For example:

I'm trying to work out 5 * 5 to the power of 3 - don't know how to write that properly with my computer. The problem is that the instructions aren't as clear as i'd hoped and it gives examples of Octal and hexadecimal calcuations, but none of the straight decimal variety

Hello,

You are victim of a confusion between terms. Base-n calculations are calculations using various numeration systems (bases) : binary (2), octal(8) hexadecimal (16) and decimal (10). What you want is how to calculate powers, logarithms, exponentials and these have also the notion of base; Log in base 10 is the common log, whereas natural logarithms, are logarithms in base e.

Back to your exemple: 5 times 5 and raise the whole to power 3?

In that case [(] 5 [x] 5 [)] [^] 3 or [X to the 3] if you have such key or

25 [^] 3, because 5x5 = 25

5 times (5 raised to power 3)

5[x] [( ] 5 [^] 3 [)] or 5 [^] 4 because of the rules about products of powers with the same base.

Hope it helps.

Posted on Sep 18, 2009

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Posted on Jan 02, 2017

You input double parenthesis in the order that you see them in the written equation. For example, ((2+ 5)*3)+6, you would enter <left parenthesis> <left parenthesis> 2 + 5 <right parenthesis> * 3 <left parenthesis> + 6 =

Try it out on some equations to make sure it is working like you expect it to work.

Good luck.

Paul

Try it out on some equations to make sure it is working like you expect it to work.

Good luck.

Paul

Aug 21, 2015 | Office Equipment & Supplies

To convert a complex number (not an equation) from rectangular coordinates to polar coordinates you do as follows

2+3[OPTN][ F3 (CLPX)][F1(i)][F6 >][F3> r ,theta][EXE]

2+3[OPTN][ F3 (CLPX)][F1(i)][F6 >][F3> r ,theta][EXE]

Nov 08, 2014 | Casio CFX-9850G Plus Calculator

Use the rules for exponents and you will calculate it by hand

(1*10^-14)/(5*10^-12)=(1/5)*10^(-14+12)= 0.2*10^-2=2*10-3

To enter powers of 10 use the short-cut key [EE]

For example to enter 6.02*10^23 type in** 6.02* [EE] 23**. When you use the EE key, the base (10) should not appear.

(1*10^-14)/(5*10^-12)=(1/5)*10^(-14+12)= 0.2*10^-2=2*10-3

To enter powers of 10 use the short-cut key [EE]

For example to enter 6.02*10^23 type in

Apr 06, 2014 | Texas Instruments TI-30XA Calculator

To enter powers of trigonometric functions you must enclose the functions in parentheses and then apply the exponent to the whole. For example X1t =(sin T)^3 , Y1t=(cos T)^3 will give you a shape similar to a rhombus with concave sides. The symbol ^syands for the operation of raising to a power. The key is the one with the caret symbol ^ , and it is wedged between the xsqure and EXIT keys (third row of keys).

As regards the cotangent, use the equivalent definition cot(X)=1/tan(X).

As regards the cotangent, use the equivalent definition cot(X)=1/tan(X).

Jul 26, 2011 | Casio FX-9750GPlus Calculator

You don't want the square root. In this example, you want the sixth root.

Press 4 5 2nd [xth-root] 6 =

The xth-root is the shifted function of the y^x key, located just above the divide key.

Press 4 5 2nd [xth-root] 6 =

The xth-root is the shifted function of the y^x key, located just above the divide key.

Jun 11, 2011 | Texas Instruments TI-30XA Calculator

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=3/2 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

Jan 10, 2011 | Texas Instruments TI-84 Plus Calculator

I am not sure what you are asking for?

Repeated multiplication (with same factor) is called raising a number (base) to a power. The exponent of the power is equal to the number of times the factor is repeated.

The calculator has several buttons to calculate powers.

For the square X*X you have a key [X^2].

For the cube X*X*X you MIGHT have a dedicated key [X^3].

However all calculators (scientific) have a universal power key marked [^] or [X^y] or [Y^x]

Example : In 5x5x5x5x5x5x5 you have a repeated multiplication, so the base is the factor (5) and the exponent is equal to 5. It is equal to the number of times the factor is repeated.

So 5x5x5x5x5x5x5 can be written much more efficiently as 5^7

Entering 5 then pressing the [^] key, followed by the exponent 7 gives the result 78125.

Repeated multiplication (with same factor) is called raising a number (base) to a power. The exponent of the power is equal to the number of times the factor is repeated.

The calculator has several buttons to calculate powers.

For the square X*X you have a key [X^2].

For the cube X*X*X you MIGHT have a dedicated key [X^3].

However all calculators (scientific) have a universal power key marked [^] or [X^y] or [Y^x]

Example : In 5x5x5x5x5x5x5 you have a repeated multiplication, so the base is the factor (5) and the exponent is equal to 5. It is equal to the number of times the factor is repeated.

So 5x5x5x5x5x5x5 can be written much more efficiently as 5^7

Entering 5 then pressing the [^] key, followed by the exponent 7 gives the result 78125.

Jun 23, 2010 | Texas Instruments TI-30X-IISTK Scientific...

Hello,

Sorry this calculator cannot solve equations. It does not have the program. It does not know how to do those things. However it can calculate.**Before you can use the calculator you must prepare all yourself. **When you have solved the problem, you ask the calculator to compute the value of what you found.

Let us clean the equation a bit

2(x-4) -5x=-5

Get rid of the parentheses 2(x-4) becomes 2x-2*4=2x-8 .Put that result where it was in equation

2x-8 -5x=-5

Group together the terms that have x in them 2x-5x=-3x

Then

2x-8-5x=-5, becomes -3x-8=-5

You want to isolate the term with x (have it on one side, and the others on the other side) -3x= -5 -(-8)= -5+8 =3.

The equation becomes

-3x=3 Thus**-x=3/3=1 and x=-1.**

**In general you would have obtained x= (some number/some other number) and that is where the calculator would intervene.**

Hope it helps.

Sorry this calculator cannot solve equations. It does not have the program. It does not know how to do those things. However it can calculate.

Let us clean the equation a bit

2(x-4) -5x=-5

Get rid of the parentheses 2(x-4) becomes 2x-2*4=2x-8 .Put that result where it was in equation

2x-8 -5x=-5

Group together the terms that have x in them 2x-5x=-3x

Then

2x-8-5x=-5, becomes -3x-8=-5

You want to isolate the term with x (have it on one side, and the others on the other side) -3x= -5 -(-8)= -5+8 =3.

The equation becomes

-3x=3 Thus

Hope it helps.

Sep 11, 2009 | Texas Instruments TI-30XA Calculator

In any scientific calculator log2(n) can be calculated with either ln or log function as
follows

Log2(n)= ln(n) / ln(2)

Or

Log2(n)=log(n) / log(2)

both will give nearly the same answers

Log2(n)= ln(n) / ln(2)

Or

Log2(n)=log(n) / log(2)

both will give nearly the same answers

Dec 08, 2007 | Casio FX-300MS Calculator

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