How Do I Find An Inverse Matrix On An Nspire

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First, make sure that your matrix is “square”. That means, it must be 2×2 or 3×3 or 4×4….

Then, press the Math Templates button (to the right of the number 9, and to the left of the “book”) and access the correct shape of the matrix you wish to enter. The first possible matrix template is for a 2×2 matrix. That is what I selected to enter my example matrix that you also see on the screen. If you wish to enter a 3×3 or larger square matrix, you will select the last matrix template shape (6th icon from the left, or the one just to the left of the sigma notation).

Now, see the image above to see the 2×2 matrix and its inverse that I typed into my TI-nspire. You will finish entering the four numbers inside the brackets, and press the ^ button, followed by -1. This does NOT mean the -1 power, it does NOT mean the reciprocal, in the context of a matrix, this symbol means INVERSE.

If you need to be able to find the inverse by hand….ask again for help!

 
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How Do I Find Capital Lease Obligations

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How do I find capital lease obligations?

six ovens were rented December 31, with $20,000 charged to rent expense. The lease runs for 6 years with an implicit interest of 5%. At the end of 6 years, Peyton will own them. Make any necessary adjusting entries.

 
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How Do I Find Half Life Rate Of Disappearance Constant And Order

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work Help With Chegg x * How do I find rates and running X+/takeAssignment/takeCovalentActivity.do?locator=assignment-take&takeAssignmentSessionLocator=assignment-take[References]Use the References to access important values if needed for this question.In a study of the decomposition of hydrogen iodide on a gold surface at 150HI(g)-21/2 H2(g) + 1/2 12(g)the following data were obtained:[HI], M0.5940.2970.1497.45×10-2seconds521781911Hint: It is not necessary to graph these data.Visited(1)The observed half life for this reaction when the starting concentration is 0.594 M iss and when the starting concentration is 0.297 M is(2)The average rate of disappearance of HI from t = 0 s to t= 521 s isM s-1>(3)The average rate of disappearance of HI from t = 521 s to t = 781 s isM s-1.(4)Based on these data, the rate constant for thisorder reaction isM s-1Submit AnswerRetry Entire Group8 more group attempts remainingCengage Learning | Cengage Technical Support

 
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How Do I Find Molecular Formula When Molar Mass And Percent Composition Are Give

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How do I find molecular formula when molar mass and percent composition are given?

(This is my question. I want to understand how to do this)A compound with the molar mass of approximately 142 g/mol has the composition 

 
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How Do I Find The Additive Identity Matrix

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The sum between two matrices can be done if and only if the two matrices are similar, that means that they have the same numbers of rows and columns.

Also the additive has to be similar to the other, so every shape of matrix has its identity matrix.

The elements, obviously, are all .

E.G.

The matrix with the shape: 3 rows and 4 columns has this identity matrix:

##0000####0000####0000##

 
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How Do I Find The Constant Term Of A Binomial Expansion

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The expansion of a binomial is given by :

##(x+y)^n=( (n), (0) )*x^n+( (n), (1) )*x^(n-1)*y^1+…+( (n), (k) )*x^(n-k)*y^k+…+( (n), (n) )*y^n = sum_(k=0)^n*( (n), (k) )*x^(n-k)*y^k ##where ##x, y in RR##, ##k, n in NN##, and ##( (n), (k) )## denotes combinations of ##n## things taken ##k## at a time.

## ( (n), (k) )*x^(n-k)*y^k ## is the general term of the binomial expansion.

We also have the formula: ##( (n), (k) )=(n!)/(k!*(n-k)!)##, where ##k! = 1*2*…*k##

We have three cases:

Case 1: If the terms of the binomial are a variable and a constant ##(y=c##, where ##c## is a constant), we have ##(x+c)^n=( (n), (0) )*x^n+( (n), (1) )*x^(n-1)*c^1+…+( (n), (k) )*x^(n-k)*c^k+…+( (n), (n) )*c^n ##

We can see that the constant term is the last one: ##( (n), (n) )*c^n## (as ##( (n), (n) )## and ##c^n## are constant, their product is also a constant).

Case 2: If the terms of the binomial are a variable and a ratio of that variable (##y=c/x##, where ##c## is a constant), we have:## (x+c/x)^n=( (n), (0) )*x^n + ( (n), (1) )*x^(n-1)*(c/x)^1+…+( (n), (k) )*x^(n-k)*(c/x)^k+…+( (n), (n) )*(c/x)^n ##

This time, we see that the constant term is not to be found at the extremities of the binomial expansion. So, we should have a look at the general term and try to find out when it becomes a constant:## ( (n), (k) )*x^(n-k)*(c/x)^k=( (n), (k) )*x^(n-k)*c^k*1/x^k = (( (n), (n) )*c^k)*(x^(n-k))/x^k = (( (n), (k) )*c^k)*x^(n-2k) ##.

We can see that the general term becomes constant when the exponent of variable ##x## is ##0##. Therefore, the condition for the constant term is: ##n-2k=0 rArr## ##k=n/2## . In other words, in this case, the constant term is the middle one (##k=n/2##).

Case 3: If the terms of the binomial are two distinct variables ##x## and ##y##, such that ##y## cannot be expressed as a ratio of ##x##, then there is no constant term . This is the general case ##(x+y)^n##

 
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How Do I Find The Directrix Of The Parabola Y 2x 2

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To find the directrix of a parabola wohse equation is ##y=ax^2+bx+c## you can use this formula:##y= (-1-(b^2-4ac))/(4a)##So in case of this parbola ##y=2x^2##the directrix is##y=(-1-(0^2-4*2*0))/(4*2)####y=-(1)/(8)##

 
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How Do I Find The Dot Product Of 1 I 2 3i And 2 I 3 2i

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how do I find the dot product of (1 – i , 2+3i) and (2 + i, 3 – 2i) ?

I know there is something with conjugates of a2, and b2 (I bolded them) I get 1 + 10 i and the book says 6 +21i… this doesn’t match….

 
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How Do I Find The Foci Of An Ellipse If Its Equation Is X 2 36 Y 2 64 1

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##x^2/36+y^2/64=1##

##a^2=64####b^2=36##

The center is ##(0,0)##.

The y-axis is the major axis. We know this because the ##y^2## term has the larger denominator.

We find the foci using the following equation and solving for ##c##.

##c^2=a^2-b^2####c^2=64-36####c^2=28####c=sqrt(2*2*7)####c=2sqrt(7)##

The coordinates for the foci are ##(0,+-c) -> (0,+-2sqrt(7))##

 
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How Do I Find The Magnitude And Direction Angle Of The Vector V 6i 6j

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Magnitude = ##|6i-6j|=sqrt(6^2+(-6)^2)=sqrt(72)=6sqrt(2)”##Angle = ##arctan(-6/6)=arctan(-1)=-pi/4=-45°##

If you draw this vector on the plane, the x-y coordinate axis, as (6,-6) because ##i## is the unit vector in the x-direction and ##j## in the y-direction. Then pictorially it might seem easier.

So if you draw the vector (so join the line between the point (6,-6) and the origin), and draw the perpendicular line between the point and the x-axis. Now, you should see a triangle, more specifically a right-angles triangle.

Now that you have you’re right-angles triangle you can use Pythagoras to find the magnitude of the vector (length of the straight line between the origin and the point (6,-6)) and the use SOHCAHTOA to find the angle between the x-axis and the vector (be careful, here if the vector is the on the left-hand side of the plane i.e. the x-coordinate is negative, then you will have to make sure the you take the angle from the positive side of the x-axis, so do ##pi/2(1-theta)##.

Some quick formulae that may be helpful to do quick calculation:magnitude of ##ai+bj=|ai+bj|=sqrt(a^2+b^2)####theta=arctan(b/a)##if ##b<0## then angle of ##ai+bj=pi/2(1-theta)##if ##bge0## then angle of ##ai+bj=theta##

 
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