MATHEMATICS COLLEGE

Rewrite the equation by completing the square. x 2 + 6 x + 9 = 0 x 2 +6x+9=0

Answers

Answer 1

Answer:

Answer:

(x +3)^2 = 0

Step-by-step explanation:

The square is already complete. We can write the equation as a square:

(x +3)^2 = 0

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At the hairdresser, jenny had 27 centimeters cut off her hair. How many decimeters of hair did jenny have cut off

Answers

cm to decimeter is division by 10
so 27 cm = 2.7 dm

27 / 10 = 2.7 decimeters.

What is the common difference in the arithmetic sequence 10, 8, 6, 4, ...?

Answers

The common difference of the given A.P. is -2.

What is a expression? What is a mathematical equation? What is Equation Modelling? What is an arithmetic sequence?

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. An arithmetic sequence is a sequence of numbers where the differences between every two consecutive terms is the same. The nth term of an A.P. is given by -

a[n] = a + (n - 1)d

We have the following sequence 10, 8, 6, 4, ... .

We have -

10, 8, 6, 4, ...

The common difference of the given A.P. is -

d = 8 - 10 = 6 - 8 = 4 - 6

d = - 2

Therefore, the common difference of the given A.P. is -2.

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Answer:

-2

Step-by-step explanation:

10-8=2

8-6=2

6-4=2

4-2=2

Determine whether [1 0 3 , −3 1 −7 , 5 −1 13] is a basis for set of real numbers R cubed 3. If the set is not a basis, determine whether the set is linearly independent and whether the set spans set of real numbers R cubed 3.

Answers

Answer:

The set is not a basis. It is not linearly independent and doesn't span the given vector space

Step-by-step explanation:

Let u = (1,0,3), v = (-3,1,-7) and w=(5,-1,13). We want to check if the set {u,v,w} is a basis for $\mathbb{R}^3$ . By definition, a basis is a linearly independent set that spans the vector space. So, if it is a basis, it automatically is linearly independent and spans the whole space. Since we have 3 vectors in

$A=\left[\begin{matrix}1 & -3 & 5 \n 0 & 1 & -1 \n 3 & -7 & 13 \end{matrix}\right]$

which is the matrix whose columns are u,v,w. To check that the set {u,v,w} is linearly independent,it is equivalent to check that the row-echelon form of A has 3 pivots.

The step by step calculation of the row-echelon form of A is ommited. However, the row-echelon form of A is

$A=\left[\begin{matrix}1 & 0 & 2 \n 0 & 1 & -1 \n 0 & 0 & 0 \end{matrix}\right]$

In this case, we have only 2 pivots on the first and second column. This means that the columns 1,2 of matrix A are linearly independent. Hence, the set {u,v,w} is not linearly independent, and thus, it can't be a basis for $\mathbb{R}^3$ . Since it is not a basis, it can't span the space.

T=v /6 + 5
a) Work out the value of T when v = 42