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Either press the "sec" button (3rd row, 3rd column) or type in "sec(   )".

you divide decimals just like you would with normal numbers!

except one thing is different:

you have to move the decimal place over in the divisor until it's a whole number, and then do the same number of spaces you moved over in the dividend.

I hope this helped! :)

Yep...I definitely like your approach better, geno....mine is only approximate.....

121m^2-144   .....I assume you might want to factor this??

It's just the difference of two squares.

Take the square roots of both terms......and write them twice. Then separate the first pair with a plus and the second pair with a minus......and it's factored !!! So we have

(11m +  12 )     ( 11m  -  12 )  ...........voila !!!!

A tangent is a line that connects with another function at a certain point without touching anywhere else in that proximity except that point. 

Well, the warning came 49 min after 8:45PM and it rained 1 hour more after that. So....the strom lasted 1hr and 49 min....in other words, it was over at 10:34PM.....11 minutes shy of 2 hours.....!!!

Wow, quinn.....that's an impressive proof.....I don't think I've seen it before.....good job!!!

Yes yours would be a reasonable approximation.    :)

You can set up a ratio:  120 mi / 8 in  =  x mi / 1 in 

Cross multiply:  120 mi x 1 in  =  x mi x 8 in

Divide both sides by  8 in  to get  15 mi = x mi   --->  x = 15 

8 inches = 120 miles

So

1 inch = (1/8) of 120 miles = 15 miles

If you allow me to use the identity sin(x + y) = sin(x)cosy + cos(x)sin(y),

then sin(2x) = sin(x + x) = sin(x)cos(x) + cos(x)sin(x) = 2sin(x)cos(x).

However, this assumes that you have the original identity.

You can find the proof of the original identity at   faculty.atu.edu/mfinan/1203/Lecture19.pdf

Melody's work was impressive......!!! Here's a slightly less sophisticated manner of looking at this if we assume that the tide rises and falls in a "linear" fashion.

The following graph illustrates the situation.........

Let x= 0 on the following graph equate to "5AM" and let x = 6 equate to "11AM"

And if we consider only the difference between the heighest and lowest tides to be 4.1 ft., then let the point (0, 4.1) be the "high" point of the tide. And the "low" point would be at y = 0 six hours later, i.e., the point (6,0).

And the ship can only enter the seaport when the tide is 1 ft above its "low point." This is represented on the graph by the function y = 1.

Therefore, the time that the ship can enter the seaport starting at 5AM would be from x = 0 to x = 4.537 hrs. later....in other words...from 5AM to about 9:32AM

And note that, the ship would not be able to enter or leave again until about 7.463 hrs after 5AM = about 12:28PM!! Thus, assuming the linear model, there is about a 3 hour period when the tide is too low for the ship to enter or leave the port....!!!

Melody's model is probably more "real-world," .....mine is probably more simplistic....(Duh!!)

in the isosceles triangle ABC, segment AD is the height of triangle from point A,and segment BE is the height of the triangle from point Learn from point B.

because triangle ABC is a isosceles triangle ,so AB=AC ,and segment AD is the height of the triangle 

so angle ADB= Angle ADC=90 degrees

In the right triangle  ADB and right triangle ADC 

AB=AC,AD=AD, so right triangle ADB is congruent to right triangle ADC ( hypotenuse-leg)

so angle BAD= Angle CAD

If angle BAD=x degrees ,then angle BAC= 2x degrees

AD=AB*cosx  DB=AB*sinx   BE=AB*sin2x

the area of triangle = BD *AD*2/2=BD*AD=BE*AC/2=BE*AB/2

BD*AD=BE*AB/2

AB*cosx *AB*sinx=AB*sin2x*AB/2 (cancel AB*AB from both side and both side times 2)

2*cosx*sinx=sin2x

this is my method,Can someone prove it algebraicallly?

To make a fraction you put the numerator, division sign, then demoninator 

I would suggest using a calculator to solve this; the one at this site works well for this problem.

r(3s)(4t)

Since the factors in multiplication can be written in any order, rewrite them as:

(3)(4)(r)(s)(t)    =  12rst

-6561 :D

A very useful equation to know is the point-slope form for a straight line:

     y - y1  =  m(x - x1)

where  m  is the slope and  (x1, y1) are the coordinates of the point.

For this problem,  m = -2,  x1 = 3,  and  y1 = -3.

Putting these values into the formula:  y - -3  =  -2(x - 3)

Simplifying:  y + 3  =  -2x + 6     --->     y  =  -2x + 3

24x - 2 + 3 = 67

Again, first add the numbers (-2) and 3 to simplify the left side:

24x + 1 = 67

Subtract 1 from both sides:

24x = 66

Divide by 24:

x = 2.75

Assuming the problem is as written:

4x - 22 + 3 = 67

(-22) + 3 = -19

4x - 19 = 67

Add 19 to both sides:

4x = 86

Divide both sides by 4:

x = 21.5

On this calculator, once you get a number you can press "Enter" again to get it in scientific notation.

(10.6-6.5)/2=2.05

the amplitude (A) of the tide is 2.05 metres.

7.5-6.5=1

For the lowest 1 metre of the tide the water will be too shallow.

Let y=0 represent the middle tide level.

So at high tide y=-2.05

At low tide y=2.05

the water will be too shallow if y<-1.05

11am-5am=6hours  The period of the wave is 6*2=12hours

I am going to use the formula   y=A*cos[n(x+s)]

A is amplitude, s is phase shift, period is 2pi/n

Now period is 12 so 

2pi/n=12

n=pi/6

If there was no phase shift then high tide would be at midnight but it is at 5am so the phase shift is 5 hours to the right      n=-5

So we have have our formula

$$y=2.06*sin[\frac{\pi}{6}(x-5)]$$             (remember that it is in radians)

we need to solve this simultaneously with y=-1.05

You can do this by hand without too much trouble but i have done it graphically.

These decimal numbers must be changed to times.  I have calculated them to the nearest minute.

x=0 is midnight      

x=0.97        0.97*60=58       This is 12:58am

x=9.03        0.03*60=2         This is 9:02am

x=12.97        This is 12.58pm

x=21.02        This is 9:02pm

So the water will be deep enough between 12:58am and 9:02am

and between  12:58pm and 9:02pm

Better leave an extra minute or 2 just to be safe.     

A linear regression model is the equation of a straight line that best approximates the data.

You need to check to see if the r-value of the model is statistically significant before you use the equation to make predictions.

96 = 12(2^x)     Divide both sides by 12:

8 = 2^x             Now, ask yourself, "2 to what power is 8?"; well, that is 3.

x = 3   because  2^3 = 8