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german
4 years ago
14

A see-saw is 25 feet long with a fulcrum in the middle of the board. If a 60 lb. child sits three feet from the fulcrum, what is

the lowest weight that will lift the child?
Mathematics
1 answer:
My name is Ann [436]4 years ago
4 0

Answer:

14.4 lb

Step-by-step explanation:

In a see-saw in equilibrium, the torque generated by one side needs to be the same generated in the other side. The torque is calculated by the product between the mass and the distance to the center of the see-saw.

The torque generated by the child is:

T1 = 60 * 3 = 180 lb*feet

So, the torque generated by the weight needs to be higher than T1 in order to lift the child.

The lowest mass is calculated when the mass is in the maximum distance, that is, 12.5 feet from the center.

So, we have that:

T2 = 180 = mass * 12.5

mass = 180/12.5 = 14.4 lb

So the lowest weight is 14.4 lb

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How do you find the domain and range of a function and write it in set notation? What is the process of finding it? Tips and tri
DerKrebs [107]

Answer:

Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis

Step-by-step explanation:

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7 0
3 years ago
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Consider a sequence whose first three
snow_tiger [21]

Answer:

Below, depends if 27 is term number 1 or term number 0.  Answered for both cases.  

Step-by-step explanation:

The most common sequences are arithmetic and geometric, so lets check those first.

Arithmetic first since its the easiest.

to go from 27 to 21 we subtract 6, if we subtract 6 from 21 again we get to 15, which is what we need, so it is indeed arithmetic.

Explicit formula is basically of the form of y=mx+b with an arithmetic sequence.  the m is the common difference and b is the first term minus the common difference.  so lets fill those in.  y = -6x + 33

Then it usually has n as the x and y f(n) so we'll just put those in

f(n) = -6n + 33

This si as long as the first term is labeled as term number 1 and not term number 0.  if you have 27 as term 0 instead just make 33 back to 27, so f(n) = -6n + 27

Let me know if this doesn't make sense.

5 0
3 years ago
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Over [174]

Answer:

15.25×3

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=40.75

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3 years ago
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BabaBlast [244]

Answer:

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3 years ago
Consider the function ​f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0
dybincka [34]

I suppose you mean

f(x)=\cos(x^2)

Recall that

\cos x=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}

which converges everywhere. Then by substitution,

\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}

(starting at n=1 because the summand is 0 when n=0)

b. Naturally, the differentiated series represents

f'(x)=-2x\sin(x^2)

To see this, recalling the series for \sin x, we know

\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}

Multiplying by -2x gives

-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}

and from here,

-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}

-2x\sin(x^2)=\displaystyle 4x\sum_{n=0}^\infty(-1)^n\frac{nx^{4n}}{(2n)!}=f'(x)

c. This series also converges everywhere. By the ratio test, the series converges if

\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}

The limit is 0, so any choice of x satisfies the convergence condition.

3 0
4 years ago
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