Question about Sentry CA 756 Calculator

Ad

Please refer to the manual and check if the chosen MODE is the correct one required to process the equation or the commutation required or else this wont work.

Check to see the different modes and study on how to apply them to the said equation.

Posted on Mar 14, 2011

Ad

Hi,

a 6ya expert can help you resolve that issue over the phone in a minute or two.

best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).

click here to download the app (for users in the US for now) and get all the help you need.

goodluck!

Posted on Jan 02, 2017

Ad

$450.00 divided by $35.00=12 hours with $30.00 left to pay the sales tax. $35.00 x 12hrs =$420.00. The deposit can be figured in the available cash as when you return the jetski you will be refunded the $50.00 deposit giving you the whole $450.00 to use to pay the bill.

Mar 15, 2015 | algebra.com Computers & Internet

Hey! The standard equation of a straight line is y=mx+b (wherein x&y are the coordinates, m is the slope of the line and, b is the y-intercept). Re-arranging your equation to follow the standard equation, you'll have y= -x - 4 or y=(-1)x + (-4).

To plot the equation, try substituting a number to the (x) coordinate to get the value of the (y) coordinate:

Ex:

If x=0, y= -4

If x=1, y= -5

If x=2, y= -6

If x=3, y= -7

Assuming you know how to plot these coordinates, you now have a graph of the equation. Hope I answered your query. Thank you.

To plot the equation, try substituting a number to the (x) coordinate to get the value of the (y) coordinate:

Ex:

If x=0, y= -4

If x=1, y= -5

If x=2, y= -6

If x=3, y= -7

Assuming you know how to plot these coordinates, you now have a graph of the equation. Hope I answered your query. Thank you.

Jun 09, 2012 | Sony HANDYCAM HIDEF 16GB MP4 FL MEM CAM 27...

Hi janeth sabay,

Have a good day!!

Here is your needed example for Homogenous differential equation:

The equation is said to be homogeneous if P and Q are homogeneous functions of x and y of the same degree.

(1)We can test to see whether this first order equation is homogeneous by substituting . If the result is in the form f(v)i.e. all the x's are canceled then the test is satisfied and the equation is Homogeneous.

Example 1:(2)(3)There are no terms in x on the right hand side and the equation is Homogereous.Example 2:(4)(5)So the original equation is not homogeneous.Methods Of Solution.A solution can be found by putting y = vx on both sides of the equation:-

Example 3:Putting y - vx

Since y is a function of x so is v

Separating the variables

Integrating

Substituting (12) in equation (10)

I can suugest you also to visit this site for more examples http://www.codecogs.com/reference/maths/differential_equations/homogeneous_differential_equations.php

Sep 12, 2011 | Toys

Press "2ND" "STAT PLOT" and then select 4:PlotsOff.

Jan 08, 2011 | Texas Instruments TI-84 Plus Calculator

"Is there anything that I can do about this to stop my answers being complex numbers when solving cubic equations?"

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

You are making an assumption that is unwarranted.Namely that a cubic equation must necessarily have 3 real roots. This is not the case.

Let your equation be

If coefficients are complex you should expect some complex roots. Right?

If the coefficients are REAL then depending on the discriminant

you can have three cases

DELTA positive : three distinct real roots

DELTA=0 , the equation has a multiple root and all roots are REAL

DELTA negative: the equation has ONE real root and 2 complex roots that are complex conjugate of each other.

It suffices to look at the Tartaglia/Cardano expressions of the roots to know that more often than not there is going to be complex roots.

May 05, 2010 | Casio CFX-9850G Plus Calculator

Rename your variable (unknown) X, because the calculator takes X as the default unknown. It will ask you "Solve for X?"

the 7, the second minus sign, 3 and 11 use the relevant keys.

For the first minus sign use the negation key or change sign (-)

For the variable X, examine the calculator keys on the row where the parentheses are. To the right corner of the key area there is a symbol in the same color as the ALPHA key. I believe the X is under the right parenthesis, or thereby.

To enter the X variable you press [ALPHA][ )]. To enter the = sign, you press [ALPHA] [CALC].

To execute the solve command, press [SHIFT] [CALC] and you will be prompted " Solve for X?"

the 7, the second minus sign, 3 and 11 use the relevant keys.

For the first minus sign use the negation key or change sign (-)

For the variable X, examine the calculator keys on the row where the parentheses are. To the right corner of the key area there is a symbol in the same color as the ALPHA key. I believe the X is under the right parenthesis, or thereby.

To enter the X variable you press [ALPHA][ )]. To enter the = sign, you press [ALPHA] [CALC].

To execute the solve command, press [SHIFT] [CALC] and you will be prompted " Solve for X?"

May 05, 2010 | Casio FX-115ES Scientific Calculator

1st equation:

11=6+a

Everything the you do to one side of the equals sign in an equation, you have to do to the other. So let's subtract 6:

5=a

------

The second equation:

a+7=10+b

Now we know from the first equation that a= 5, so let's put that in:

5+7=10+b

Or, simplifying:

12=10+b

Let's take away 10 from both sides:

2=b

So your answer is: b=2

11=6+a

Everything the you do to one side of the equals sign in an equation, you have to do to the other. So let's subtract 6:

5=a

------

The second equation:

a+7=10+b

Now we know from the first equation that a= 5, so let's put that in:

5+7=10+b

Or, simplifying:

12=10+b

Let's take away 10 from both sides:

2=b

So your answer is: b=2

Apr 04, 2010 | Vivendi ADI English and Maths Year 5 Full...

Put the figure which you have to multiply in cell A1, Put the figure in cell A2 by which you have to multiply by, type =A1*A2 in cell A3. You will get the right answer.

You may substract, multiply or add by this method by appropriate mark in cell A3 in between A1 and A2.

npbhoge@rediff.com

You may substract, multiply or add by this method by appropriate mark in cell A3 in between A1 and A2.

npbhoge@rediff.com

Jul 29, 2008 | Microsoft Excel for PC

Here is the phone number for equator, 1800-935-1955 or parts and service 800-776-3538. They sould be able to help you.

Their address is 10067 Timber Oak Dr. Houston, TX 77080-7045

Their address is 10067 Timber Oak Dr. Houston, TX 77080-7045

Jun 27, 2008 | Equator EZ 2512 CEE Front Load Washer /...

try a simple equation like x^2-1 which should come out with x = 1,-1

no sign change would indicate that the graph of the curve never crosses the x axis, if you graphed the above equation and calculated the zeros, it would come out with -1, 1 which it should, but if you put in x^2+1, the graph would never cross and therefore have no sign change from pos to neg or vice versa, the left-rt are the left and right bounds or the lower and upper bounds which was corrected in later calculators, Please let me know the outcome of graphing both of the above equations and then putting them in the solver. you could also use the quadratic equation to prove the answers if you liked, hope this helps

no sign change would indicate that the graph of the curve never crosses the x axis, if you graphed the above equation and calculated the zeros, it would come out with -1, 1 which it should, but if you put in x^2+1, the graph would never cross and therefore have no sign change from pos to neg or vice versa, the left-rt are the left and right bounds or the lower and upper bounds which was corrected in later calculators, Please let me know the outcome of graphing both of the above equations and then putting them in the solver. you could also use the quadratic equation to prove the answers if you liked, hope this helps

Mar 13, 2008 | Texas Instruments TI-83 Plus Calculator

Feb 10, 2014 | Sentry CA 756 Calculator

Jan 13, 2014 | Sentry CA 756 Calculator

Jan 06, 2014 | Sentry CA 756 Calculator

Nov 04, 2013 | Sentry CA 756 Calculator

47 people viewed this question

Usually answered in minutes!

×