Question about Texas Instruments TI-84 Plus Calculator

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Posted on Jan 02, 2017

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SOURCE: ti-89 titanium SIN!!!!!

The answer that you have received in actuality is correct. However, it is in exact form when the answer that you were looking for was an approximation (decimal form) answer. To achieve this all you have to do is hit the alpha button (the yellow button with a diamond in the center) and then enter which will give you your desired answer :-)

Posted on Jan 15, 2008

SOURCE: error :invalid DIM

This is my first post on this site, but I've had friends at school ask me about this. Apparently I'm like a calculator genius or something...

One solution I know of is clearing the RAM. Sounds strange, doesn't it? But it fixed three different calcs.

Be sure to back up or archive important data before doing this.

Posted on Oct 21, 2008

SOURCE: ERR: No Sign change...

-0.171x+43.43 and -0.146x+39.83

i tried that and i got the same thing, but i noticed that if you change your window so it fits more of the graph on the right, were you can actually see the lines intersect, then you can get the intersection.

oh, and by the way, the intersection is (144, 18.806)

Posted on Mar 01, 2009

SOURCE: Graphing radical functions

I have a TI-89 Titanium and I have experienced the same problem. HERE IS HOW TO FIX IT:

Goto " **MODE** " and scroll down the options until you get to

" **COMPLEX FORMAT** " change that option to " * RECTANGULAR* " and press

Posted on Mar 15, 2009

SOURCE: TI-15 square root Function

Hello,

You should enter it as follows [SQRT]36 ) [ENTER/=].

If you define some function f of a variable x you write that f(x), where the parentheses enclose the so-called argument (objet on wich the function acts). It seems that on this calculator the opening parenthesis is implicit: the calculator supplies it when you press [SQRT] but does not display it (a design flaw?). However the closing parenthesis must be entered by you to signifie to the function [SQRT] that you have finished entering the argument. Weird but one can live with it.

Hope it helps.

Posted on Sep 07, 2009

In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions,** if it exists, **is a point (x_i,y_i) such that **f(x_i)=g(x_i)**.

Graphically this means that**(x_i, f(x_i))=(x_i, g(x_i))**: The two curves pass through the same point in the Cartesian plane.

Now consider**f(x_i)=g(x_i)**. That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or **f(x)-g(x)=0**.

If you can feed it to the SOLVER in**EQN**, the equation mode, and the calculator gives you the roots of the equation, then for each root x_i found you have an intersection point (x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to calculate the y-value since the two functions are supposed to be equal at the root x_i.

**However it is much simpler to do with a graphing calculator.**

Graphically this means that

Now consider

If you can feed it to the SOLVER in

Oct 05, 2013 | Casio Office Equipment & Supplies

In theory yes. If you have a function y=f(x) and another y=g(x), an intersection point of the graphs of the two functions,** if it exists, **is a point (x_i,y_i) such that **f(x_i)=g(x_i)**.

Graphically this means that**(x_i, f(x_i))=(x_i, g(x_i))**: The two curves pass through the same point in the Cartesian plane.

Now consider**f(x_i)=g(x_i)**. That is an equation in x_i, or lest us just drop the _i to shorten and write f(x)=g(x), or **f(x)-g(x)=0**.

If you can feed it to the SOLVER in**EQN**,
the equation mode, and the calculator gives you the roots of the
equation, then for each root x_i found you have an intersection point
(x_i, f(x_i)), or (x_i, g(x_i)). You can use either function to
calculate the y-value since the two functions are supposed to be equal
at the root x_i.

**However it is much simpler to do with a graphing calculator.**

Graphically this means that

Now consider

If you can feed it to the SOLVER in

Oct 05, 2013 | Casio FX-115ES Scientific Calculator

Draw the graphs of the functions you are interested in. While the graphs are displayed Press the Menu key. In the drop down window locate the submenu Settings (number 8 or 9) depending on your calculator

The following screen is displayed with at the top the setting Float.

Follow the right pointing arrow on the same line as float and select Float 3 or 4 as on the screen capture above.

Point to the graph of one of the functions select it then press the menu select option Analyze graph then intersection. You will be asked to select the second function, then the lower bound and upper bound. As you see on the previous screen capture, the coordinates of the intersection point are given with 3 decimal digits.

The following screen is displayed with at the top the setting Float.

Follow the right pointing arrow on the same line as float and select Float 3 or 4 as on the screen capture above.

Point to the graph of one of the functions select it then press the menu select option Analyze graph then intersection. You will be asked to select the second function, then the lower bound and upper bound. As you see on the previous screen capture, the coordinates of the intersection point are given with 3 decimal digits.

Apr 27, 2013 | Texas Instruments TI-Nspire Graphic...

You have to use **Y Editor** to define functions on the both sides of the equation and then graph it. Finally option is **F5: Intersection**.

See captured images for equation**5x+10=-2x-7**, **y1=5x+10**, **y2=-2x-7** and **x=-17/7**

See captured images for equation

Feb 25, 2012 | Texas Instruments TI-89 Calculator

We can write this polynomial as:

You can see this polynomial in following picture:

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

- (x-(-2))*(x-1)*(x-3)=
- (x+2)(x-1)(x-3)=
- (x+2)[x*(x-3)-1*(x-3)]=
- (x+2)*(x^2-3x-x+3)=
- (x+2)(x^2-4x+3)=
- x*(x^2-4x+3)+2*(x^2-4x+3)=
- x^3-4x^2+3x+2x^2-8x+6=
- x^3-2x^2-5x+6

You can see this polynomial in following picture:

Notice that it intersects x axis for x=-2, 1 and 3 (because these are roots of polynomial).

Oct 03, 2011 | Office Equipment & Supplies

Yes, there is shortcut because this is right triangle, so you can use Pythagorean theorem (see picture).

If this was helpful please rate 4 thumbs :)

- Length of hypotenuse is square root of sum of squares of lengths of other two sides of triangle, which is equal to square root of 30^2+10^2=31.6 cm.
- Sin(a)=longer cathetus/hypotenuse=0.949 so a=arcsin(0.949)=71.6 degrees
- Finally b=90-a=18.4 degrees.

If this was helpful please rate 4 thumbs :)

Sep 05, 2011 | Texas Instruments TI-30XA Calculator

the curve is refering to the function. So you have two different "lines". You choose intersect then one line, then it will ask for the other line, then a guess. your guess may or may not matter if there is only 1 intersection it will only calculate faster. If there is more than one intersection it will let you choose which one you want.

Basically you can graph 1,2,3,4,5 or how ever many lines it will take, but you may only want the intersect of 2 of them, thats why it asks.

hope this helps

Basically you can graph 1,2,3,4,5 or how ever many lines it will take, but you may only want the intersect of 2 of them, thats why it asks.

hope this helps

Apr 24, 2010 | Texas Instruments TI-84 Plus Calculator

Half of the work involved in solving a problem is being able to formulate it so that it can be solved. I am afraid your formulation leaves too many details in the dark.

But I think I figured out what you want.

"You want to solve the equation graphically, ie find the zeros of a function"

FIRST METHOD

I assume you know how to use the [2nd][TRACE] (CALC) [2:Zero] function to find the zeros.

Here is the negative one Here is the positive one

SECOND METHOD

The second method entails

As you see, one root does not show and you have either to Zoom out or move the Yrange downward (as seen on the right picture for which I set Ymin=-15, Ymax=5

I assume you know how to find the intersection of the two curves, and I will show you only one point.

To find the intersection you use the [2nd][TRACE] (CALC) [5:Intersect] command

But I think I figured out what you want.

"You want to solve the equation graphically, ie find the zeros of a function"

FIRST METHOD

- Create a single expression from the equation : gather all terms on one side so as to make "expression"=0
- One such expression is X^2 +X -14=0
- Draw the function y=X^2+X-14
- Find the X-coordinates of the points where y=0

I assume you know how to use the [2nd][TRACE] (CALC) [2:Zero] function to find the zeros.

Here is the negative one Here is the positive one

SECOND METHOD

The second method entails

- defining two functions, Y1=-X^2+5 and Y2=X-9,
- Graphing the two functions.
- Finding their intersections.
- The X-Values of the two points of intersection of Y1 and Y2 are the solutions of the equation.

As you see, one root does not show and you have either to Zoom out or move the Yrange downward (as seen on the right picture for which I set Ymin=-15, Ymax=5

I assume you know how to find the intersection of the two curves, and I will show you only one point.

To find the intersection you use the [2nd][TRACE] (CALC) [5:Intersect] command

Jan 29, 2010 | Texas Instruments TI-84 Plus Calculator

Hi,

You should check your understanding of what a function is. You are drawing two functions the ranges of which do not overlap, since one branch is positive and the other is negative. You know that the** only two points where **there could be overlapping are the points where y=0 for both functions. Why would you need the calculator to confirm to you what you already know.

To define the two branches you had to take the square root of some expression say y= SQRT(5-x^2). That is a circle centered on O(0,0) with radius SQRT(5). The two points where the positive branch intersects the negative one are for y=0, meaning x1= SQRT(5) or x2= -SQRT(5).

What do you think is the exact value of SQRT(5): 2.236067977....? No, because SQRT(5) is an irrational number that has an infinite number of digits and no matter how many additional digits you may align to determine it will not make that representation the EXACT value of SQRT(5).

That does not mean you will never be able to find an intersection of the two curves. Maybe, if you take y=SQRT(4-x^ 2) the calculator will be able to find the intersections but that will remain one case.In general the calculator will not find the intersection.

I hope that I convinced that it is futile to seek,**with the help of the calculator,** the intersection of two irrational functions ( for they are irrational not rational as you claim) that share only two points.

Hope it helps.

You should check your understanding of what a function is. You are drawing two functions the ranges of which do not overlap, since one branch is positive and the other is negative. You know that the

To define the two branches you had to take the square root of some expression say y= SQRT(5-x^2). That is a circle centered on O(0,0) with radius SQRT(5). The two points where the positive branch intersects the negative one are for y=0, meaning x1= SQRT(5) or x2= -SQRT(5).

What do you think is the exact value of SQRT(5): 2.236067977....? No, because SQRT(5) is an irrational number that has an infinite number of digits and no matter how many additional digits you may align to determine it will not make that representation the EXACT value of SQRT(5).

That does not mean you will never be able to find an intersection of the two curves. Maybe, if you take y=SQRT(4-x^ 2) the calculator will be able to find the intersections but that will remain one case.In general the calculator will not find the intersection.

I hope that I convinced that it is futile to seek,

Hope it helps.

Dec 03, 2009 | Texas Instruments TI-84 Plus Calculator

No need for a manual. Just hold it so that the two edges of the "vee" line up with the circumference of the circle and scribe a line along the steel rule. Then turn approx. 90 degrees and scribe another line along the rule that intersects the first. The point of intersection is exactly the center of the circle. This tools only function is to find the center of circles.

Mar 05, 2009 | Measuring Tools & Sensors

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