Solve A=Q + ms for Q.

timofeeve [1]

The question above was not written properly

Complete Question

Lacey and Haley are rewriting expressions in an equivalent, simpler form.

a. Haley simplified x³⋅ x² and got

x⁵

Lacey simplified x³ + x² and got the same result! However, their teacher told them that only one simplification is correct. Who simplified correctly and how do you know?

b. Haley simplifies 3⁵⋅ 4⁵ and gets the result 12^10, but Lacey is not sure.

Is Haley correct? Be sure to justify your answer.

Answer:

a) Haley is correct, Lacey simplified wrongly.

b) Haley is incorrect

Step-by-step explanation:

a. Haley simplified x³⋅ x² and got

x⁵

Lacey simplified x³ + x² and got the same result! However, their teacher told them that only one simplification is correct. Who simplified correctly and how do you know?

For Question a, when it comes to simplifying algebraic expression that has to do with powers, there are certain rules that should be followed.

For example

x^a × x^b = x^(a + b)

For Haley, she simplified x³⋅ x² and got

x⁵

She is correct because this follows the product rule of powers or exponents above

= x³⋅ x² = x³+² = x⁵

For Lacey she is wrong because:

x³ + x² ≠ x⁵

x³ + x² when simplified as quadratic equation = x²(x + 1)

b. Haley simplifies 3⁵⋅ 4⁵ and gets the result 12^10, but Lacey is not sure.

Is Haley correct? Be sure to justify your answer.

For question b, when we have two distinct or different numbers with the same power(exponents) the rule states that:

x^a × y^a = (x × y)^a = (xy)^a

Haley is simplified wrongly. She did not apply the rule above

Haley simplified 3⁵⋅ 4⁵ = (3 × 4) ⁵+⁵

= 12^10, this is wrong.

The correct answer according to the rule =

3⁵⋅ 4⁵ = (3 × 4) ⁵ = 12⁵

Therefore,

3⁵⋅ 4⁵ ≠ 12^10

3⁵⋅ 4⁵ = 12⁵

Haley is wrong.

3 0

3 years ago

If x= eighth root of y and y= z^16 how is x proportionate to z ?

ankoles [38]

$x= \sqrt[8]{y}
\\ y = z^{16}

$

Plug "y" into the first equation.

$x = \sqrt[8]{z^{16}} = (z^{16})^{1/8} = z^{16/8} = z^2$

8 0

3 years ago

Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the

Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A= $\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]$ , to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B= $\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$

the system Bx=0 can be represented in matrix form as:

$\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have $B_{0}$ :

$B_{0}$ = $\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]$ . The rank of $B_{0}$ can be found by using the second column and third column pair as follows:

| $B_{0}$ |=(3*0)-(0*2)=0 i.e, $B_{0}$ is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is $\neq$ 0.

Comparing the rank of both B and $B_{0}$ , it is obvious that

Rank of B $\neq$ Rank of $B_{0}$ since (-2)<1.

Therefore, we can conclude that equation(1) is inconsistent and thus has no solution.

(2) If B= $\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ is the transpose of matrix A= $\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right]$ , then

Then the equation Bx=0 is represented as:

$\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.

$B_{0}$ = $\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right]$ ,

| $B_{0}$ |=(5*0)-(0*10)=0-0=0 i.e $B_{0}$ has a rank of order 1.

we can therefor conclude that since

rank B=rank $B_{0}$ =1, equation(2) is consistent and has 2 solutions for the 2 unknown ( $X_{1}$ and $X_{2}$ ).

Summary:

Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
Determine the rank of both the coefficients matrix B and $B_{0}$ which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.

5 0

3 years ago

Samira is playing a game with the spinner shown.

liraira [26]

Answer:

2/10 or 1/5

Step-by-step explanation:

there are 10 equal 'slices'

then count those with a J

that would be 2

7 0

3 years ago

At the zoo for every 6 adult camelsthere are 5 baby camels there are a total of 44 adult and baby camels at the zoo how many bab

slavikrds [6]

6adults = 5 babies
12 adults =10 babies
18 adults= 15 babies
24 adults= 20 babies
= 44 total camels
20 babies

7 0

3 years ago