x-b)(x-c) \\ &=& x^3\underbrace{-(a+b+c)}_{=a}x^2\underbrace{+(ab+ac+bc)}_{=b}x\underbrace{-abc}_{=c} \\ \hline \mathbf{-abc} &=& \mathbf{c} \\ -ab &=& 1 \\ \mathbf{ab} &=& \mathbf{-1} \\ \mathbf{b} &=& \mathbf{-\dfrac{1}{a}} \\\\ \mathbf{-(a+b+c)} &=&\mathbf{ a } \\ -a-b-c &=& a \\ b+c &=& -2a \quad | \quad \cdot a \\ \mathbf{ab+ac} &=& \mathbf{-2a^2} \quad | \quad ab=-1 \\ -1+ac &=& -2a^2 \\ \mathbf{ ac }&=& \mathbf{1-2a^2} \\\\ \mathbf{ab+ac+bc} &=& \mathbf{b} \quad | \quad ab+ac= -2a^2 \\ -2a^2+bc &=& b \\ -2a^2 &=& b- bc \\ -2a^2 &=& b(1-c) \quad | \quad b=-\dfrac{1}{a} \\ -2a^2 &=&-\dfrac{1}{a} (1-c) \\ 2a^2 &=& \dfrac{1}{a} (1-c) \\ 2a^3 &=& 1-c \\ \mathbf{c} &=& \mathbf{1-2a^3} \\\\ \mathbf{ ac }&=& \mathbf{1-2a^2} \quad | \quad c=1-2a^3 \\ a(1-2a^3)&=& 1-2a^2 \\ -2a^4+a &=& 1-2a^2 \\ \mathbf{-2a^4+2a^2+a-1} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|lcll|} \hline \mathbf{-2a^4+2a^2+a-1=0},\ b=-\dfrac{1}{a},\ c=1-2a^3 \\\\ a = 1,\ b = -1,\ c = -1 \\ \text{triplet}_1 ~(a,b,c) = (1,\ -1,\ -1 ) \\\\ a =0.56519771738363939644,\ b=-1.7692923542386314152,\ c=0.6388969194713526224 \\ \text{triplet}_2 ~(a,b,c) = (0.56519771738363939644,\ -1.7692923542386314152,\ 0.6388969194713526224 ) \\ \hline \end{array} \)

check:

\(\begin{array}{|lcll|} \hline \mathbf{x^3+ x^2-x-1=0} \\ x = -1 \\ x = 1 \\ \hline \mathbf{x^3+ 0.56519771738363939644x^2-1.7692923542386314152x+0.638896919471352622=0} \\ x=-1.76929235423863142 \\ x=0.5651977173836394 \\ x=0.6388969194713526 \\ \hline \end{array}\)