Compare your response to the sample response. Which of these did your response include? Check all of the boxes that apply.

The real number is on the number line, but the complex number is in the complex plane.

Both are distances.

Both are positive values.

The distance formula or the Pythagorean theorem is used to calculate the absolute value of a complex number.

Answer:

-2n -32

Step-by-step explanation:

Acc. to BODMAS

we multiply first

⇒-6n - -8 x [(+4) (n)]

⇒ 6n  - 8n x -32

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textbf{Equation:}[/tex]

[tex]\mathsf{6n - 8(n + 4)}[/tex]

[tex]\huge\textbf{Solve: }[/tex]

[tex]\mathsf{6n - 8(n + 4)}[/tex]

[tex]\huge\textbf{Distribute \boxed{\bf -8} within the parenthesis:}[/tex]

[tex]\mathsf{= 6n - 8(n) - 8(4)}[/tex]

[tex]\mathsf{= 6n - 8n - 32}[/tex]

[tex]\huge\textbf{Combine the like terms: }[/tex]

[tex]\mathsf{= (6n - 8n) - (32)}[/tex]

[tex]\mathsf{= 6n - 8n - 32}[/tex]

[tex]\mathsf{= -2n - 32}[/tex]

[tex]\huge\textbf{Therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\frak{{ -2n - 32}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}}[/tex]

The area below the mean compares to the area above the mean in a normal distribution as the areas are always equal regardless of the mean. Option A This is further explained below.

Generally, The normal distribution, also known as the Gaussian distribution, is a kind of probability distribution that is symmetric around the mean. This means that it demonstrates that data that are closer to the mean are more likely to occur than data that are farther away from the mean. When represented graphically, the normal distribution takes the shape of a "bell curve."

In conclusion, In a normal distribution, the area below the mean is compared to the area above the mean since the areas are always equal regardless of the mean. This is true even if the mean is different.

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Complete Question

How does the area below the mean compare to the area above the mean in a normal distribution?

A. the areas are always equal regardless of the mean

B. the areas are sometimes equal depending upon the standard deviation of the distribution

C. the area above the mean is larger since the values are larger as you move above the mean

D. the areas are sometimes equal depending upon the value of the mean

19 trees should be planted to maximize the total

From the question, we have the following parameters:

Number of apples, x = 18

Yield, f(x) = 80 per tree

When the number of apple trees is increased (say by x).

We have:

Trees = 18 + x

The yield decreases by four apples per tree.

So, we have

Yield = 80 - 4x

So, the profit function is

P(x) = Apples * Yield

This gives

P(x) = (18 + x) *(80 - 4x)

Expand the bracket

P(x) = 1440 - 72x + 80x - 4x^2

Differentiate the function

P'(x) = 0 - 72 + 80 - 8x

Evaluate the like terms

P'(x) = 8 - 8x

Set P'(x) to 0

8 - 8x = 0

Divide through by 8

1 - x = 0

Solve for x

x = 1

Recall that:

Trees = 18 + x

So, we have

Trees = 18 + 1

Evaluate

Trees = 19

Hence, 19 trees should be planted to maximize the total

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The ski club has 44 members in all

What is population?

Population refers to the total number of items or individuals that form a group.

In this case, the population means the total  number of all members in the  Mogul Runners ski club.

What is sample?

Sample means a fraction of the total population, in other words, the 25% of the members, whose number is 11, that signed up to go for the planned trip.

We can convert 25% to 100% by equation 25% to 11 members that signed up to go for the planned trip.

25% of club members=11 members

divide both sides by 25%

25%/25% of club members=11/25% members

club members=44 members

The ski club has 44 members altogether as computed above.

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Answer:

bro it had been decreased

The information given regarding the computation of the profit or loss shows that there's a loss of RS 500,000.

From the information given, it was stated that the real estate agent bought two triangular Kittas' having a common side as shown in the figure from two people for Rs 2000,000 altogether.

It was further depicted that he sold the land in the shape of a quadrilateral at Rs 1,500,000. In their case, the sale is less than the cost and therefore, there's a loss.

The loss will be:

= 2,000,000 - 1,500,000

= 500,000

Therefore, the information given regarding the computation of the profit or loss shows that there's a loss of RS 500,000.

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Answer:

Square Pyramid

Step-by-step explanation:

Thats what a Square Pyramid is

Answer: The given integral is impossible to solve.

Step-by-step explanation: I tried solving it right now with integration by parts and trigonometric substitution, and I was unable to solve it.

So I looked at two math solvers and tried inputting the question, and it either says "There is nothing more you can do with this problem," or "We are unable to solve this problem."

First, let's substitute [tex]x=\frac w4[/tex] and [tex]dx=\frac{dw}4[/tex] to get rid of the fraction.

[tex]\displaystyle \int \csc^6\left(\dfrac w4\right) \cot^4\left(\dfrac w4\right) \, dw = 4 \int \csc^6(x) \cot^4(x) \, dx[/tex]

Recall that

[tex]\cot^2(x) + 1 = \csc^2(x)[/tex]

[tex]\dfrac{d}{dx} \cot(x) = -\csc^2(x)[/tex]

We can the rewrite the integrand as

[tex]\displaystyle 4 \int \csc^6(x) \cot^4(x) \, dx = 4 \int \left(\cot^2(x) + 1\right)^2 \cot^4(x) \csc^2(x) \, dx[/tex]

then substitute [tex]y=\cot(x)[/tex] and [tex]dy=-\csc^2(x)\,dx[/tex] to get

[tex]\displaystyle -4 \int (y^2 + 1)^2 y^4 \, dy[/tex]

Expand the integrand.

[tex]\displaystyle -4 \int (y^4 + 2y^2 + 1) y^4 \, dy = -4 \int (y^8 + 2y^6 + y^4) \, dy[/tex]

Now integrate with the power rule.

[tex]\displaystyle -4 \left(\frac{y^9}9 + \frac{2y^7}7 + \frac{y^5}5\right) + C[/tex]

[tex]\displaystyle -\frac{4y^9}9 - \frac{8y^7}7 - \frac{4y^5}5 + C[/tex]

Put everything back in terms of [tex]x[/tex], then [tex]w[/tex].

[tex]\displaystyle -\frac{4\cot^9(x)}9 - \frac{8\cot^7(x)}7 - \frac{4\cot^5(x)}5 + C[/tex]

[tex]\displaystyle \boxed{-\frac49 \cot^9\left(\dfrac w4\right) - \frac87 \cot^7\left(\dfrac w4\right) - \frac45 \cot^5\left(\dfrac w4\right) + C}[/tex]

Answer:

True

Step-by-step explanation:

This is true in an equilateral triangle.

See below for the values of the probabilities

The probabilities of the whole events

For event A, the probability of the whole event is calculated using

P(A) = n(A)/Total

Using the table of values, we have:

P(F) = 130/200 = 0.65

P(M) = 70/200 = 0.35

P(L) = 60/200 = 0.30

P(S) = 40/200 = 0.20

The probabilities of the intersection events

For events A and B, the probability of the intersection events is calculated using

P(A n B) = n(A n B)/Total

Using the table of values, we have:

P(F n E) = P(E n F) = 60/200 = 0.30

P(F n L) = P(L n F) = 40/200 = 0.20

P(F n S) = P(S n F) = 30/200 = 0.15

P(M n E) = P(E n M) = 40/200 = 0.20

P(M n L) = P(L n M) = 20/200 = 0.10

P(M n S) = P(S n M) = 10/200 = 0.05

The probabilities of the Union (OR) disjoint events

For events A and B, the probability of the union (OR) disjoint events is calculated using

P(A u B) = P(A) + P(B)

Using the table of values, we have:

P(E u S) = P(E) + P(S) = 100/200 + 40/200 = 0.70

P(E u L) = P(E) + P(L) = 100/200 + 60/200 = 0.80

P(E u L u S) = P(E) + P(L) + P(S) = 100/200 + 60/200 + 40/100 = 1

P(F u M) = P(F) + P(M) = 130/200 + 70/200 = 1

The probabilities of the Union (OR) joint events

For events A and B, the probability of the union (OR) joint events is calculated using

P(A u B) = P(A) + P(B) - P(A n B)

Using the table of values, we have:

P(E u F) = P(E) + P(F) - P(E n F) = 100/200 + 130/200 - 60/100 = 0.85

P(L u M) = P(L) + P(M) - P(L n M) = 60/200 + 70/200 - 20/100 = 0.55

The probabilities of the conditional probabilities

For events A and B, the conditional probability is calculated using

P(A/B) = P(A n B)/P(B)

Using the table of values, we have:

P(F/S) = P(F n S)/P(S) = 0.15/0.20 = 0.75

P(F/E) = P(F n E)/P(E) = 0.30/0.50 = 0.60

P(M/S) = P(M n S)/P(S) = 0.05/0.20 = 0.25

P(M/L) = P(M n L)/P(L) = 0.10/0.30 = 0.33

P(S/F) = P(F n S)/P(F) = 0.15/0.65 = 0.23

P(E/F) = P(F n E)/P(F) = 0.30/0.65 = 0.46

P(S/M) = P(M n S)/P(M) = 0.05/0.35 = 0.14

P(L/M) = P(M n L)/P(M) = 0.10/0.35 = 0.29

The multiplication of dependent events

For events A and B, the conditional probability is calculated using

P(A n B) = P(A) * P(B/A)

Using the table of values, we have:

P(F n L) = P(F) * P(L/F)

This gives

P(F n L) = 0.65 * (0.20/0.30)

Evaluate

P(F n L) = 0.43

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The correct option regarding whether the relation is a function is:

B. The relation is a function. Each input value is paired with one output value.

A relation represent a function if each value of the input is paired with one value of the output.

In this problem, when the input - output mappings are given by:

{(–2, 3), (–1, 3), (0, 2), (1, 4), (5, 5)}.

Which means that yes, each input value is paired with one output value, hence the relation is a function and option B is correct.

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Answer:

Point A:  (2, 10)

Point B:  (-3, 0)

Point C:  (-5, -4)

Point D:  (-5, -32)

Step-by-step explanation:

Points A and B are the points of intersection between the two graphs.

Therefore, to find the x-values of the points of intersection, substitute one equation into the other and solve for x:

[tex]\implies 2x+6=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-12=0[/tex]

[tex]\implies 2(x^2+x-6)=0[/tex]

[tex]\implies x^2+x-6=0[/tex]

[tex]\implies x^2+3x-2x-6=0[/tex]

[tex]\implies x(x+3)-2(x+3)=0[/tex]

[tex]\implies (x-2)(x+3)=0[/tex]

[tex]\implies x=2, -3[/tex]

From inspection of the graph:

To find the y-values, substitute the found x-values into either of the equations:

[tex]\begin{aligned} \textsf{Point A}: \quad 2x+6 & =y\\2(2)+6 & =10\\ \implies & (2, 10)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point B}: \quad -2x^2+18 & =y\\-2(-3)^2+18 & =0\\ \implies & (-3,0)\end{aligned}[/tex]

Therefore, point A is (2, 10) and point B is (-3, 0).

If the distance between points C and D is 28 units, the y-value of point D will be 28 less than the y-value of point C.  The x-values of the two points are the same.

Therefore:

[tex]\textsf{Equation 1}: \quad y=2x+6[/tex]

[tex]\textsf{Equation 2}: \quad y-28=-2x^2+18[/tex]

As the x-values are the same, substitute the first equation into the second equation and solve for x to find the x-value of points C and D:

[tex]\implies 2x+6-28=-2x^2+18[/tex]

[tex]\implies 2x^2+2x-40=0[/tex]

[tex]\implies 2(x^2+x-20)=0[/tex]

[tex]\implies x^2+x-20=0[/tex]

[tex]\implies x^2+5x-4x-20=0[/tex]

[tex]\implies x(x+5)-4(x+5)=0[/tex]

[tex]\implies (x-4)(x+5)=0[/tex]

[tex]\implies x=4,-5[/tex]

From inspection of the given graph, the x-value of points C and D is negative, therefore x = -5.

To find the y-value of points C and D, substitute the found value of x into the two original equations of the lines:

[tex]\begin{aligned} \textsf{Point C}: \quad 2x+6 & =y\\2(-5)+6 & =-4\\ \implies & (-5,-4)\end{aligned}[/tex]

[tex]\begin{aligned} \textsf{Point D}: \quad -2x^2+18 & = y \\ -2(-5)^2+18 & =-32\\ \implies & (-5, -32)\end{aligned}[/tex]

Therefore, point C is (-5, -4) and point D is (-5, -32).

Answer:

a) A = (2, 10) and B = (-3, 0)

b) C = (-5, -4) and D = (-5, -32)

Explanation:

a) To determine the coordinates of A and B, find the intersection points of  the line "y = 2x + 6" and curve "y = -2x² + 18".

Solve the equation's simultaneously:

y = y

⇒ 2x + 6 =  -2x² + 18

⇒ 2x² + 2x + 6 - 18 = 0

⇒ 2x² + 2x - 12= 0

⇒ 2x² + 6x - 4x - 12 = 0

⇒ 2x(x + 3) - 4(x + 3) = 0

⇒ (2x - 4)(x + 3) = 0

⇒ 2x - 4 = 0, x + 3 = 0

⇒ x = 2, x = -3

Then find value of y at this x points,

at x = 2, y = 2(2) + 6 = 10

at x = -3, y = 2(-3) + 6 = 0

Intersection points: A(2, 10) and B(-3, 0)

b) Given that CD = 28 units. Also stated parallel to y axis so x coordinates for both will be same but differ in y coordinate.

[tex]2x + 6 = -2x^2 + 18 + 28[/tex]

[tex]-2x^2 + 18 + 28-2x - 6 = 0[/tex]

[tex]-2x^2-2x+40=0[/tex]

[tex]-2x^2-10x+8x+40=0[/tex]

[tex]-2(x+5)+8(x+5)=0[/tex]

[tex](-2x+8)(x+5)=0[/tex]

[tex]x = -5, 4[/tex]

[tex]\leftrightarrow \sf C(-5, y_2), \ D(-5, y_2)[/tex]

Find y value for Point C : 2x + 6 = 2(-5) + 6 = -4

Find y value for Point D :  -2x² + 18 =  -2(-5)² + 18 = -32

[tex]\sf \rightarrow Point \ C = (-5, -4)\\ \\\rightarrow Point \ D = (-5, -32)[/tex]

[tex] \qquad \qquad \bf \huge\star \: \: \large{ \underline{Answer} } \huge \: \: \star[/tex]

The straight angles are angles that form a straight line, and their measure = 180°

In the given figure, the Straight angles is :

[tex] \qquad \large \sf {Conclusion} : [/tex]

There are 151200 distinct ways to arrange the letters of the word CONNECTION

The word is given as:

CONNECTION

In the above word, we have the following parameters:

Total number of characters, n = 10

The repeated letters are:

C's = 2

O's = 2

N's = 3

The number of arrangements of the letters is then calculated as:

Arrangements = n!/(C! * O! * N!)

Substitute the known values in the above equation

Arrangements =  10!/(2! * 2! * 3!)

Evaluate the expression

Arrangements =  151200

Hence, there are 151200 distinct ways to arrange the letters of the word CONNECTION

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Answer: 15

Step-by-step explanation: Using the pythagorean theorem A^2 + B^2 = C^2 where C is the hypotenuse of the triangle and any A and B are any sides of the triangle that are not the hypotenuse; we can rewrite this as A^2 = C^2 - B^2 to find out the missing side. I picked A to be the unknown side but you could also choose B. We know the hypotenuse is 39, and side B is 36. 39^2 - 36^2, or 1521 - 1296, is 225.

Now we have A^2 = 225. To get rid of the square, we have to take the square root of both sides. The A^2 cancels to be A, and sqrt(225) is 15. Thus, side A, or the missing side is 15.

Hope this helped!

The effect of  Edible a portion Price on your profit margin , if we only use the as purchased price to determine our cost and selling price is that it will maximize the profit because it will account for every part of the production.

The portion cost can be calculated by multiplying the cost of a usable kg  with the  portion size.

This can be represented as : portion cost = (portion size x cost of usable kg)

It should be noted that Edible portion (EP)  serves as the  portion of food  which will be given top the   customer after  the preparation.

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Answer:

Step-by-step explanation:

Step-by-step explanation:

The value of b will be 43° as it is vertically opposite angle.. !!

Using the arrangements formula, it is found that 1680 arrangements can be made with these numbers.

The number of possible arrangements of n elements is given by the factorial of n, that is:

[tex]A_n = n![/tex]

When there are repeated elements, repeating [tex]n_1, n_2, \cdots, n_n[/tex] times, the number of arrangements is given by:

[tex]A_n^{n_1, n_2, \cdots, n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}[/tex]

For the number 28535852, we have that:

Hence the number of arrangements is:

[tex]A_8^{3,2,2} = \frac{8!}{3!2!2!} = 1680[/tex]

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The interpretation of the p-value is that there is a 0.006 = 0.6% probability of the sample means differing by the amount that they differed on the sample.

At the null hypothesis, it is tested if the means are equal, that is:

[tex]H_0: \mu_1 = \mu_2[/tex]

At the alternative hypothesis, it is tested if the means are different, that is:

[tex]H_1: \mu_1 \neq \mu_2[/tex]

Since we are testing if the means are different, we have a two-tailed test, which means that the p-value is the probability of the means differing by the amount they different on the sample, or a greater amount.

Hence, the interpretation is that there is a 0.006 = 0.6% probability of the sample means differing by the amount that they differed on the sample.

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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 equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.

To write the equation that represents the total cost of a given scenario like the one given, you can use variables to represent each component that makes up the total cost in the given situation.

Thus, let:

a = apple = 3a

s = can of soup = 2s

b = box of cereal = b

c = total cost

We know that unit price of each items are:

Apple = $0.89

Can of soup = $2.98

Box of cereal = $4.99

Equation would be:

c = 3a + 2s + b

Plug in the values

c = 3(0.89) + 2(2.98) + 4.99

Therefore, the equation that represents the total cost, c, of the purchases made by Jake is: c = 3(0.89) + 2(2.98) + 4.99.

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The area of the sector of the circle is: A. 50/3π in.².

The area of a sector that is bounded by two radii of a circle is calculated using the formula, ∅/360 × πr², where we have the following parameters:

Thus, we are given the following regarding the sector of the circle:

Central angle (∅) = 60 degrees

Radius (r) = 10 inches.

Plug in the values into ∅/360 × πr²:

Area of sector = 60/360 × π(10²)

Area of sector = 1/6 × π(100)

Area of sector = 100/6 × π

Area of sector = 50/3 × π

Area of sector = 50/3π in.²

Thus, the area of the sector of the circle is: A. 50/3π in.².

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Answer:

a company powder large group of people to assess the company's reputation in the surrounding

a. Given the 2nd order ODE

[tex]y''(x) = 4y(x) + 4[/tex]

if we substitute [tex]z(x)=y'(x)+2y(x)[/tex] and its derivative, [tex]z'(x)=y''(x)+2y'(x)[/tex], we can eliminate [tex]y(x)[/tex] and [tex]y''(x)[/tex] to end up with the ODE

[tex]z'(x) - 2y'(x) = 4\left(\dfrac{z(x)-y'(x)}2\right) + 4[/tex]

[tex]z'(x) - 2y'(x) = 2z(x) - 2y'(x) + 4[/tex]

[tex]\boxed{z'(x) = 2z(x) + 4}[/tex]

and since [tex]y(0)=y'(0)=1[/tex], it follows that [tex]z(0)=y'(0)+2y(0)=3[/tex].

b. I'll solve with an integrating factor.

[tex]z'(x) = 2z(x) + 4[/tex]

[tex]z'(x) - 2z(x) = 4[/tex]

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

[tex]\left(e^{-2x} z(x)\right)' = 4e^{-2x}[/tex]

[tex]e^{-2x} z(x) = -2e^{-2x} + C[/tex]

[tex]z(x) = -2 + Ce^{2x}[/tex]

Since [tex]z(0)=3[/tex], we find

[tex]3 = -2 + Ce^0 \implies C=5[/tex]

so the particular solution for [tex]z(x)[/tex] is

[tex]\boxed{z(x) = 5e^{-2x} - 2}[/tex]

c. Knowing [tex]z(x)[/tex], we recover a 1st order ODE for [tex]y(x)[/tex],

[tex]z(x) = y'(x) + 2y(x) \implies y'(x) + 2y(x) = 5e^{-2x} - 2[/tex]

Using an integrating factor again, we have

[tex]e^{2x} y'(x) + 2e^{2x} y(x) = 5 - 2e^{2x}[/tex]

[tex]\left(e^{2x} y(x)\right)' = 5 - 2e^{2x}[/tex]

[tex]e^{2x} y(x) = 5x - e^{2x} + C[/tex]

[tex]y(x) = 5xe^{-2x} - 1 + Ce^{-2x}[/tex]

Since [tex]y(0)=1[/tex], we find

[tex]1 = 0 - 1 + Ce^0 \implies C=2[/tex]

so that

[tex]\boxed{y(x) = (5x+2)e^{-2x} - 1}[/tex]

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

The given ordinary differential equation is y''(x) = 4y(x) + 4, y(0) = y'(0) = 1 ... (d).

We are also given a substitution function, z(x) = y'(x) + 2y(x) ... (z).

Putting x = 0, we get:

z(0) = y'(0) + 2y(0),

or, z(0) = 1 + 2*1 = 3.

Rearranging (z), we can write it as:

z(x) = y'(x) + 2y(x),

or, y'(x) = z(x) - 2y(x) ... (i).

Differentiating (z) with respect to x, we get:

z'(x) = y''(x) + 2y'(x),

or, y''(x) = z'(x) - 2y'(x) ... (ii).

Substituting the value of y''(x) from (ii) in (d) we get:

y''(x) = 4y(x) + 4,

or, z'(x) - 2y'(x) = 4y(x) + 4.

Substituting the value of y'(x) from (i) we get:

z'(x) - 2y'(x) = 4y(x) + 4,

or, z'(x) - 2(z(x) - 2y(x)) = 4y(x) + 4,

or, z'(x) - 2z(x) + 4y(x) = 4y(x) + 4,

or, z'(x) = 2z(x) + 4y(x) - 4y(x) + 4,

or, z'(x) = 2z(x) + 4.

The initial value of z(0) was calculated to be 3.

6.a) The differential equation for z(x) is z'(x) = 2z(x) + 4, z(0) = 3.

Transforming z(x) = dz/dx, and z = z(x), we get:

dz/dx = 2z + 4,

or, dz/(2z + 4) = dx.

Integrating both sides, we get:

∫dz/(2z + 4) = ∫dx,

or, {ln (z + 2)}/2 = x + C,

or, [tex]\sqrt{z+2} = e^{x + C}[/tex],

or, [tex]z =Ce^{2x}-2[/tex] ... (iii).

Substituting z = 3, and x = 0,  we get:

[tex]3 = Ce^{2*0} - 2\\\Rightarrow C - 2 = 3\\\Rightarrow C = 5.[/tex]

Substituting C = 5, in (iii), we get:

[tex]z = 5e^{2x} - 2[/tex].

6.b) The value of z(x) is [tex]z(x) = 5e^{2x} - 2[/tex].

Substituting the value of z(x) in (z), we get:

z(x) = y'(x) + 2y(x),

or, 5e²ˣ - 2 = y'(x) + 2y(x),

which gives us:

[tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex] for the initial condition y(x) = 0.

6.c) The value of y(x) is [tex]y(x) = \frac{5e^{2x}}{4} - \frac{1}{4e^{2x}} -1[/tex].

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Answer:

6840

Step-by-step explanation:

multiply combinations

n!/(r!(n − r)!)

n choose r

n (objects) = |n

r (sample) = r

(4 teachers, CHOOSE 2)*(20 students, CHOOSE 3)

C(4,2) times C(20,3) =

C(n,r)=?

C(n,r) = C(4,2)

4! / (2!(4-2)!)

4! /2! x 2!

= 6

C(n,r)=?

C(n,r) = C(20,3)

20! / (3!(20-3)!)

20! /  3! x 17!

= 1140

6 * 1140 = 6840

alegbracom Edwin McCravy

The option A is correct because the value of the statement is 6+3/n-8.

According to the statement

we have given that the statement and we have to write the statement in the numerical form.

So, For this purpose

we know that the given statement is

"Six more than the quotient of three and a number, decreased by eight".

In this statement "Six more than" indicates the sum between the quotient and 6.

And "The quotient of three and a number" indicates a fraction where the numerator is 3 and the denominator is , a number.

"Decreased by eight" indicates a subtraction between the quotient an 8 units".

If we unit all parts of the statements, we would have

6+3/n-8

So, From the given statement, The option A is correct because the value of the statement is 6+3/n-8.

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Step-by-step explanation:

the total sale price for the magazines were

2×4.95 = $9.90

so,

100% = 9.9

1% = 100%/100 = 9.9/100 = 0.099

6.75% = 6.75×1% = 6.75×0.099 = 0.66825 ≈ $0.67

so, he paid in total

$9.90 + $0.67 = $10.57

Answer:

Step-by-step explanation:

D because the quadratic formula is basically [tex]\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex]±

so if we place the equation D into the formula

a = 1

b = +3

c = +24

we can clearly say that D eq is correctly used and can be.

Answer:

80

Step-by-step explanation:

Let the number of adult tickets sold be a and the number of student tickets sold be s.

Substituting the first equation into the second,

[tex]2a+a-20=220 \\ \\ 3a-20=220 \\ \\ 3a=240 \\ \\ a=80[/tex]