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Using I = Prt , we have

I = 250(.072)(5) = $90

.05(1664+x)=x       simplify

83.2 + .05x = x      subtract .-05x from both sides

83.2 = .95x            divide both sides by .95

83.2/.95 = x = 87.5789473684210526

√97 = 9.8488578017961047

Use the following to find the principal, (P)

500,000 = P(1 + .12/2)^2*10 =

500,000 = P(1.06)²⁰      divide both sides by (1.06)²⁰

500,000/[1.06²⁰]  = P = about $155,902.36

Well he still has 3.  One of them is just in his stomach.

0 because the calculator is broken.

0 because the calculator is broken.

0 because he was so hungry that he ate all three apples and went to the store to buy more and he ate those apples as well. He WAS REALLY!!!!!!!!!!!!!!!!!!!!!!!!!!!! hungry.

If a function is continuous at "a,"  then, these three conditions have to be met......

Notice that the second and third are met, but not the first one.....as geno has said, this function does not exist at x = 2, so it isn't continuous.......(note that we can see this just from the way the function is defined in the problem itself !!!   x < 2 and x > 2  means that we don't know what happens at 2....)

N

NN

The question is:  Why is it not continuous at x = 2?

You gave the answer to that when you said that it is not defined at that location.

The limit, approaching from both the left and the right is 4; thus, it has a limit at x = 2, even though it is not defined at that location; even though it is not defined at that location. Indeed, there is a 'hole' in the graph at that location, which makes is discontinuous, even though it has a limit at that point.

My understanding is a little different:

A linearization gives approximate y-values near a given point on a graph, while Newton's method is used to find a zero of a function by repeated approximations, starting at a given point.

A linearization will give an approximate y-value anywhere on a function while Newton's method is used to find an x-value that gives a y-value of zero. 

If something is "10 feet shorter than y" ....we can write this as "y - 10"

And if y -10 = 8, just "solve" for y!!!

Does that help ???

You can write

width = height - 2

Rearrange this to get height =, then replace width by 2.

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if you have n values: x1, x2, ... xn, the geometric mean is the n'th root of the product of all the values.  That is:

geometric mean = ⁿ√(x1*x2*....*xn)

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$$\\a^2\times a=a^3\\
and\\
a^2*1=a^2\\
so\\
$The the Highest common factor is $a^2$$

your question is incomplete.   

If mass density is ρ then you need to find ∫ρ.dV  where dV is infinitesimal volume and the integral goes from zero to infinity.

dV = 4pi.s².ds, so

 $$mass=\int_0^{\infty}\frac{4\pi s^2 }{s(1+s)^3}ds$$ 

or

$$mass = 2\pi$$

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The number 81 is 3 raised to which power :  3^4 = 81

$$\small{\text{
\begin{array}{rcl}
3^x &=& 81 \quad | \quad \ln{()} \\\\
\ln{(3^x)} &=& \ln{(81)} \\\\
x*\ln{(3)} &=& \ln{(81)} \\\\
x &=& \frac{ \ln{(81)} } {\ln{(3)} } \\\\
x &=& 4
\end{array}
}}$$

$$\boxed{3^4 = 81}$$

Do you mean quadrature?

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Look at the bottom two pictures. Imagine the bolt starts to drop just as the floor of the elevator is at the same level as the observer's feet  By the time the bolt hits the floor of the elevator, the floor will have moved upwards.  

Imagine the observer has a large vertical ruler 3m long.  He positions the top of the ruler at the height of the top of the elevator, when the elevator floor is at his feet.  The bottom of the ruler, level with the floor of the elevator at the instant the bolt starts to drop, would be 3m below the top.  Because the elevator is rising, by the time the bolt hits the elevator floor, the floor, and hence the bolt, will be at a level above the bottom of the ruler (which isn't in the elevator and so doesn't move).  This means the ruler will measure a distance less than 3m for the drop.

The [2nd] button is in the top left corner of the calc.

45 with  4 zeros

450000

In pard d .. ehy the distance is not 3 m .. the tall of the elevator ? 

Yes, the linear approximation they are talking about isn't the bisection method I've illustrated above (Newton's method is the same).  

 Their linear method uses a tangent line, as does Newton's method, but it's just used to find an approximation to the value of the function at some "ugly" number (where "ugly" means it is not easy to calculate the function at that point) by using a nearby "non-ugly" number!  I suspect this is a very old-fashioned method, where calculations were done either in your head, or with a slide rule, or with log tables.  In these days of calculators and computers, there aren't really any such "ugly" numbers! 

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That makes good sense.  Thanks Alan.

I still haven't looked at the article properly but it seems to be talking about something different.