How to solve in the calculator 5000e^.15? - Casio FX-260 Calculator

It is easy to find solution for this problem. Type solve(0,1Q+118,5 = 0,02Q+15,Q)
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On this calculator the exponential function requires the argument be entered first. To calculate the exponential of a number, you enter the exponent (0.08) then press [2nd][LN].
For your calculation, you should calculate e^(0.08) then multiply the result by (5000x2=10000). The result should be 10 832.87068

x/2=5/15 This implies that x=2*5/15=10/15=2/3

For this problem you have to type solve(5(a+3)-a = -(4a-6) +1,a)
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Do you really need a calculator to solve a linear equation?
Let us see how to solve it without a calculator.
Remove the parentheses fronm the second term on the left : y-3
Use the distributive property on the last term on the right side -2y-6
Take the term -3 to the right side, changing its sign in the process (becomes +3)
Take the -2y term to the left making it +2y.
3y +2y=18+3-6 or 5y=15 and the answer is straightforward .

As to your Casio FX300ES, it cannot solve equations because it does not have an EQUATION calculation mode. The Casios FX-115ES and FX-991ES have an Equation Mode. If you use any of the two to solve an equation in one variable (linear, quadratic or other) the unknown is taken to be x by default. You would have to change the name of the variable from y to x.

This calculator does not support the modulus operation directly, you have to simulate by a divide followed by a subtraction and a multiplication. Suppose you want to calculate 256 mod 17:

1. Key in 256 [÷] 17 [=] and get the result 15,0588...
2. Note the 15 in front of the decimal separator and key in [−] 15 [=]
3. Get the modulus: [×] 17 [=], result should be 1.

Using elementary algebria in the binomial theorem, I expanded the power (x + y)^n into a sum involving terms in the form a x^b y^c. The coefficient of each term is a positive integer, and the sum of the exponents of x and y in each term is n. This is known as binomial coefficients and are none other than combinatorial numbers.

Combinatorial interpretation:

Using binomial coefficient (n over k) allowed me to choose k elements from an n-element set. This you will see in my calculations on my Ti 89. This also allowed me to use (x+y)^n to rewrite as a product. Then I was able to combine like terms to solve for the solution as shown below.
(x+y)^6= (x+y)(x+y)(x+y)(x+y)(x+y)(x+y) = x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6

This also follows Newton's generalized binomial theorem:



Now to solve using the Ti 89.

Using sigma notation, and factorials for the combinatorial numbers, here is the binomial theorem:

The summation sign is the general term. Each term in the sum will look like that as you will see on my calculator display. Tthe first term having k = 0; then k = 1, k = 2, and so on, up to k = n.
Notice that the sum of the exponents (n ? k) + k, always equals n.






The summation being preformed on the Ti 89. The actual summation was preformed earlier. I just wanted to show the symbolic value of (n) in both calculations. All I need to do is drop the summation sign to the actual calculation and, fill in the term value (k), for each binomial coefficient.





This is the zero th term. x^6, when k=0. Notice how easy the calculations will be. All I'm doing is adding 1 to the value of k.




This is the first term or, first coefficient 6*x^5*y, when k=1.
Solution so far = x^6+6*x^5*y






This is the 2nd term or, 2nd coefficient 15*x^4*y^2, when k=2.
Solution so far = x^6+6*x^5*y+15*x^4*y^2







This is the 3rd term or, 3rd coefficient 20*x^3*y^3, when k=3.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3







This is the 4th term or, 4th coefficient 15*x^2*y^4, when k=4.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4






This is the 5th term or, 5th coefficient 6*x*y^5, when k=5.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5





This is the 6th term or, 6th coefficient y^6, when k=6.
Solution so far = x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6





Putting the coefficients together was equal or, the same as for when I used the expand command on the Ti 89.

binomial coefficient (n over k) for (x+y)^6
x^6+6*x^5*y+15*x^4*y^2+20*x^3*y^3+15*x^2*y^4+6*x*y^5+y^6

5 0 0 0 2ND [e^x] . 9 6 ENTER

computes 5000 e^.96

If you want (5000e)^.96 then it's

( 5 0 0 0 2ND [e^x] 1 ) ^ . 9 6

e^x is just to the left of the 4 key.

There are a few things this could be, most of them are motherboard related, but there could be a cooling issue or a power supply issue. Check the BIOS log for "thermal events". Things like CPU overheating can cause weird reactions in the machine, called System hysteresis. Basically, the computer goes crazy, opening things, crashing, rebooting, etc. Based on what youve said, "restarts normally", I dont think there is a CPU issue, nor a memory issue.

The first thing I would try is replacing the power supply and remove the battery (just as a test), and see if the machine starts behaving better or exhibits the same issues.

I believe the issue you have described can be most likely resolved with a new motherboard, especially if the power supply route does not yield positive results. Bad news is, a p3 motherboard is rather hard to get, and hard to replace, and rather pricey. I would sooner consider getting another notebook pc before replacing the board.

I want to Benq 5000E model no.6661-9UI soft ware