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3 years ago

Find f(-3) PLEASE HELP

Mathematics

2 answers:

irga5000 [103]3 years ago

8 0

Answer: 2

Step by step explanation

Artemon [7]3 years ago

6 0

Answer:

Step-by-step explanation:

When the point on the x-axis is on -3, the point on the y-axis is 2. Therefore, f(-3) is equal to 2.

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Adam rides his bike from his home to the school's bike lot at point F each day. Assuming he takes the same route home, how far d

Wewaii [24]

I'm saying the answer is D.

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3 years ago

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Show that 2 1/3 × 3 3/4 = 8 3/4

alexgriva [62]

Answer:

Here you go :)

Step-by-step explanation:

(2 1/3)(3 3/4)

= 7/3(3 3/4)

= 7/3(15/4)

= 35/4

= 8 3/4

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2 years ago

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Solve these recurrence relations together with the initial conditions given. a) an= an−1+6an−2 for n ≥ 2, a0= 3, a1= 6 b) an= 7a

8_murik_8 [283]

Answer:

a) 3/5·((-2)^n + 4·3^n)
b) 3·2^n - 5^n
c) 3·2^n + 4^n
d) 4 - 3 n
e) 2 + 3·(-1)^n
f) (-3)^n·(3 - 2n)
g) ((-2 - √19)^n·(-6 + √19) + (-2 + √19)^n·(6 + √19))/√19

Step-by-step explanation:

These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

$r^n=c_1\cdot r^{n-1}+c_2\cdot r^{n-2}$

Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

$r^2-c_1r-c_2=0$

The quadratic formula tells us values of r that satisfy this are ...

$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

$p=\dfrac{a[0]r_2-a[1]}{r_2-r_1}\\\\q=\dfrac{a[1]-a[0]r_1}{r_2-r_1}$

_____

Using these formulas on the first recurrence relation, we get ...

$c_1=1,\ c_2=6,\ a[0]=3,\ a[1]=6\\\\r_1=\dfrac{1+\sqrt{1^2+4\cdot 6}}{2}=3,\ r_2=\dfrac{1-\sqrt{1^2+4\cdot 6}}{2}=-2\\\\p=\dfrac{3(-2)-6}{-5}=\dfrac{12}{5},\ q=\dfrac{6-3(3)}{-5}=\dfrac{3}{5}\\\\a[n]=\dfrac{3}{5}(-2)^n+\dfrac{12}{5}3^n$

The rest of (b), (c), (e), (g) are solved in exactly the same way. A spreadsheet or graphing calculator can ease the process of finding the roots and coefficients for the given recurrence constants. (It's a matter of plugging in the numbers and doing the arithmetic.)

_____

For problems (d) and (f), the quadratic has one root with multiplicity 2. So, the formulas for p and q don't work and we must do something different. The generic solution in this case is ...

$a[n]=(p+qn)r^n$

The initial condition equations are now ...

$\left[\begin{array}{cc}1&0\\r&r\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

and the solutions for p and q are ...

$p=a[0]\\\\q=\dfrac{a[1]-a[0]r}{r}$

Using these formulas on problem (d), we get ...

$c_1=2,\ c_2=-1,\ a[0]=4,\ a[1]=1\\\\r=\dfrac{2+\sqrt{2^2+4(-1)}}{2}=1\\\\p=4,\ q=\dfrac{1-4(1)}{1}=-3\\\\a[n]=4-3n$

And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

_____

<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

6 0

3 years ago

Given x=5 + √16 , select the value(s) of x. -11 1 9 21

Schach [20]

Answer:

The answers would be 9 and 1.

Step-by-step explanation:

x= 5±√16 can be separated into two equations.

x=5+√16 and x=5-√16

1. 5+√16 can be solved by simplify √16, which is 4. So then it would become 5+4 which equals 9.

2. 5-√16 can be solved by simplifying √16 which is 4. So then it would become 5-4 which equals 1.

So the values of x are 9 and 1.

6 0

3 years ago

Read 2 more answers

Please help thank you

Allushta [10]

The average rate of change of credit card is $ 401.79 /year.

<h3>What is Average Rate?</h3>

A single rate that is a weighted average of the different rates that are applicable to property in various locations.
An average is a single number calculated as the average of a set of numbers, typically calculated as the sum of the numbers divided by the total number of numbers in the set (the arithmetic mean).

<u>Solution</u>

The difference of credit card debt between the year 2006 and 1992 = 8900 - 3275 = $5625

The difference of years between the year 2006 and 1992 = 2006 - 1992 = 14

average rate of change of credit card debt with respect to time = $\frac{difference of credit card debt}{difference of years}$

average rate = $\frac{5625}{14}$ = $ 401.79 / year

know more about average numerical brainly.com/question/18296887

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2 years ago