Questions on Algebra: Matrices, determinant, Cramer rule answered by real tutors!

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I think this is your problem.

If not, please let me know.





The answer is A.

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I'm guessing on the make up of the matrices.
First off let's look at [C][F].
[C]=
[F]=
[C][F]=
where each element of [C][F] comes from multiplying a row of [C] with a column of [F].
Example: First element is product of first row and first column.

.
.
.
Now that we have [C][F], we can subtract it from [B], element by element,
[B]-[C][F]=
[B]-[C][F]=
.
.
.
If this is not how the matrices look,please re-state the problem and be more specific about the make up of the matrices (rows x columns).
Here's an example.
[A] is a 2x2 matrix. A=[1,2,3,4].
The assumption is that [A] looks like this,
[A]=
[B] is a 3x2 matrix. B=[5,6,7,8,9,10]
[B]=

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need a problem..thanks

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I am going to assume that .6 is a typo and call it 6 it I am right you have your answer if wrong just follow the method and replace it with .6
:

the R1C1 element is found by 2(3)-8(2)=-10
the R1C2 element is found by 2(0)-8(-1)=8
the R2C1 element is found by 6(3)-3(2)=12
the R2C2 element is found by 6(0)-3(-1)=9
:

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start with augmented matrix
:
(R1)/3-6R1+R2R2/-11-(5/3)R2+R1
:
answer is 

remember that R1 is Row 1
.................R2 is Row 2
:
and when you see something like R1/3 that is all the elements in Row 1 being divided by 3. Maybe you'll see 6R1+R2..thats elements in Row 1 multiplied by 6 and added to elements in Row 2.

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solve the system of equations using the Gauss-Jordan elimination?
3x+5y=7
6x-y=-8
-------------
Multiply thru the 2nd equation by 5 to get:
30x - 5y = -40
Add that to the 1st equation to solve for "x":
33x = -33
x = -1
-------------
Substitute into 3x+5y=7 to solve for "y":
3*-1 + 5y = 7
5y = 10
y = 2
==================
Cheers,
Stan H.

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Solved by pluggable solver: Finding the Determinant of a 3x3 Matrix

If you have the general 3x3 matrix: the determinant is: Which further breaks down to: Note: , and are determinants themselves. If you need help finding the determinant of 2x2 matrices (which is required to find the determinant of 3x3 matrices), check out this solver -------------------------------------------------------------- From the matrix , we can see that , , , , , , , , and Start with the general 3x3 determinant. Plug in the given values (see above) Multiply Subtract Multiply Combine like terms. ====================================================================== Answer: So , which means that the determinant of the matrix is

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Determinant of A = 4*(-2)-3*(-7)=-8+21=13
I hope that this helps

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Determinant of G = 12(2)-0*(-6)=24-0=24
I hope that this helps

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detG=det= 12(2)-0(-6)=24

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How do I find the element of a(base 1),(base 2)?

You shouldn't put the word "base" there. You should put "row 1" 
instead of "base 1" and "column 2" instead of "base 2". The 
first number in , the , is the row number 
going from top to bottom, and the second number  is the 
column number going from left to right.  You can remember which
is which by remembring that  looks like the word 
"arc".  

          C C C 
          O O O 
          L L L
          U U U
          M M M
          N N N
          # # # 
          1 2 3 
          ¯ ¯ ¯ 
ROW #1 ® 1 2 3 
ROW #2 ® 4 5 6 
ROW #3 ® 7 8 9

The 1 is in ROW #1 and COLUMN #1, so 1 is the element 
The 2 is in ROW #1 and COLUMN #2, so 2 is the element 
The 3 is in ROW #1 and COLUMN #3, so 3 is the element 
The 4 is in ROW #2 and COLUMN #1, so 4 is the element 
The 5 is in ROW #2 and COLUMN #2, so 5 is the element 
The 6 is in ROW #2 and COLUMN #3, so 6 is the element 
The 7 is in ROW #3 and COLUMN #1, so 7 is the element 
The 8 is in ROW #3 and COLUMN #2, so 8 is the element 
The 9 is in ROW #3 and COLUMN #3, so 9 is the element 

Your answer is the red one.

Edwin

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Matrix addition and subtraction is performed by adding/subtracting comparable components of each matrix.
A+B = [3+2,0+8,2+.6,-1+3]=[5,8,2.6,2]
I hope that this helps

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I am assuming that A is a 2x2 since you cannot multiply a 1x2 and a 1x4 matrix
:
=
:
now multiply by 2 =

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4c=4(12)=48
4c=4(0)=0
4c=4(3/2)=6
4c=4(1)=4
4c=4(-6)=-24
4c=4(7)=28
:

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4C is just the matrix that has all the same elements as C, only multiplied by 4. For example, the first element would be 4(12) = 48.

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Determinants can be used to solve a linear system of equations using Cramer’s Rule.
Cramer’s Rule is used for Two Equations in Two Variables, which is what you have: variables x and y.
2x+3y = 41
2x-3y = -13
To find the determinant, we use:
D = 2(-3) - 2(3)
D = -6 -6
D = -12
To find x use:
x = Dx/D
We already know D, right?
We need to find Dx.
We find Dx = 41(-3) - 3(-13)
Dx = -123 + 39
Dx = -84
We can now find x.
So, x = -84/-12
Then x = 7
Lastly, we need to find y.
y = Dy/D
We know D to be -12, right?
To find y, we need to first find Dy.
We can find Dy = 2(-13) - 2(41)
Dy = -26 - 82
Dy = -108
We can now find y.
y = -108/-12
y = 9
The solution to the above system of equations in two variables is the point
(7, 9)
The two equations given to you meet or cross each other at the point (7, 9) and so, this is why that particular point is the solution.
Understood?
The answer is choice (A).

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If you plug in the x and y values for b you will find it incorrect.
x=-9,y=7
2x+3y=41 -18+21=39
2x-3y=-13 -18-21=-39
2x+3y=41
2x-3y=-13
Add each column
4x+0=28
divide both sides by 4
x=7
Plug in the value of 7 for x to get the y value
14+3y=41
Subtract 14 from both sides
3y=27
divide both sides by 3
y=9
Plug in the values of x and y in the second equation
(2*7)-(3*9)=-13
-13=-13

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can you be more specific...you can name a matrix anything you choose matrix A,
matrix B, matrix (Keanu Reeves)......or anything else you can think of. Naming a matrix or variable within math, is simply a process of identifying that item so that it doesnt get confused with something else.

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A=

:
adjA= in a 2x2 matrix...the elements of the main diagonal of the matrix A are switched whereas we take the negatives of the remaining elements to obtain the adjugate of A
:
detA=1(4)-2(-3)=10
:
/10=
:
=

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Solve for what?

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There is no UNIQUE solution to this equation because you have 4 unknowns & only one equation.
However you can make up a table & find many combinations od a, c, e & f that satisfus this equation thus:
a---2c---3e---2f=270
50+2*25+3*10+2*70=270
50+50+30+140=270
OR
100+2*10+3*20+2*45=270
100+20+60+90=270
270=270
And there are many additional combinations of a,c,e & f that are solutions.

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The problem asks you to multiply matrices. A^2 =A*A and A^3=A*A*A

1,2,-1
A = 2,1,4
1,-2,1
The first row, first column of A*A = A^2 is calculated by multiplying each element of the first row of A by each element of the first column of A and then each of the products are added together.
The second row, first column of A*A = A^2 is calculated by each element of the second row of A by each element of the first column of A and then each of the products are added together.
The third row, first column element of A*A = A^2 is calculated by each element of the third row of A by each element of the first column of A and then each of the products are added together.
The general rule that is being used is the i'th row and j'th column element of the result is found by multiplying each element in the i'th row of one matrix(multiplier) by each element of the j'th column of the other matrix (multiplican) and then the products are added to together.
So to calculate A^2 you have to calculate 9 numbers using the rule that is described above.
Since A^3 = (A^2)*A, you have to calculate 9 numbers using the rule described above with A^2 being one matrix and A the other matrix.
I hope that this helps but it does take some practice to understand and be able to do matrix multiplication.
Good luck.

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we need the rest of the information for this problem

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[6-a 4] * [-2] = [2a]
[-3 -3(d-2] * [3] = [-3]
----------------------------------------
(6-a)(-2) + 4*3 = 2a
-12+2a + 12 = 2a
2a = 2a
a is indeterminate
-------------------------
-3*-2 + -2(d-2)(3) = -3
6 - 6(d-2) = -3
-6d+12 = -9
-6d = -21
d = 7/2
-------------------------
Cheers,
Stan H.

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The perimeter of a rectangle is 28 cm. The perimeter of each of the triangles is 24 cm. The diagonal of the rectangle is 2 cm longer than the longer side of the rectangle.
Let shorter side be a; let longer side be b: let diagonal be c
Equations:
2a + 2b = 28
a + b + c = 24
c = b + 2
write a system of three questions in three unknowns
---------------
Rearrange the equations:
2a + 2b + 0 = 28
a + b + c = 24
0 + b - c = -2
-----------------------------
simplify to a system of two equations in two unknowns?
let b = c+2
Then you get:
2a + 2(c+2) + 0 = 28
a + c+2 + c = 24
----
Rearrange:
2a + 2c = 24
a + 2c = -2
----
write an augmented matrix for the system?

2...2..|..24
1...2..|..-2
=====================
Cheers,
Stan H.

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let L and W be the sides of the rectangle and Z be the diagonal
:
2L+2W=28.....eq 1
L+W+Z=24.....eq 2
L+2=Z........eq 3
:
take Z's value of L+2 from eq 3 and plug it into eq 2
:
L+W+(L+2)=24--->2L+W=22....revised eq 2
:
now we have 2 eq with 2 unknowns
:
augmented matrix

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lets call the time spent at soup kitchen,trash pickup, and toy collecting
s, t, and c, respectively
:
c=4t...........eq 1
s=t-2..........eq 2
s+t+c=40.......eq 3
:
rewrite equations
:
s+t+c=40......eq 3
s-t+0c=-2.....eq 2
0s-4t+c=0......eq 1
:
now writing out the main matrix
:
...lets call this matrix X
:
we need to find the determinant of this matrix to begin with
:
we can choose any row or column to determine this ..I will work with column 1 since it has a zero in it
:
det X=(1) det-(1)det+0=
.....1(-1)-1(5)=-6
:
s=det s/-6 where matrix s is matrix X with the 1st column replaced by constant coefficents 40,-2,0. I will use column one for our determinant
:
det s = 40 det-(-2)det
=40(-1)+2(5)=-30
:
s=det s/-6=hours in the soup kitchen
:
:
t=det t/-6 where matrix t is matrix X with the 2nd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det t=0+0+(1)det=1(-42)=-42
:
t=det t/-6=hours picking up trash
:
:
c=det c/-6 where matrix c is matrix X with the 3rd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det c= 0-(-4)+0=4(-42)=-168
:
c=det c/-6=hours toy collecting

If you forgot the number of hours you volunteered you could not determine the hours you spend doing each activity....
:
you would only have 2 equations and 3 unknowns

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1. Write a system of equations to represent the given information and solve using the substitution method.
Please help and try to show every bit of work that you can to help me better understand the problem, Thank You! =)
:
lets call the time spent at soup kitchen,trash pickup, and toy collecting
s, t, and c, respectively
:
c=4t...........eq 1
s=t-2..........eq 2
s+t+c=40.......eq 3
:
rewrite equations
:
s+t+c=40......eq 3
s-t+0c=-2.....eq 2
0s-4t+c=0......eq 1
:
now writing out the main matrix
:
...lets call this matrix X
:
we need to find the determinant of this matrix to begin with
:
we can choose any row or column to determine this ..I will work with column 1 since it has a zero in it
:
det X=(1) det-(1)det+0=
.....1(-1)-1(5)=-6
:
s=det s/-6 where matrix s is matrix X with the 1st column replaced by constant coefficents 40,-2,0. I will use column one for our determinant
:
det s = 40 det-(-2)det
=40(-1)+2(5)=-30
:
s=det s/-6=hours in the soup kitchen
:
:
t=det t/-6 where matrix t is matrix X with the 2nd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det t=0+0+(1)det=1(-42)=-42
:
t=det t/-6=hours picking up trash
:
:
c=det c/-6 where matrix c is matrix X with the 3rd column replaced by constant coefficents 40,-2,0. I will use row 3 for our determinant as it has 2 zeros
:
det c= 0-(-4)+0=4(-42)=-168
:
c=det c/-6=hours toy collecting

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120g of fat, 220g of carbohydrates, and 80g of protein.
_______________ A______B_______C
Fat____________ 10 g____ 4g_____ 12 g
Carbohydrates___11 g___77 g_____0g
Protein_________4 g_____1 g_____16 g
:
:
A)10A+4B+12C=120
11A+77B =220
4A+ 1B+16C=80
:
B)using cramers rule
:
find determinant of main matrixusing column 3---> 12-0+16---->12(-297)+16(726)=8052
:
A=det/8052---->so using the 3rd column--->det = 12-0+16--->12(-5940)+16(8360)=62480
:
:
:
B=det/8052---->so using the 3rd column--->det = 12-0+16---->0+16(880)=14080
:
:
C=det/8052---->using column 1--->det= 10-11+4--->10(5940)-11(200)+4(-8360)--->59400-2200-33440=
23760
:
:
:
C)if C was no longer available we would have the equation
:
10A+4B+0C=120
11A+77B+0C=220
4A+ B+0C=80
:
we cannot use Cramers rule because there would be no unique solution as the determinant you would be dividing by would be zero.
:according to this graph there is no unique solution to this scenario.
It appears you could reach 2 of the conditions in any order but not all 3 at once.

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What is the solution to:

The solution to that system is:



If you want to know how to get that solution,
post again stating which method you are
studying.  There are a number of different
ways of solving it, so please be sure to
state which method.  But, of course, all
ways lead to the same solution.

Edwin

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I have a feeling no one is helping you on this because we are all stumped by it too!! I'm retired, and particularly bored today, so I thought I would give it a looksee! I would say first of all, for you to go with your calculator solution!!

I haven't worked with matrices for a VERY long time, but, if your equations are correct and I think they are correct, you need to convert from this system of equations to one in which you have eliminated coefficients to leave 0s and 1s. You may have to try a few times and I must admit that I didn't get the calculator answers the first time. This is what I did. First, reduce the equations to smaller numbers:
10a + 4b + 12c = 120 (Divide by 2)
11a + 77b + 0c = 220 (Divide by 11)
4a + 1b + 16c = 80

5a + 2b + 6c = 60
a + 7b + 0c = 20 (Mult by -4 and add to eq #3)
4a + 1b + 16c = 80

5a + 2b + 6c = 60
a + 7b + 0c = 20 (Mult by -5 and add to eq #1)
0a - 27b + 16c = 0

5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a - 27b + 16c = 0

Next, eliminate the b coefficients by multiplying Eq #2 times -27 and Eq #3 times 33 and add together. (The first two equations remain the same!)
5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a + 0b + 366c = 1080


Next, divide the Eq #3 by 366:
5a + 2b + 6c = 60
0a - 33b + 6c = -40
0a + 0b + 1c = 1080/366

So, c= 1080/366 = 2.9508 approximately, which is the calculator value that I got. I'm sure you can finish this now. By the way, my other calculator values were a=7.759 and b= 1.7486 approximately.

Good luck!!

R^2

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Let = the cost per gallon of red paint
Let = the cost per gallon of yellow paint
Given is:
A hardware store mixes paints in a ratio of two parts red to six parts yellow to make pumpkin orange.
Therefore, The gallons of red per gallon of punpkin orange is
gal
gal
The gallons of yellow per gallon of pumpkin orange is
gal
gal
Therefore,
(1)
Also given:
A ratio of five parts red to three parts yellow makes red-pepper red.
The gallons of red in a gallon of red-pepper red is
gal
gal
The gallons of yellow in a gallon of red-pepper red is
gal
Therefore,
(2)
(A)
Multiply both sides of (1) by
(1)
Multiply both sides of (2) by
(2)
(B)
Subtract (1) from (2)


Substitute this in (1)



The cost per gallon of red paint is $31
The cost per gallon of yellow paint is $23
(C) There are 4 quarts in a gallon,so
red paint is
/qt
yellow paint is
/qt

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Let x = the amount invested at 10%, then x-500 = the amount invested at 8%, so that 2x-500 would be the amount invested at 7%.
The sum of the three invested amounts (x)+(x-500)+(2x-500) is $5,000.
Now let's look at the amounts of interest earned on these amounts:
is the amount earned at 10 %.
is the amount earned at 8%.
is the amount earned at 7%.
The sum of these three amounts is given as $405.
So we can write the following equation in x.
Simplify and solve for x.

Add 75 to both sides.
Divide both sides by 0.32

$1,500 was invested at 10%
Check:

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:

take terms and  and switch their positions. This is simply switching the elements on the main diagonal 

 

take terms and  and change those numbers to their opposites keeping there positions.





:now we find the determinant of this matrix

:

product of the main diagonal-product of the other diagonal

:



:

we take that result and divide every element in the matrix and the result is our inverse matrix

:



the way you can check is to multiply the two together and see if your result is the identity matrix....if it is you can rest easy you have the correct answer.

:



:

:

:

2)and 3)

:

now I will let you do the steps to 2 and 3

:

answers to 2 is 

...........3 is no solution as the determinant = 0

:

4)X or=

3a+4b=6

2a+3b=5

:

3R2-2R1(Row 2)---->R1-4R2(row 1)---->1/3R1---> so we end with 

a=-2

b=3

so X=

:

please re write equation 5  making it clear what their after

:

6)6.[1 -6 0]      [1}

    [0 1 -7] [X} =[4]

    [3 0 2]       [11]



In order to go from a 3x3 to a 3x1 X also has to be a 3x1

:

call X=

:

a-6b=1.....eq 1

b-7c=4.....eq 2

3a-2c=11...eq 3

:

lets solve by substution method

:

revise eq 1 to a=1+6b and plug that value into eq 3

:

3(1+6b)-2c=11

:

3+18b-2c=11

:

18b-2c=8....eq 3 revised

  b-7c=4....eq 2

:

multiply eq 2 by -18 and add the results to revised eq 3

:

18b-2c=8.......eq 3 revised

-18b+126c=-72..eq 2 revised

:

as you can see the b terms are eliminated. We are left with -2c+126c=8-72

:

124c=-64

:

c=-64/124=-32/62=-16/31

:

b-7(-16/31)=4

:

b+112/31=4...multiply by 31

:

31b+112=124

:

31b=12

:

b=12/31

:

a=1+6(12/31)

a=31/31+72/31

:

a=103/31

:

X=

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:

We must find the inverse of matrix A in order to solve for x

:

A= [6,5]    

   [4,-2]

:

take terms and  and switch their positions. This is simply switching the elements on the main diagonal 

 

take terms and  and change those numbers to their opposites keeping there positions.



-revised matrix

:now we find the determinant of this matrix

:

product of the main diagonal minus the product of the off diagonal

:



:

we take that result and divide every element in the revised matrix  and the result is our inverse matrix

:



: 

Now we have to multiply each side of the equation AX=B by the matrix A's inverse

On the left hand side we will end up with the indentity matrix multiplied by X which is equal to just X. So 

:

A= [6,5]    B= [18,49]

   [4,-2]      [-20,6]

:

so 

X=*---->

:

=

:

2) I will leave the details and steps of # 2 to you 

:

the answer is 

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:
take terms and  and switch their positions. This is simply switching the elements on the main diagonal 
 
take terms and  and change those numbers to their opposites keeping there positions.


:now we find the determinant of this matrix
:
product of the main diagonal-product of the other diagonal
:

:
we take that result and divide every element in the matrix and the result is our inverse matrix
:

the way you can check is to multiply the two together and see if your result is the identity matrix....if it is you can rest easy you have the correct answer.
:

:
:
:
2)and 3)
:
now I will let you do the steps to 2 and 3
:
answers to 2 is 
...........3 is no solution as the determinant = 0
:
4)X or=
3a+4b=6
2a+3b=5
:
3R2-2R1(Row 2)---->R1-4R2(row 1)---->1/3R1---> so we end with 
a=-2
b=3
so X=
:
5) is done the same as 4 but this time you have 3 variables....have fun

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First find the inverse of  by

1. find the determinant 



2. Swap the elements is the upper left to lower right diagonal.
   Change the signs in the other diagonal.




3. Divide the matrix through by the determinant

 

That is the inverse matrix. So left-multiply both sides
of the original equation by this inverse:










Edwin

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1. Write the equation of a polynomial that has zeros at –3 and 2.
Equation: f(x) = (x+3)(x-2)
---------------------------------------
Write each polynomial function in standard form. Then classify it by degree and by the number of terms.
2. n = 4m2 – m + 7m4
n = 4m3 + 4m – 2; cubic trinomial
n = 7m4 + 4m2 – m; quartic trinomial---Correct
n = 4m4 + 8m2 – m ; quartic trinomial
n = 3m3 + 2m – 5; cubic trinomial
-----------------------------------------
3. f(t) = 4t + 3t3 + 2t – 7
f(t) = 3t3 + 6t – 7; cubic trinomial-----Correct
f(t) = 2t3 + 4t – 1; cubic trinomial
f(t) = 5t3 + 2t – 7; cubic trinomial
f(t) = 7t3 + 3t – 4; cubic trinomial
------------------------------------------
4. f(r) = 5r + 7 + 2r2
f(r) = 8r3 + 5r + 1; cubic trinomial
f(r) = r2 + 5r + 7; quadratic trinomial
f(r) = 2r3 + 5r + 7; cubic trinomial
f(r) = 2r2 + 5r + 7; quadratic trinomial----Correct
------------------------------------------
For each function, determine the zeros and their multiplicity.
5. y = (x + 2)^2(x – 5)^4
–2, multiplicity 2; 5, multiplicity 4----Correct
2, multiplicity 2; 5, multiplicity 3
–2, multiplicity 4; 5, multiplicity 2
2, multiplicity 2; 7, multiplicity 4
------------------------------------------
6. y = (3x + 2)^3(x – 5)^5
4, multiplicity 2; 5, multiplicity 2
3, multiplicity 4; 2, multiplicity 4
-2/3, multiplicity 3; 5, multiplicity 5----Correct
2, multiplicity 3; 5, multiplicity 5
-------------------------------------------
7. y = x^2(x + 4)^3(x – 1)
2, multiplicity 2; –4, multiplicity 2; 1, multiplicity 1
0, multiplicity 2; –4, multiplicity 3; 1, multiplicity 1----Correct
3, multiplicity 4; –3, multiplicity 2; 2, multiplicity 3
0, multiplicity 2; –3, multiplicity 2; 2, multiplicity 3
-------------------------------------------

8. (x^3 + 3x^2 – x – 3) ÷ (x – 1)
1)....1....3....-1....-3
.......1....4.....3....|..0
Quotient: x^2 + 4x + 3
Remainder: 0
--------------------------------------------
10. Use synthetic division to find P(–3) for P(x) = –2x^4 – 3x^3 – x + 4
-3)....-2....-3....-1....4
.........-2....3.....-8..|..28
Quotient: -2x^2 + 3x -8
Remainder: 28
====================
Cheers,
Stan H.

You can put this solution on YOUR website!
at a local farmers market,Jane sold 27 squash,31 tomatoes,24 peppers,and 18 melons.Jose sold 48 squash,72 tomatoes,61 peppers, and 25 melons.
a)Create a 2x4 matrix of this data.Name this matrix P.

Or if your teacher doesn't want headings, then just use





b)What is the address of the number of peppers that jane sold?

1,3, or "Row 1, Column 3"

c)What is the address of the data stored in the second row and first column?

2,1

What does this entry represent?

The number of squash that Jose sold.


d)Could you have created a matrix with different dimensions from the created part a?

Yes, this 4x2 matrix:

 

or without headings 

Edwin