All categories

2 years ago

45. Which sequence is represented by

Mathematics

1 answer:

Sidana [21]2 years ago

6 0

A sequence can be arithmetic, geometric or neither.

The sequence is (c) -19, -9, 1, 11

In this case, the sequence is an arithmetic sequence, and it is represented with the following formula

$\mathbf{y = 10x - 29}$

When x = 1, we have:

Substitute 1 for x in $\mathbf{y = 10x - 29}$

$\mathbf{y = 10 \times 1 - 29}\\$

$\mathbf{y =- 19}$

When x = 2, we have:

Substitute 2 for x in $\mathbf{y = 10x - 29}$

$\mathbf{y = 10 \times 2 - 29}$

$\mathbf{y = - 9}$

When x = 3, we have:

Substitute 3 for x in $\mathbf{y = 10x - 29}$

$\mathbf{y = 10 \times 3 - 29}$

$\mathbf{y = 1}$

When x = 4, we have:

Substitute 4 for x in $\mathbf{y = 10x - 29}$

$\mathbf{y = 10 \times 4 - 29}$

$\mathbf{y = 11}$

So, the corresponding y values when x = 1, 2, 3 and 4 are -19, -9, 1 and 11.

Hence, the sequence is (c) -19, -9, 1, 11

You might be interested in

What is the perimeter of a rhombus whose diagonals measure 42cm and 56cm?

Trava [24]

Answer:

676

Step-by-step explanation:

d1×d2/2

42×56/2

1352/2=676

please check it by doing again

3 0

3 years ago

Read 2 more answers

Tim needs 3 pounds of vegetables for a vegetable soup recipe.

klio [65]

Answer:

25/12 - pounds more vegetables

Step-by-step explanation:

first we need to put these variables into like terms

36/12 - 3

8/12 - 2/3

3/12 - 1/4

add the last two up and subtract from the first and you get how much you need left over

8 0

3 years ago

What is 3 3/8 +3 1/2 using common denominator

horrorfan [7]

1/2=4/8. 3 and 4/8+ 3 and 3/8 is 6 and 7/8.

8 0

3 years ago

Read 2 more answers

Find an explicit rule for the nth term of a geometric sequence where the second and fifth terms are -6 and 162, respectively. Pl

kicyunya [14]

Hello,

$u_{0} =a\\
 u_{1} =a*r\\
 u_{2} =a*r^{2}=-6\\
 u_{3} =a*r^{3}\\
 u_{4} =a*r^{4}\\
 u_{5} =a*r^{5}=162\\
....\\
\boxed{ u_{n} =a*r^{n}} \\

\dfrac{ 162}{-6} = \dfrac{ u_{5} }{ u_{3}} = \dfrac{ a*r^{5}}{ a*r^{2}} =r^3= -27\\
==\ \textgreater \ r=-3\\

 u_{2} =a*r^{2}=-6=a*(-3)^2 \ ==\ \textgreater \ a=-\frac{6}{9} =-\frac{2}{3}\\


\boxed{ u_{n} =-\frac{2}{3}*(-3)^{n}=(-1)^{n+1}*2*3^{n-1}} \\

$

8 0

4 years ago

Read 2 more answers

Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.

Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]$

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

$\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]$

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

$\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right]$ ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

$\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]$

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0

3 years ago