If a b c d and m 2 is increased by 20 degrees how must m 3 be changed to keep the segments parallel?

The wholesale cost for the pianos that Tim pays the manufacturer, wages and utility bills that Tim pays are counted as explicit cost. On the other hand, the salary that Tim could earn if he worked as an accountant and the rental income that Tim could earn from renting his showroom fall under implicit cost.

Difference between Implicit Cost and Explicit Cost:

These two expenses, implicit cost and explicit cost, both represent company activity. The connection between cost and business is what distinguishes these two.

An explicit cost is one that the owner bears directly; an implicit cost is one that the owner bears indirectly without paying or utilizing its own resources. However, the organization or corporation takes into account both expenses while making management decisions.

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In calculating the factor A variation, there are two degrees of freedom. In determining the variation of factor B, there are two degrees of freedom.

A two-factor factorial design is an experiment that collects data for all potential values of the two factors of the study. The design is a balanced two-factor factorial design if equivalent sample sizes are used for every of the possible factor combinations.

Suppose we have two components, A and B, each of which has a high number of levels of interest. We will select a random level of component A and a random level of factor B, and n observations will be taken for each experimental combination.

From the data given:

a.

In calculating the factor A variation, there are two degrees of freedom.

In determining the variation of factor B, there are two degrees of freedom.

b.

Finding the degree of freedom using the interaction variation, there are four degrees of freedom.

c.

In finding the random variable, there are 9(4-1) = 27 degrees of freedom.

d.

In calculating the total variable, there are 9*4-1 =35 degrees of freedom.

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Answer:

1.

Step-by-step explanation:

10C0 = 1.

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

Question 13

Part (a)

[tex]\frac{3840-3905}{110}=\boxed{0.5\overline{90}}[/tex]

By similar logic, the other answers are

(b) 2.6818181....

(c) 3.5909090...

(d) 1.1363636...

Question 14

Part (a)

[tex]\frac{46800-41400}{4700} \approx \boxed{1.15}[/tex]

Part (b)

[tex]-2.57=\frac{x-41400}{4700} \\ \\ -2.57(4700)=x-41400 \\ \\ x=41400-2.57(4700) \\ \\ x \approx \boxed{29300}[/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

Answer:

47°

Step-by-step explanation:

If angles are complementary, that means those angles are added up to 90°.

Therefore,

43 + x = 90

Take 43 to the right side to make x as the subject.

x = 90 - 43

x = 47°

Note that a complementary angle adds up to 90 degrees.

In this problem, we get the equation:

43° + x = 90°

So, we need to find the value of x by doing:

90° – 43° = 47°

x = 47°

43° + 47° = 90°

We get the final result as:

x = 47°

Hope this helps :)

Answer:

22

Step-by-step explanation:

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

The total cost to run Val’s AC costs for a summer month of 30 days is $94.5.

Quantity of electricity used in kilowatts

1000 watts = 1 kilowatts

2,500 watts = 2.5 kilowatts

Cost of Val’s AC to run for a summer month of 30 days = Quantity of electricity used × Number of hours Val runs AC per day × Number of days × Cost of electricity

= 2.5 × 7 × 30 × 0.18

= $94.5

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The measure of angle a is 70 degrees

The angle in a semicircle is a right angle.

This means that:

a + d = 90

Where O is the center of the circle

From the figure, we have:

Angle d and the angle with a measure of 20 degrees are corresponding angles

This means that

d = 20

Substitute d = 20 in a + d = 90

a + 20 = 90

Subtract 20 from both sides of the equation

a = 70

Hence, the measure of angle a is 70 degrees

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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]

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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Answer:

4, 6, 4

6, 12, 8

8, 12, 6

Step-by-step explanation:

Tetrahedron:

4 faces, 6 edges, 4 vertices

Cube:

6 faces, 12 edges, 8 vertices

Octahedron:

8 faces, 12 edges, 6 vertices

you can also use the v - e + f = 2 to check

5 and 6 thousandths in standard form is 5.006

The number is given as:

5 and 6 thousandths

This can be rewritten as

5 +  6 thousandths

Express as numbers

5 +  0.006

Evaluate the sum

5.006

Hence, 5 and 6 thousandths in standard form is 5.006

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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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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: B

Step-by-step explanation:

[tex]\frac{4+5i}{4i}\\ \\ =\frac{1}{4} \left(\frac{4+5i}{i} \right)\\\\=\frac{1}{4}((-i)(4+5i))\\\\=\frac{1}{4}(-4i-5i^2)\\\\=\frac{5-4i}{4}[/tex]

Answer: 25 in³

Step-by-step explanation:

We can calculate the volume of the pyramid by first calculating the volume of a prism with the same dimensions, and then dividing by 3 (all pyramids a volume that's [tex]\frac{1}{3}[/tex] of the volume of a prism with the same dimensions).

The volume of a prism is its base area times its height. The base would be a square, so its area is 5², which is 25. The height is 3 inches, making the prism's volume 75 in³.

The volume of the pyramid would be one-third of this value, which is 75[tex]75\div3[/tex] which is 25 in³.

46) True

47) False

These are true by definition.

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]

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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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:

the perpendicular bisector

Explanation:

The perpendicular bisector will intersect the segment at its midpoint.

Answer: 10,204 feet

Step-by-step explanation: i would assume you would add the two together, seeing as if the mountain is 10,093 above sea level and the valley is 111 below, the difference in elevation is also the distance between each other.

The particle will stop at D.

If an event can occur in m different ways and if following it, a second event can occur in n different ways, then the two events in succession can occur in m × n different ways.

In pentagon ABCDE shown above, each side is 1 cm. If a particle starts at point A and travels clockwise 723 cm along ABCDE, then

Number of sides = 5

Let x be the one complete revolution.

5 x = 725

x = 145

Now, 145 - 2 = 143

Hence, it will stop at D which is 2 less than A.

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Answer:

2 and 1/16th inch

Step-by-step explanation:

the half inch mark would be 2 and 1/2 and you know half of half is 1/4 so the best answer is A.

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.

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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