What is the perimeter of angle DEF to the nearest tenth of a unit?
helppppp
tysm

Answer: well one is

Bc its the smallest besides zero, but zero is neither odd or even

Step-by-step explanation:

Answer:

22

Step-by-step explanation:

Answer:

-181 - 79x

Step-by-step explanation:

I am not sure your comment fits the given problem.

I get the following approach and solution :

T(1 + 4x) = 1×a + 4x×b = 2 + 2x

a = 2 + 2x - 4xb

T(4 + 15x) = 4×a + 15x×b = -3 + 3x

4×(2 + 2x - 4xb) + 15xb = -3 + 3x

8 + 8x - 16xb + 15xb = -3 + 3x

11 + 5x - xb = 0

11 + 5x = xb

b = (11 + 5x)/x

a = 2 + 2x - 4x(11 + 5x)/x = 2 + 2x - 44 - 20x = -42 - 18x

T(3 - 5x) = 3×a - 5xb = 3(-42 - 18x) - 5x(11 + 5x)/x =

= -126 - 54x - 55 - 25x = -181 - 79x

control :

T(1 + 4x) = a + 4xb = -42 - 18x + 44 + 20x = 2 + 2x

T(4 + 15x) = 4a + 15xb = -168 - 72x + 165 + 75x = -3 + 3x

Answer:

So g(6) comes out to be 2.

Step-by-step explanation:

So f(6)=5 means when f acts on 6, the result is 5....so the inverse would take 5 back to 6...or g(5)=6

So f(2)=6 means that f reassigns 2 to 6....so the inverse would take 6 back to 2....or g(6)=2

hope this helps :)

Answer: 3

Step-by-step explanation:

If f(x) = y, then invf(y) = x
So, f(5) = invg(5) = 3

Step-by-step explanation:

sin(90-A) = sin90cosA-sinAcos90

= cosA*1-0*sinA

= cos90

hence proved

Step-by-step explanation:

sin(90-A)=cosA

sin(90-A)=sin(90-A)

90-A=90-A

-A+A=90-90

=0

Answer:

Simplified: 4X^3

Step-by-step explanation:

Simplify the expression.

9. The curve passes through the point (-1, -3), which means

[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]

Compute the derivative.

[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]

At the given point, the gradient is -7 so that

[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]

Eliminating [tex]b[/tex], we find

[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]

Solve for [tex]b[/tex].

[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]

10. Compute the derivative.

[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]

Solve for [tex]x[/tex] when the gradient is 2.

[tex]x^2 - 5x + 6 = 2[/tex]

[tex]x^2 - 5x + 4 = 0[/tex]

[tex](x - 1) (x - 4) = 0[/tex]

[tex]\implies x=1 \text{ or } x=4[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]

[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]

11. a. Solve for [tex]x[/tex] where both curves meet.

[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]

[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]

[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]

[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]

[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]

[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]

[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]

11. b. Compute the derivative for the curve.

[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]

Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.

[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]

[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]

[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]

12. a. Compute the derivative.

[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]

12. b. By completing the square, we have

[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]

so that

[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]

13. a. Compute the derivative.

[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]

13. b. Complete the square.

[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]

Then

[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]

No the parents do not feel differently as they used to feel twenty years ago

Given :-

Po: p=0.46

Pa: p NE 0.46

alpha=0.01

Test statistic is one sample proportion [tex]z=(phat-p)/\sqrt{{ {0.46)(0.54)/800}}[/tex]

Critical value z> |2.576|

z=0.025/0.0176

=1.418

This is a p-value of 0.078

Fail to reject Po  and insufficient evidence to support the claim that parents feel differently.

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Answer: 3

Step-by-step explanation:

Substituting into vertex form, the equation is

[tex]y=a(x+2)^2 +2[/tex]

Substituting in the coordinates (-3, 5),

[tex]5=a(-3+2)^2 +2\\\\5=a+2\\\\a=3[/tex]

Answer:

the perpendicular bisector

Explanation:

The perpendicular bisector will intersect the segment at its midpoint.

Answer:

-40x

Step-by-step explanation:

f(x) =-3x-5

g(x) =4x-2

(f+g)(x) =?

now,

f(g(x))

=f(4x-2)

=-3×4x-2-5

=-10×4x

=-40x

Considering it's discriminant, it is found that:

A. The classmate is wrong, as the discriminant is of zero, hence the equation has one solution.

B. The quadratic equation has 1 x-intercept.

A quadratic equation is modeled by:

y = ax^2 + bx + c

The discriminant is:

[tex]\Delta = b^2 - 4ac[/tex]

The solutions are as follows:

In this problem, the equation is:

y = 9x² - 6x + 1.

The coefficients are a = 9, b = -6 and c = 1, hence the discriminant is:

[tex]\Delta =(-6)^2 - 4(9)(1) = 36 - 36 = 0[/tex]

Since the discriminant is zero, the classmate is wrong, as it means that the equation has one solution = one x-intercept.

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[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$132.90\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ t=years \end{cases} \\\\\\ 132.90 = (6000)(0.055)(t)\implies \cfrac{132.90}{(6000)(0.055)}=t\implies \cfrac{443}{1100}=t \\\\\\ \stackrel{\textit{converting that to days}}{\cfrac{443}{1100}\cdot 365} ~~ \approx ~~ 147~days[/tex]

now, if we move back from August 14th by 147 days backwards, that'd put us on March 20th.

The limit of j (x) as x approaches 3 is 4.

According to the question, A line begins at an open circle (2, 6) on a coordinate plane and descends through ( -2, 2). At, a complete circle is (3, 6). From a solid circle (2, 3), a curve leads to an open circle (3, 4). From the open circle to the closed circle, a line runs (6, 5).

From the graph, it can be seen that the limit of the function j(x) as the value of x approaches 3 is 4.

A diagram or pictorial representation that organizes the depiction of data or values is known as a graph.

The relationships between two or more items are frequently represented by the points on a graph.

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Answer:

4

Step-by-step explanation:

egg 2023

Answer:

Length = 30ft, Width = 5ft

Step-by-step explanation:

Let x be the width.

Area of Rectangle = Length * Width

Given from the information in the question,

Length = 6x ft

Width = x ft

Substitute the values into the formula:

6x * x = 150

[tex]6x^{2}[/tex] = 150

[tex]x^{2} =\frac{150}{6}[/tex]

[tex]x^{2} =25[/tex]

[tex]x=\sqrt{25}[/tex]

x = 5 ft.

Therefore,

Length = 6 * 5 = 30ft

Width = 5 ft.

The point estimate of difference of the sample his will be -20.

Based on the information given, the. following can be depicted:

n1 = 13

x1 = 141

s1 = 13

n2 = 9

x2 = 161

s2 = 12

The point estimate of difference will be:

= 141 - 161

= -20

The margin of error to be used in constructing the confidence interval will be calculated by multiplying the standard error which is 5.467 and the critical value. This will be:

= 5.467 × 2.528

= 13.822

The margin of error is 13.822.

The confidence interval will now be:

= (-20 + 13.822) and (-20 - 13.822)

= -6.178 and -33.822

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Answer:

C

Step-by-step explanation:

the answer is c

The equation that can be used to find the cost of each binder n in dollars is 861=7n.

Given that the cost of 7 binders is $8.61.

We are required to form an equation that represents the total cost of each binder n in dollars.

Equation is like a relationship between all the variables that are expressed in equal to form.It may be linear equation or may be more types.

Suppose the cost of 1 binder is n dollar.

We know that the total cost is basically the product of price of 1 unit and number of quantities of units.

Total cost=Price of 1 unit* number of units

8.61=n*7

8.61=7n

Hence the equation that can be used to find the cost of each binder n in dollars is 861=7n.

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See below for the distance between the points and the lines

Question 2

The line and the points are given as:

x = y

P = (4, -2)

Rewrite the equation as:

y = x

The slope of the above equation is

m = 1

The slope of a line perpendicular to it is

m = -1

A linear equation is represented as:

y = mx + b

Substitute m = -1

y = -x + b

Substitute (4, -2) in y = -x + b

-2 = -4 + b

Solve for b

b = 2

Substitute b = 2 in y = -x + b

y = -x + 2

So, we have:

x = y and y = -x + 2

Substitute x for y

x = -x + 2

Solve for x

x = 1

Substitute x = 1 in y = x

y = 1

So, we have the following points

(1, 1) and (4, -2)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(1 - 4)² + (1 + 2)²

Evaluate

d = 3√2

Hence, the distance between x = y and P = (4, -2) is 3√2 units

Question 3

The line and the points are given as:

y = 2x + 1

Q = (2, 10)

The slope of the above equation is

m = 2

The slope of a line perpendicular to it is

m = -1/2

A linear equation is represented as:

y = mx + b

Substitute m = -1/2

y = -1/2x + b

Substitute (2, 10) in y = -1/2x + b

10 = -1/2 * 2 + b

Solve for b

b = 11

Substitute b = 11 in y = -1/2x + b

y = -1/2x + 11

So, we have:

y = 2x + 1 and y = -1/2x + 11

Substitute 2x + 1 for y

2x + 1 = -1/2x + 11

Solve for x

x = 4

Substitute x = 4 in y = 2x + 1

y = 9

So, we have the following points

(4, 9) and (2, 10)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(4 - 2)² + (9 - 10)²

Evaluate

d = √5

Hence, the distance between the line and the point is √5 units

Question 4

The line and the points are given as:

y = -x + 3

R = (-5, 0)

The slope of the above equation is

m = -1

The slope of a line perpendicular to it is

m = 1

A linear equation is represented as:

y = mx + b

Substitute m = 1

y = x + b

Substitute (-5, 0) in y = x + b

0 = 5 + b

Solve for b

b = -5

Substitute b = 5 in y = x + b

y = x + 5

So, we have:

y = x + 5 and y = -x + 3

Substitute x + 5 for y

x + 5 = -x + 3

Solve for x

x = -1

Substitute x = -1 in y = x + 3

y = 2

So, we have the following points

(-1, 2) and (-5, 0)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(-1 + 5)² + (2 - 0)²

Evaluate

d = 2√5

Hence, the distance between the line and the point is 2√5 units

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Answer:

a=4.40

Step-by-step explanation:

To find value of a use pythagoras theorem:

c^2=b^2+a^2

Rearrange the equation:

a^2=c^2-b^2

Substitute the values:

a^2=(4.58)^2-(1.26)^2

After calculation:

a=4.40

Answer:

-25

Step-by-step explanation:

[tex] \dfrac{(20 - 5^2)(16 + 2^2)}{-2^3 + (3 \times 2^2)} = [/tex]

First, do all exponents.

[tex] = \dfrac{(20 - 25)(16 + 4)}{-8 + (3 \times 4)} [/tex]

Now do each operation in parentheses.

[tex] = \dfrac{(-5)(20)}{-8 + 12} [/tex]

Multiply in the numerator. Add in the denominator.

[tex] = \dfrac{-100}{4} [/tex]

Divide the numerator by the denominator.

[tex] = -25 [/tex]

Answer:

-25

Step-by-step explanation:

PEMDAS

The PEMDAS rule is an acronym representing the order of operations in math:

Given expression:

[tex]\sf \dfrac{(20-5^2)(16+2^2)}{-2^3+(3 \times 2^2)}[/tex]

As the given expression is a fraction, carry out the operations in the numerator and denominator first before finally dividing them.

Following PEMDAS, carry out the calculations inside the parentheses first, then carry out the rest of the calculations following the order of operations:

Parentheses

Calculate the exponents inside the parentheses:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(3 \times 4)}[/tex]

Multiply:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(12)}[/tex]

Add and subtract:

[tex]\implies \sf \dfrac{(-5)(20)}{-2^3+(12)}[/tex]

Exponents

Calculate the exponent:

[tex]\implies \sf \dfrac{(-5)(20)}{-8+(12)}[/tex]

Multiply and Divide

Multiply:

[tex]\implies \sf \dfrac{-100}{-8+(12)}[/tex]

Add and Subtract

Add:

[tex]\implies \sf \dfrac{-100}{4}[/tex]

Finally, divide the numerator by the denominator:

[tex]\implies \sf -25[/tex]

Answer:

g(0)⁻¹ = 6

Step-by-step explanation:

First, you must find the inverse of the function. Remember, another way of representing g(x) is with "y".  To find the inverse, you must swap the positions of the "x" and "y" variables in the equation. Then, you must rearrange the equation and isolate "y".

g(x) = 18 - 3x                                    <----- Original function

y = 18 - 3x                                        <----- Plug "y" in for g(x)

x = 18 - 3y                                        <----- Swap the positions of "x" and "y"

x + 3y = 18                                        <----- Add 3y to both sides

3y = 18 - x                                         <----- Subtract "x" from both sides

y = (18 - x) / 3                                    <----- Divide both sides by 3

y = 6 - (1/3)x                                     <----- Divide both terms by 3

Now that we have the inverse function, we need to plug x = 0 into the equation and solve for the output. In the inverse function, "y" is represented by the symbol g(x)⁻¹.

g(x)⁻¹ = 6 - (1/3)x                                  <----- Inverse function

g(0)⁻¹ = 6 - (1/3)(0)                               <----- Plug 0 in for "x"

g(0)⁻¹ = 6 - 0                                       <----- Multiply 1/3 and 0

g(0)⁻¹ = 6                                             <----- Subtract

a) The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function g(x) in factored form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no slant asymptotes.

c) There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

a) In the first part of this question we need to determine the equation of a polynomial in expanded form, derived from its factor form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the slant asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the slant asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no slant asymptotes.

c) There is vertical asymptote at a x-point if the denominator is equal to zero. There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

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The length of the diagonal that he cut to the nearest whole inch is 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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well, with the assumption that a year has 365 days, that means one day is really just 1/365th of a year, so then 54 days will be 54/365 of a year.

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$17100\\ r=rate\to 9.65\%\to \frac{9.65}{100}\dotfill &0.0965\\ t=years\dotfill &\frac{54}{365} \end{cases} \\\\\\ I = (17100)(0.0965)(\frac{54}{365})\implies \stackrel{\textit{interest owed}}{I\approx 244.13}~\hfill \underset{amount~owed}{\stackrel{17100~~ + ~~244.13}{\approx 17344.13}}[/tex]

Answer:

below

Step-by-step explanation:

1) slope = rise / run

2 coordinates are (-4, 0), (0, 2).

2 - 0 = 2

0 -- 4 = 4

2 / 4 = 1/2 so the slope is 0.5 or ½

2) it crosses the y axis at the average of the origin and 4.

4 + 0 = 4 / 2 = 2 so y intercept is 2.

3) in y= mx + b form

f(x) = ½x + 2, or, f(x) = 0.5x + 2

Answer: 19/40

Step-by-step explanation:

First Write 0.475 as 0.4751Multiply both numerator and denominator by 10 for every number after the decimal point0.475 × 10001 × 1000 = 4751000. Reducing the fraction gives The answer

The nth term of the sequence is 2n - 8

The nth term of an arithmetic progression is expressed as;

Tn = a + (n - 1)d

where

a is the first term

d is the common difference

n is the number of terms

Given the following parameters

a = f(1)=−6

f(2) = −4

Determine the common difference

d = f(2) - f(1)

d = -4 - (-6)
d = -4 + 6

d = 2

Determine the nth term of the sequence

Tn = -6 + (n -1)(2)

Tn = -6+2n-2
Tn = 2n - 8

Hence the nth term of the sequence is 2n - 8

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By definition, we have

[tex]f(n) = f(n - 1) + f(n - 2)[/tex]

so that by substitution,

[tex]f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)[/tex]

[tex]f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)[/tex]

[tex]f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)[/tex]

[tex]f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)[/tex]

and so on.

Recall the Fibonacci sequence [tex]F(n)[/tex], whose first several terms for [tex]n\ge1[/tex] are

[tex]\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}[/tex]

Let [tex]F_n[/tex] denote the [tex]n[/tex]-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.

[tex]f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)[/tex]

[tex]f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)[/tex]

[tex]f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)[/tex]

[tex]f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)[/tex]

[tex]f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)[/tex]

and so on. After [tex]k[/tex] iterations of substituting, we would end up with

[tex]f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))[/tex]

so that after [tex]k=n-2[/tex] iterations,

[tex]f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}[/tex]

A 0.13 sec and a 86 kVp technical changes are the changes that would have to produce the radiographic density and the contrast that are more similar to the original.

The radio graphic density can de defined to be the total amount or the overall darkening that can be found in a particular radiograph. This is usually known to have a certain type of density range that lies between  0.3 to 2.0 density.

When there is a density that is less than 0.3, the problem is basically because there is the issue of the density which would be found at the base and the fact that the film that is being used has fog in it.

It is known that based on the values that we have here the density and contrast would have to be of the given kvp of 86 and 0.13 seconds in order to be said to be similar to the original.

Read more on the radiographic density here:

https://brainly.com/question/14332593

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