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3 years ago

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Given the recursive formula find the first 4 terms for each. (not multiple choice) (MAFS.912.F-BF.1.3) A. An=an-1 3 when a1=5 B.

An=an-1 + 2/3 when a1=1 C. An=an-1 + 12 when a1=30

1 answer:

Neko [114]3 years ago

3 0

Answer:

<h3>See explanations below</h3>

Step-by-step explanation:

1) Given the recursive function An=an-1 + 3 when a1 = 5, we are to find the first four terms;

First term a1 = 5

a2 = a1 +3

a2 = 5 + 3

a2 = 8

a3 = a2 + 3

a3 = 8+3

a3 = 11

a4 = a3 + 3

a4 = 11 + 3

a4 = 14

<em>The first four terms are 5, 8, 11 and 14</em>

<em></em>

<em>2) </em>For the recursive function An=an-1 + 2/3 when a1 = 1

a2 = a1 + 2/3

a2 = 1 + 2/3

a2 = 5/3

a3 = a2 + 2/3

a3 = 5/3 + 2/3

a3 = 7/3

a4 = a3 + 2/3

a4 = 7/3 + 2/3

a4 = 9/3

a4 = 3

<em>Hence the first four terms of the sequence are 2/3, 5/3, 7/3, 3</em>

<em></em>

3) For the recursive function An=an-1 + 12 when a1=30

a2 = a1 + 12

a2 = 30 + 12

a2 = 42

a3 = a2 +12

a3 = 42 + 12

a3 = 54

a4 = a3 + 12

a4 = 54+12

a4 = 66

<em>Hence the first four terms of the sequence are 30, 42, 54, 66</em>

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The following 3 points are on a parabola defining the edge of a ski.

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a. The linear equation for the first point (-4,1) is 16A-4B+C=1

b. The linear equation for the second point (-2, 0.94) is 4A-2B+C=0.94

c. The linear equation for the third point (0,1) is 0A+0B+C=1

d. The matrix equation looks like this:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

e. The inverse of the coefficient matrix looks like this:

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]$

f. The equation of the parabola is: $\frac{3}{200}x^{2}+\frac{3}{50}x+1=y$

a. In order to build a linear equation from the given points, we need to substitute them into the general form of the equation.

Let's take the first point (-4,1). When substituting it into the general form of the quadratic equation we end up with:

$(-4)^{2}A+(-4)B+C=1$

which yields:

$16A-4B+C=1$

b. Let's take the second point (-2,0.94). When substituting it into the general form of the quadratic equation we end up with:

$(-2)^{2}A+(-2)B+C=0.94$

which yields:

$4A-2B+C=0.94$

c. Let's take the third point (0,1). When substituting it into the general form of the quadratic equation we end up with:

$(0)^{2}A+(0)B+C=1$

which yields:

$0A+0B+C=1$

d. A matrix equation consists on three matrices. The first matrix contains the coefficients (this is the numbers on the left side of the linear equations). Make sure to write them in the right order, this is, the numbers next to the A's should go on the first column, the numbers next to the B's should go on the second column and the numbers next to the C's should go on the third column.

The equations are the following:

16A-4B+C=1

4A-2B+C=0.94

0A+0B+C=1

So the coefficient matrix looks like this:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]$

Next we have the matrix that has the variables, in this case our variables are the letters A, B and C. So the matrix looks like this:

$\left[\begin{array}{c}A\\B\\C\end{array}\right]$

and finally the matrix with the answers to the equations, in this case 1, 0.94 and 1:

$\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

so if we put it all together we end up with the following matrix equation:

$\left[\begin{array}{ccc}16&-4&1\\4&-2&1\\0&0&1\end{array}\right]*\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}1\\0.94\\1\end{array}\right]$

e. When inputing the coefficient matrix in our graphing calculator we end up with the following inverse matrix:

$A^{-1}=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]$

Inputing matrices and calculating their inverses depends on the model of a calculator you are using. You can refer to the user's manual on how to do that.

f. Our matrix equation has the following general form:

AX=B

where:

A=Coefficient matrix

X=Variables matrix

B= Answers matrix

In order to solve this type of equations, we can make use of the inverse of the coefficient matrix to end up with an equation that looks like this:

$X=A^{-1}B$

Be careful with the order in which you are doing the multiplication, if A and B change places, then the multiplication will not work and you will not get the answer you need. So when solving this equation we get:

$\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{ccc}\frac{1}{8}&-\frac{1}{4}&\frac{1}{8}\\\frac{1}{4}&-1&\frac{3}{4}\\0&0&1\end{array}\right]*\left[\begin{array}{c}1\\\frac{47}{50}\\1\end{array}\right]$

(Notice that I changed 0.94 for the fraction 47/50 you can get this number by dividing 94/100 and simplifying the fraction)

So, in order to do the multiplication, we need to multiply each row of the coefficient matrix by the answer matrix and add the results. Like this:

$\frac{1}{8}*1+(-\frac{1}{4})(\frac{47}{50})+\frac{1}{8}*1$

$\frac{1}{8}-\frac{47}{200}+\frac{1}{8}=\frac{3}{200}$

So the first number for the answer matrix is $\frac{3}{200}$

$\frac{1}{4}*1+(-1)(\frac{47}{50})+\frac{3}{4}*1$

$\frac{1}{4}-\frac{47}{50}+\frac{3}{4}=\frac{3}{50}$

So the second number for the answer matrix is $\frac{3}{50}$

$0*1+0(\frac{47}{50})+1*1$

$0+0+1=1$

So the third number for the answer matrix is 1

In the end, the matrix equation has the following answer.

$\left[\begin{array}{c}A\\B\\C\end{array}\right]=\left[\begin{array}{c}\frac{3}{200}\\\frac{3}{50}\\1\end{array}\right]$

which means that:

$A=\frac{3}{200}$

$B=\frac{3}{50}$

and C=1

so, when substituting these answers in the general form of the equation of the parabola we get:

$Ax^{2}+Bx+C=y$

$\frac{3}{200}x^{2}+\frac{3}{50}x+1=y$

For further information, you can go to the following link:

brainly.com/question/12628757?referrer=searchResults

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3 years ago