Question about Texas Instruments TI-84 Plus Calculator

Hello,

I have no idea what can of data you have, so I am taking a theoretical stance. In a radioactive decay the period of half life is the time after which half the nuclei have desintegrated. Let us call that period tau τ.

The constant, usually called lambda, λ appears in the expoential. It is the constant term that multiplies the time variable, t.

If you have y1 nuclei at a time t, and y2 nuclei at time t+ τ., the ratio y2/y1 =1/2. It is also equal to e^(-λ* τ). Thus

**e^(-λ* τ) =1/2. **

and ln(e^(-λ* τ)) = ln(1/2)=-ln(2). But since **ln(e¨(x)) = x, ** we get

-λ* τ= - ln(2) and

**λ= ln(2)/ τ**

In the enclosed screen capture the equation **e^(-λ* τ) =1/2** is solved for λ

If you have experimental data, or a radioactive decay curve drawn you can obtain the constant λ , as the negative of the slope of the function at t=0. See on following graph. Disregard the values of the function before t=0 ( I should have restricted the window to the first quadrant.) The function drawn is e^(-0.365t)

Hope it helps.

Thank you for using FixYa

Posted on Nov 20, 2009

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Posted on Jan 02, 2017

Lithium ION batteries are usually rechargeable. The use mode for these batteries is that there are a set number of recharges that equate to the life of the battery. Usually that life is approximately 1,000 charges.

If you use your battery to 50% then recharge, you can do this ~2,000 times. As the life of the battery is used up, the maximum charge lessens which means where you may have had 8 hours of play time when the battery was new, you will only have 6 hours as the battery life is used up.

In order for a Lithium ION battery to recharge there must be at least a small amount of charge left as a "seed" for that recharge. If you drain a L-ION battery down to zero, it may never recharge again or the max charge will be reduced to an insignificant amount of the original full charge. eg instead of 8 hours you may only get 15 minutes, or 5 minutes, or no minutes.

You are much better off recharging when 50% depleted than using it up til your product stops.

L-ION batteries cannot be repaired only replaced. Most cheap products using L-ION batteries cannot have their batteries replaced.

Places like Batteries + are able to replace some tablet batteries.

If you use your battery to 50% then recharge, you can do this ~2,000 times. As the life of the battery is used up, the maximum charge lessens which means where you may have had 8 hours of play time when the battery was new, you will only have 6 hours as the battery life is used up.

In order for a Lithium ION battery to recharge there must be at least a small amount of charge left as a "seed" for that recharge. If you drain a L-ION battery down to zero, it may never recharge again or the max charge will be reduced to an insignificant amount of the original full charge. eg instead of 8 hours you may only get 15 minutes, or 5 minutes, or no minutes.

You are much better off recharging when 50% depleted than using it up til your product stops.

L-ION batteries cannot be repaired only replaced. Most cheap products using L-ION batteries cannot have their batteries replaced.

Places like Batteries + are able to replace some tablet batteries.

Jan 20, 2015 | Amazon Kindle Fire HD

The half life of Actinium 226 is about 1.2 days (Check with the source of the question). Let the half life be T_1/2.

The equation of the radioactive decay is

N(t)=N_0*exp(-ln2*t/T_1/2)

For the mass

m(t)=m_0*exp(-(ln2)*t/T_1/2)

There remains 6.16 mg after 87 hours.

The equation of the radioactive decay is

N(t)=N_0*exp(-ln2*t/T_1/2)

For the mass

m(t)=m_0*exp(-(ln2)*t/T_1/2)

There remains 6.16 mg after 87 hours.

Jun 26, 2014 | Sharp EL-531VB Calculator

Definition

A mathematical statement used to evaluate a value. An equation can use any combination of mathematical operations, including addition, subtraction, division, or multiplication. An equation can be already established due to the properties of numbers (2 + 2 = 4), or can be filled solely with variables which can be replaced with numerical values to get a resulting value. For example, the equation to calculate return on sales is: Net income ÷ Sales revenue = Return on Sales. When the values for net income and sales revenue are plugged into the equation, you are able to calculate the value of return on sales.

There are many types of mathematical equations.

1. Linear Equations y= mx + b (standard form of linear equation)

2. Quadratic Equations y= ax^2+bx+c

3. Exponential Equations y= ab^x

4. Cubic Equations y=ax^3+ bx^2+cx+d

5. Quartic Equations y= ax^4+ bx^3+ cx^2+ dx+ e

6. Equation of a circle (x-h)^2+(y-k)^2= r^2

7. Constant equation y= 9 (basically y has to equal a number for it to be a constant equation).

8. Proportional equations y=kx; y= k/x, etc.

Jun 14, 2011 | Computers & Internet

I can help you there :) In the upper-left hand corner of your keyboard, below the touch pad, there should be a button labelled '=' Enter the first part of your equation press '=' then enter the second half. After you have entered the equation hit enter and it will tell you if it is true or false.

Feb 07, 2011 | Texas Instruments TI-Nspire Graphic...

For this type of problem, a equals the constant cost of operation (monthly rent..etc) and b equals the cost to manufacture the items.

first equation: 320 = a + b(10)

second equation: 520 = a +b(20)

320 = a + 10b

520 = a + 20b subtract equation 2 from equation 1

-200 = -10b divide by -10

20 = b plug b into one of the equations

320 = a +(20)(10) multiply

320 = a +200 subtract 200

120 = a

Let me know if i need to explain further.

first equation: 320 = a + b(10)

second equation: 520 = a +b(20)

320 = a + 10b

520 = a + 20b subtract equation 2 from equation 1

-200 = -10b divide by -10

20 = b plug b into one of the equations

320 = a +(20)(10) multiply

320 = a +200 subtract 200

120 = a

Let me know if i need to explain further.

Dec 29, 2009 | Mathsoft StudyWorks! Mathematics Deluxe...

by keeping the player constantly plugged in when of will have seriously shortened the battery life, rechargeable batteries ( exept the very latest ones) perform best by charging then using until fully discharged and charging again. this will ensure your battery lasts a long time, constantly charging a half charged battery will shorten the batteries life.

Nov 23, 2009 | Audio Players & Recorders

Hello,

First of all, you have to set the mode to EQUATION by pressing [MODE][5:Equation]. When you do that, you are offered 4 choices of equation types to solve.

PART 1 a_n*X+b_n*Y=C_n.

In this type, you have only 2 unknows, X, Y, the others are coefficients. When this mode is selected, you are given access to a what looks like a matrix.

The top row, inaccessible to you, displays headings a, b, c.

The first column, also inaccessible to you numbers the lines as (1 and 2)

First line

Enter the coefficient of X in the first equation, a_1, and press [=] Cursor moves to the dark rectangle under the b heading.

Enter the coefficient of Y, (b_1) in the first equation and press [=] Cursor moves to the dark rectangle under heading c.

Enter the constant term in the first equation, C_1, and press [=]

At this stage you have finished entering the coefficients and constant term of the first linear equation of your system, namely

a_1*X+b_1*Y= C_1

After you enter C_1 and press [ENTER] the cursor moves to the

Second Line

On this line, just as in the previous one, you enter a_2, b_2 and c_2 Once you finish entering all coefficients and constant terms you press [ENTER].

Calculator does its thing and returns with a solution and displays at the top left of the screen

X=

and on the bottom right the value for X ( a fraction or a decimal number)

Press [=] again and the screen displays the value of the second unknown Y.

If you press [=] a third time the claculator takes you back to the entry screen to modifiy coefficients and constant terms.

Exemple a_1=6 b_1=3 c_1=7 a_2=5 b_2=8.3 c_2=1

X= 19/12 To convert it to decimal press [S<->D] to get 1.58333 Y=-5/6 To convert it to decimal press [S<->D] to get -0.8333333

PART B a_n*X+b_n*Y+c_n*Z =D_n This is a linear system in 3 unknowns X, Y, X.

It requires

a_1, b_1, c_1, and d_1

a_2, b_2, c_2, and d_2

a_3, b_3, c_3 and d_3.

Procedure to enter the coefficients and constant terms is the same as above, but due to the limited screen, you have to use right arrow to access the places where you enter d_1, d_2. d_3.

When finished you have to press [=] 1 time to get X, a second time to get Y, and a third time to get Z.

PART C aX^2 + bX +c=0 (quadratic equation)

You enter a, b, and c in the template.

When finished press[ENTER] once to get X1, and a second time to get X2.

PART D aX^3 +bX^2+cX+d =0 (cubic equation)

Same procedure as Part C, except that you have to enter a, b, c, and d in template and press [=] 1 time to get X1, a second time to get X2, and a third time to get X3.

Hope it helps.

PS to the power that be of this site, if you are listening: We ought to have prompts to confirm the post before validating it. We ought to be able to correct typos, errors; to amend what we entered.

First of all, you have to set the mode to EQUATION by pressing [MODE][5:Equation]. When you do that, you are offered 4 choices of equation types to solve.

PART 1 a_n*X+b_n*Y=C_n.

In this type, you have only 2 unknows, X, Y, the others are coefficients. When this mode is selected, you are given access to a what looks like a matrix.

The top row, inaccessible to you, displays headings a, b, c.

The first column, also inaccessible to you numbers the lines as (1 and 2)

First line

Enter the coefficient of X in the first equation, a_1, and press [=] Cursor moves to the dark rectangle under the b heading.

Enter the coefficient of Y, (b_1) in the first equation and press [=] Cursor moves to the dark rectangle under heading c.

Enter the constant term in the first equation, C_1, and press [=]

At this stage you have finished entering the coefficients and constant term of the first linear equation of your system, namely

a_1*X+b_1*Y= C_1

After you enter C_1 and press [ENTER] the cursor moves to the

Second Line

On this line, just as in the previous one, you enter a_2, b_2 and c_2 Once you finish entering all coefficients and constant terms you press [ENTER].

Calculator does its thing and returns with a solution and displays at the top left of the screen

X=

and on the bottom right the value for X ( a fraction or a decimal number)

Press [=] again and the screen displays the value of the second unknown Y.

If you press [=] a third time the claculator takes you back to the entry screen to modifiy coefficients and constant terms.

Exemple a_1=6 b_1=3 c_1=7 a_2=5 b_2=8.3 c_2=1

X= 19/12 To convert it to decimal press [S<->D] to get 1.58333 Y=-5/6 To convert it to decimal press [S<->D] to get -0.8333333

PART B a_n*X+b_n*Y+c_n*Z =D_n This is a linear system in 3 unknowns X, Y, X.

It requires

a_1, b_1, c_1, and d_1

a_2, b_2, c_2, and d_2

a_3, b_3, c_3 and d_3.

Procedure to enter the coefficients and constant terms is the same as above, but due to the limited screen, you have to use right arrow to access the places where you enter d_1, d_2. d_3.

When finished you have to press [=] 1 time to get X, a second time to get Y, and a third time to get Z.

PART C aX^2 + bX +c=0 (quadratic equation)

You enter a, b, and c in the template.

When finished press[ENTER] once to get X1, and a second time to get X2.

PART D aX^3 +bX^2+cX+d =0 (cubic equation)

Same procedure as Part C, except that you have to enter a, b, c, and d in template and press [=] 1 time to get X1, a second time to get X2, and a third time to get X3.

Hope it helps.

PS to the power that be of this site, if you are listening: We ought to have prompts to confirm the post before validating it. We ought to be able to correct typos, errors; to amend what we entered.

Oct 11, 2009 | Casio FX-115ES Scientific Calculator

your best bet in any kind of exponential equation is to set up the general form of the equation and then solve for the constants with the numbers they gave you. With P being population, t being time, you know it is going to look like P=Ae^(b*t) and you are given 2 P's for 2 t's.

So for the first set :

Eq 1 P(26)=40,000=Ae^(b*26)

and the second being a time t=0 P was half the current value:

Eq 2 P(0)=20,000=Ae^(b*0)

20,000=A*1 so A=20,000

now b from Eq 1

40,000=20,000e^(b*26)

So for the first set :

Eq 1 P(26)=40,000=Ae^(b*26)

and the second being a time t=0 P was half the current value:

Eq 2 P(0)=20,000=Ae^(b*0)

20,000=A*1 so A=20,000

now b from Eq 1

40,000=20,000e^(b*26)

Aug 04, 2009 | Texas Instruments TI-30XA Calculator

i dont know the solution...

Jul 04, 2008 | Computers & Internet

A positive number is 5 times another number.If 21 is added to both the numbers,then one of the new numbers becomes twice the other number.What are the numbers

Nov 29, 2007 | Computers & Internet

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