Question about Texas Instruments TI-84 Plus Calculator

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Assuming you want the solution graphically.

You press the Y= button (top left)

Enter the first side of the equation after Y1=

Press the down arrow

Enter the other side of the equation after Y2=

Set the setting by pressing the window button (set xmin to -10; xmax to 10; ymin to -10; ymax to 10 )

Press the graph button

You will see now the two functions potted.

Now press Calc ( 2nd and after that Trace)

Select the correspondant number for intersect.

Set min to -10

Press enter

Set max to 10

Press enter

Guess something between -10 and 10 (eg. 0)

The answer will come out quickly (x=-3,5)

Hope it helps.

Posted on Mar 16, 2011

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Posted on Jan 02, 2017

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Use FOIL, first, outside, inside and last.

First 2x * x = 2x^2

Outside 2x * -1 = -2x

Inside -10* x = -10x

Last -10 * -1 = 10

Thus, we have 2x^2 -2x -10x + 10

Collecting like terms, we get 2x^2 -12x + 10

Good luck,

Paul

First 2x * x = 2x^2

Outside 2x * -1 = -2x

Inside -10* x = -10x

Last -10 * -1 = 10

Thus, we have 2x^2 -2x -10x + 10

Collecting like terms, we get 2x^2 -12x + 10

Good luck,

Paul

Nov 29, 2016 | Office Equipment & Supplies

Now is the equation supposed to be y=2x+1? If so,

The graph of y = 2x+1 is a straight line

When x increases, y increases twice as fast, hence 2x

When x is 0, y is already 1. Hence +1 is also needed

So: y = 2x + 1

Here are some example values:

xy = 2x + 1

-1y = 2 × (-1) + 1 = -1

0y = 2 × 0 + 1 = 1

1y = 2 × 1 + 1 = 3

2y = 2 × 2 + 1 = 5

Check for yourself that those points are part of the line above!

The graph of y = 2x+1 is a straight line

When x increases, y increases twice as fast, hence 2x

When x is 0, y is already 1. Hence +1 is also needed

So: y = 2x + 1

Here are some example values:

xy = 2x + 1

-1y = 2 × (-1) + 1 = -1

0y = 2 × 0 + 1 = 1

1y = 2 × 1 + 1 = 3

2y = 2 × 2 + 1 = 5

Check for yourself that those points are part of the line above!

Jan 16, 2015 | SoftMath Algebrator - Algebra Homework...

I assume the 10x2 means 10 squared. (I will show it as 10x**2)

(22x+11)(10x**2+5x) factors into"

11(2x+1)*5(2x**2+1x)

=55(2x+1)(2x**2+1x)

=(55x)(2x+1)(2x+1)

Multiplied out, it becomes 55x(4x**2+4x+1)

= 220x**3 + 220x**2 + 55x

(22x+11)(10x**2+5x) factors into"

11(2x+1)*5(2x**2+1x)

=55(2x+1)(2x**2+1x)

=(55x)(2x+1)(2x+1)

Multiplied out, it becomes 55x(4x**2+4x+1)

= 220x**3 + 220x**2 + 55x

Dec 11, 2014 | Computers & Internet

Melissa. The 2X-1 is for use on a M42 screw connection, like on Pentax The 2X-3 is for Nikon.

Dec 19, 2013 | Vivitar Auto 2x Custom Tele-converter...

This has nothing to do with laptops, please remove it from this section. The answer is
2x+y=1
so y=1-2x
therefore 1-(2X-2)
=1--4
Y= 5
so do the same thing again
1-(2X0)
1-0
Y=1
and again
1-(2X1)
1-2
Y=-1

Jan 25, 2011 | Computers & Internet

=>2x+3=0 ; taken x as common and 4-2=2 i.e. x(4-2) = 2x

=>2x=-3 ; +3's moved on the right-hand side, so it becomes -3

=>x=-3/2 ; 2 has been moved to the right-hand side and divided with -3 i.e 1/2 multiplied by -3 = -3/2

Good luck.

Thanks for using

Aug 13, 2010 | SoftMath Algebrator - Algebra Homework...

Hi,

Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:

a = 2x-1, b = 2x+1, c = 2x+3

Now, your problem looks like this:

2(2x-1)(2x+1) = (2x+1)(2x+3) + 7

2(4x^2-1) = 4x^2+6x+2x+3 + 7

8x^2-2 = 4x^2+8x+10

4x^2-8x-12 = 0

4(x^2-2x-3) = 0

x^2-2x-3 = 0

The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:

[1] a = 5, b = 7, c = 9

[2] a = -3, b = -1, c = 1

Take care,

Alex

Let's say your 3 numbers are a, b and c. Because they are consecutive and odd, you can write them like this:

a = 2x-1, b = 2x+1, c = 2x+3

Now, your problem looks like this:

2(2x-1)(2x+1) = (2x+1)(2x+3) + 7

2(4x^2-1) = 4x^2+6x+2x+3 + 7

8x^2-2 = 4x^2+8x+10

4x^2-8x-12 = 0

4(x^2-2x-3) = 0

x^2-2x-3 = 0

The solutions for this quadratic equation ( see more about this kind of equations here: http://en.wikipedia.org/wiki/Quadratic_equation ) are x=3 or x=-1. So, you've got two sets of solutions:

[1] a = 5, b = 7, c = 9

[2] a = -3, b = -1, c = 1

Take care,

Alex

Oct 15, 2009 | Google Computers & Internet

sec^4X- sec^2X = 1/cot^4X + 1/cot^2X

RHS

1/cot^4X + 1/cot^2X

=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)

=Sin^4X/Cos^4X + Sin^2X/Cos^2X

=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X

=Sin^2X/Cos^4(Sin^2X + Cos^2X)

=Sin^2X/Cos^4X

=(1-Cos^2X)/Cos^4X

=1/Cos^4X - Cos^2X/Cos^4X

=1/Cos^4X - 1/Cos^2X

=Sec^4X - Sec^2X

=LHS

RHS

1/cot^4X + 1/cot^2X

=1/(Cos^4X/Sin^4X) + 1/(Cos^2X/Sin^2X)

=Sin^4X/Cos^4X + Sin^2X/Cos^2X

=Sin^4X/Cos^4X + Cos^2X.Sin^2X/Cos^4X

=Sin^2X/Cos^4(Sin^2X + Cos^2X)

=Sin^2X/Cos^4X

=(1-Cos^2X)/Cos^4X

=1/Cos^4X - Cos^2X/Cos^4X

=1/Cos^4X - 1/Cos^2X

=Sec^4X - Sec^2X

=LHS

Feb 02, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

(sinX-cosX)(sinX+cosX)
=(sin^2 - Cos^2X)

=Sin^2X - (1-Sin^2X)

=Sin^2X -1 + Sin^2X

=2Sin^2X -1

Zulfikar Ali

ali_zulfikar@yahoo.com

9899780221

=Sin^2X - (1-Sin^2X)

=Sin^2X -1 + Sin^2X

=2Sin^2X -1

Zulfikar Ali

ali_zulfikar@yahoo.com

9899780221

Jan 26, 2009 | Super Tutor Trigonometry (ESDTRIG) for PC

B. 2((2-Y)/3)-Y=3

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

2(2-Y)-Y=9

4-2Y-Y=9

-3Y=9-4

-3Y=5

-Y=5/3

Y=-1.67

A. 3X+(-1.67)=2

3X-1.67=2

3X=2+1.67

3X=3.67

X=3.67/3

X=1.22

Nov 05, 2007 | SoftMath Algebrator - Algebra Homework...

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