Find the missing segment in the image below

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]

[tex]~\hspace{7em}\textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} a^{-\frac{ n}{ m}} \implies \cfrac{1}{a^{\frac{ n}{ m}}} \implies \cfrac{1}{\sqrt[ m]{a^ n}} \\\\[-0.35em] ~\dotfill\\\\ a^{\frac{1}{5}}\implies \sqrt[5]{a^1}\implies \sqrt[5]{a}[/tex]

Answer: [tex]y= x-8[/tex]

Step-by-step explanation:

[tex]y-x=-8[/tex]

[tex]y-x+x=-8+x[/tex]

Final Answer: [tex]y=x-8[/tex]

The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

Given the data in the question;

To find the equation, we use the equation of the parabola that opens up or down since the directrix ( y = -13 ) is vertical.

The equation is expressed as;

( x - h )² = 4p( y - k )

First, we find the distance from the focus to the vertex.

|p| is the distance rom the focus to the vertex and from the vertex to the directrix.

p = -9 + 13

p = 4

We substitute the values into the equation;

( x - h )² = 4p( y - k )

( x - (-5) )² = 4(4)( y - (-9) )

( x + 5 )² = 16( y + 9 )

Multiply both side by 1/16

(1/16)( x + 5 )² =  y + 9

Make y the subject of the formula

(1/16)( x + 5 )² - 9 =  y

y = (1/16)( x + 5 )² - 9

The equation of the parabola with the given vertex and directrix in vertex form is y = (1/16)( x + 5 )² - 9.

Hence, option C is the correct answer.

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It looks like the function is

[tex]f(x,y) = 3 + 4x^{3/2}[/tex]

We have

[tex]\dfrac{\partial f}{\partial x} = 6x^{1/2} \implies \left(\dfrac{\partial f}{\partial x}\right)^2 = 36x[/tex]

[tex]\dfrac{\partial f}{\partial y} = \left(\dfrac{\partial f}{\partial y}\right)^2 = 0[/tex]

Then the area of the surface over [tex]R[/tex] is

[tex]\displaystyle \iint_R f(x,y) \, dS = \iint_R \sqrt{1 + 36x + 0} \, dA \\\\ ~~~~~~~~ = \int_0^5 \int_0^2 \sqrt{1+36x} \, dx \, dy \\\\ ~~~~~~~~ = 5 \int_0^2 \sqrt{1+36x} \, dx \\\\ ~~~~~~~~ = \frac5{36} \int_1^{73} \sqrt u \, du \\\\ ~~~~~~~~ = \frac5{36}\cdot \frac23 \left(73^{3/2} - 1^{3/2}\right) = \boxed{\frac5{54} (73^{3/2} - 1)}[/tex]

Answer:

It is the second answer

Step-by-step explanation:

The standard form of an ellipse is

x^2/a^2 + y^2/b^2  or x^2/b^2 + y^2/a^2 = 1

If the x is the main axis we use the first form.  If the y is the main axis we use the second form.  We will use the second form.

our a is 3 and our b is 4

x^2/3^2 + y^2/4^2

Answer: 70 and 60 degrees

Step-by-step explanation:

Angle AOB = 180 - 50 - 60 = 70 degrees so x is 70 degrees

Angle OCD = 180 - 130 = 50 so y = 180 - 70 - 50 = 60 degrees

Answer:

A:The Reqd. No. of Permutations=3360

Step-by-step explanation:

I couldn’t find the answer to B..sorry

The two different ways of finding the area are,
Case 1 = assume horizontal rectangles,
Case 2 = assume vertical rectangles.

the rectangle is a four-sided polygon whose opposites sides are equal and has an angle of 90° between its sides.

Here,

case 1,
As shown in the image
Area = sum of horizontal rectangles
Area = 10 * 3 + 2 * 5 + 2 * 1
Area  = 30 + 10 + 2
Area = 42    

Case II,
As shown in right figure,
Area of the vertical rectangles
Area = 3 * 5 + 5 * 3 + 2 * 6
Area = 15 + 15 + 12
Area = 42
Here, the area in case 1 is equal to case 2.

Thus, the two different ways of finding the area have shown above.

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Answer: -221

Step-by-step explanation:

31 is smaller than 252. Therefore, since 31 MINUS x equals 252, x needs to be a negative number in order to complete the equation (recall that a negative number times a negative number equals a positive number).

Therefore, if we subtract 31 on both sides, in other words transpose, we get,

-x = 221

The coefficient of -x is -1, however it is not written as it's implied that if there is not written coefficient in from of a variable then the coefficient of the variable is 1 or -1, depending on its sign.

Therefore, dividing -1 on both sides, we get,

x = -221

Hence, the desired answer is -221.

Answer:

3%, 32%

Step-by-step explanation:

winning 1 game only: two possibilities

a. winning balltoss, losing dart, which is 20%*85% = 17%

b. winning dart, losing ball toss, which is 15%*80% = 12%

so winning 1 game only: 29%

winning both games:

20% * 15% = 3%

winning either one: winning both games+winning 1 game only

29% + 3% = 32%

Answer:

B (2nd option)

Step-by-step explanation:

Factor each one.

6x^2+39x-21 is divisible by 3 -> 3(2x^2+13x-7) -> 3(2x-1)(x+7)

6x^2+54x+84 is divisble by 6 -> 6(x^2+9x+14) -> 6(x+2)(x+7)

The greatest common factor is 3(x+7), so taking that out of each polynomial, we have (2x-1) and 2(x+2). The least common multiple is the greatest common factor*(2x-1)*2(x+2) which, simplifying, is 12x^3+102x^2+114x-84, or B.

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]

Answer:

84

Step-by-step explanation:

Let the number of hamburgers be h and the number of cheeseburgers be c.

This means that:

Substituting c=3h into the first equation, it follows that 4h=336, and thus h=84.

Answer: Logarithmic

Explanation:

This curve is a reflection of the exponential curve over the line y = x, to show that it is the inverse of exponentials. We use logs to help isolate the exponent among other useful properties.

The relation that is also a function is described as follows:

A nonlinear function on a coordinate plane vertex at (minus 3, minus 2) passes through (minus 7, 0), and (minus 6, minus 4).

A relation is a function if each value of the input is mapped to only one value of the output.

In this problem, we have that:

Hence a function is given by:

A nonlinear function on a coordinate plane vertex at (minus 3, minus 2) passes through (minus 7, 0), and (minus 6, minus 4).

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Answer:

  y = -1/2n +40

Step-by-step explanation:

We are given two ordered pairs (stalks, ounces) and asked for the slope-intercept form equation of the line through them.

The slope of the desired line can be found from the formula ...

  m = (y2 -y1)/(x2 -x1)

For the given points (30, 25) and (32, 24), the slope is ...

  m = (24 -25)/(32 -30) = -1/2

The y-intercept of the desired line can be found from the formula ...

  b = y -mn

  b = 25 -(-1/2)(30) = 25 +15 = 40

The slope-intercept equation of a line is ...

  y = mn +b . . . . . line with slope m and y-intercept b

  y = -1/2n +40 . . . . . . line with slope -1/2 and y-intercept 40

The linear relationship between stalks (n) and yield (y) is ...

  y = -1/2n +40

Answer:

25.2

Step-by-step explanation:

I'll assume you wrote:

[tex]f(x) = 4e^{x-2} - 4[/tex]

So when x = 4:

[tex]4e^{4-2} - 4 = 4e^2 - 4[/tex] ≅ [tex]4\cdot 7.3 - 4 = 25.2[/tex]

Answer: [tex]\boxed{30^{\circ}, 60^{\circ}, 90^{\circ}}[/tex]

Step-by-step explanation:

Let the smallest angle be x.

Then, the middle angle is x+30.

The largest angle is 2x+30.

Angles in a triangle add to 180 degrees, so:

[tex]x+x+30+2x+30=180\\\\4x+60=180\\\\4x=120\\\\x=30[/tex]

So, the angles measure [tex]\boxed{30^{\circ}, 60^{\circ}, 90^{\circ}}[/tex]

The measures of all three angles are 30°,60°, and 90°.

A triangle is a 3-sided polygon occasionally (though not frequently) referred to as the trigon. Every triangle has three sides and three angles, some of which may be the same.

Find the measures of all three angles:

The three angles be a, b and c.

Angle “a" = x

Angle “b" = x + 30°

Angle “c" = “a" + “b"

                = x + x + 30°

                = 2x + 30°

Total angle in a triangle = 180°

Therefore;

“a" + “b" + “c" = 180°

Births; (x + (x + 30°) + (2x + 30°)) = 180°

So,

4x + 60° = 180°

x = (180° - 60°) ÷ 4

x = 30°

Plugging this value of x into the earlier equations for angles “a,” “b,” and “c."

“a" = x = 30°

“b" = x + 30° = 60°

“c" = 2x + 30° = 90°

The measures of all three angles are 30°,60°, and 90°.

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Using a system of equations, it is found that Allison bought 2 bottles of juice and 9 bottles of soda.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, the variables are:

Allison purchased a total of 11 bottles of juice, hence:

x + y = 11 -> x = 11 - y.

These 11 bottles contain 445 grams of sugar, hence, considering the amounts of each bottle, we have that:

20x + 45y = 445

Since x = 11 - y:

20(11 - y) + 45y = 445

25y = 225

y = 225/25

y = 9.

x = 11 - y = 11 - 9 = 2.

She bought 2 bottles of juice and 9 bottles of soda.

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I'm not sure if the last two apostrophes are part of the quote - "Solve ... " - or if you mean the second derivative [tex]x''[/tex]. I think you mean the first interpretation, but I'll include both cases since they are both solvable.

If the former is correct, separate variables to solve.

[tex]x' = -9tx \implies \dfrac{dx}{dt} = -9tx \implies \dfrac{dx}x = -9t\,dt[/tex]

Integrate both sides to get

[tex]\ln|x| = -\dfrac92 t^2 + C[/tex]

Solve for [tex]x[/tex].

[tex]e^{\ln|x|} = e^{-9/2\,t^2 + C} \implies \boxed{x = Ce^{-9/2\,t^2}}[/tex]

If you meant the latter, then the ODE can be rewritten as

[tex]9t x'' + x' = 0[/tex]

Reduce the order of the equation by substituting [tex]y(t) = x'(t)[/tex] and [tex]y'(t) = x''(t)[/tex].

[tex]9t y' + y = 0[/tex]

Solve for [tex]y'[/tex] and separate variables.

[tex]y' = -\dfrac y{9t} \implies \dfrac{dy}{dt} = -\dfrac y{9t} \implies \dfrac{dy}y = -\dfrac{dt}{9t}[/tex]

Integrate.

[tex]\ln|y| = -\dfrac19 \ln|t| + C[/tex]

Solve for [tex]y[/tex].

[tex]e^{\ln|y|} = e^{-1/9 \,\ln|t| + C} \implies y = Ct^{-1/9}[/tex]

Solve for [tex]x[/tex] by integrating.

[tex]x' = Ct^{-1/9} \implies x = C_1 t^{8/9} + C_2[/tex]

Answer:

96 degrees

Step-by-step explanation:

The angle bisectors splits the two angles mentioned down the middle so the 2 angles are equal to each other.

4x + 4 = 2(x + 13)  Distribute the 2

4x + 4 = 2x + 26  Subtract 2x from both sides

2x + 4 = 26  Subtract 4 from both sides

2x = 22  Divide both sides by 2

x = 11  Plug that back into either the right side or the left side of the original equation

4x + 4

4(11) + 4

44 + 4

48.  Each angle is 48 degrees.  48 + 48 is 96

Step-by-step explanation:

(7m - 3)² = 49m² - 42m + 9

play it through and do the actual multiplication behind the square :

(7m - 3)² = (7m - 3)(7m - 3) =

= 7m×7m + 7m×(-3) + (-3)×7m + (-3)(-3) =

= 49m² - 2×21m + 9 = 49m² - 42m + 9

Answer:

-32x^5

Step-by-step explanation:

using binomial expression we have (1-2x)^6

9x - 4(x - 2) =x + 20

We move all terms to the left:

Multiply

We get rid of the parentheses.

We add all the numbers and all the variables.

We move all terms containing x to the left hand side, all other terms to the right hand side

Answer:

No solution

Step-by-step explanation:

[tex]\sqrt{6x-3}=2\sqrt{x} \\ \\ 6x-3=4x \\ \\ -3=2x \\ \\ x=-\frac{3}{2} [/tex]

However, this would make the right hand side of the equation undefined over the reals, so there is no solution.

Using translation concepts, it is found that pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

In this problem, the change is given as follows:

x -> x - pi

It means that the change is in the domain, in which pi is the phase shift of the graph, and since it is negative, the graph is shifted right pi units.

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The missing value in the logarithm are as follows:

Using logarithm rule,

logₐ b - logₐ c = logₐ (b / c)

logₐ b + logₐ c = logₐ (b × c)

Therefore,

log₃ 7 - log₃ 2 = log₃ (7 / 2)

log₉ 7 + log₉ 4 = log₉ (7 × 4) =  log₉ 28

log₆ 1 / 81 = log₆ 81⁻¹ = log₆ 3⁻⁴ =  - 4 log₆ 3

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The relationship of the graph g(x) = (x − 2)^3 + 6 compare to the parent function of f(x) = x^3 is that g(x) is shifted 2 units to the right and 6 units up.

Translations is a transformation technique that changes the position of an object from one point on the plane to another.

Given the function below

g(x) = (x − 2)^3 + 6

The function compared to f(x) = x^3, shows a translation of f(x) by 2 unit to the right along the horizontal and vertical translation of the function 6 units up

Hence the relationship of the graph g(x) = (x − 2)^3 + 6 compare to the parent function of f(x) = x^3 is that g(x) is shifted 2 units to the right and 6 units up.

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Answer:

g(x) is shifted 6 units to the left and 2 units down.

Step-by-step explanation:

took the test