plsss help and please dont be scammer

The maximum height of the firework is; 387.78 ft

The time for which the firework is in the air before it explodes is; 1 second

We are given the function;

h = -(500/9)t² + (1000/3)t + 10

where;

h is height after time of t seconds

At height of zero, the firework would either be about to launch or when it has come down.

Thus, let us set h = 0 to find the times.

0 = -(500/9)t² + (1000/3)t + 10

Using quadratic equation calculator, we have;

t ≈ 2 seconds

Now the  firework explodes at the highest point which will be the mid point of where the launching started and the endpoint where it stopped.

This means time at midpoint = (0 + 2)/2 = 1 sec

It explodes after 1 second.

Thus maximum height at this time is;

h = -(500/9)(1)² + (1000/3)(1) + 10

h = 387.78 ft

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The slope of the line is 7/9

It is important to note that the equation of a line is represented as;

y = mx + c

Where;

The formula for calculating the slope of a line is expressed as;

Slope, m = y₂ - y₁/x₂ - x₁

Now, let's substitute the values into the formula from the points given we have;

Slope, m =1 -(-6)/ -3 - (-6)

expand the bracket

Slope, m = 1 + 6/ 3 + 6

add the values

Slope, m = 7/9

Hence, the value is 7/9

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Step-by-step explanation:

4/x + 4 = x - 2/x

multiply everything by x

4 + 4x = x² - 2

x² - 4x - 6 = 0

the general solution to such a quadratic equation

ax² + bx + c = 0

is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

x = (4 ± sqrt((-4)² - 4×1×-6))/(2×1) =

= (4 ± sqrt(16 + 24))/2 = (4 ± sqrt(40))/2 =

= (4 ± sqrt(4×10))/2 = (4 ± 2×sqrt(10))/2 =

= 2 ± sqrt(10)

x1 = 2 + sqrt(10) = 5.16227766...

x2 = 2 - sqrt(10) = -1.16227766...

Table

Top right: [tex]5+194=199[/tex]

Middle right: [tex]217+13=230[/tex]

Bottom left: [tex]5+217=222[/tex]

Bottom middle: [tex]194+13=207[/tex]

Bottom right: [tex]222+207=429[/tex]

Question 1

[tex]\frac{207}{429} \approx \boxed{0.4825}[/tex]

Question 2

[tex]\frac{13}{429} \approx \boxed{0.0303}[/tex]

Question 3

[tex]\frac{194+217}{429} \approx \boxed{0.9580}[/tex]

The correct interpretation of the confidence interval is: "We are 95% confident that between 50.6% and 55.4% of California adult residents say that marijuana should be legal." (Option A)

The confidence interval represents the range of values within which we are 95% confident that the true population parameter (in this case, the percentage of California adult residents who say that marijuana should be legal) lies. It is specific to the population of California adult residents that was surveyed, and cannot be generalized to other states or to all American adults.

Therefore, Option A is the correct answer that we are 95% confident that between 50.6% and 55.4% of California adult residents say that marijuana should be legal.

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The confidence interval denotes the range of values within which we are 95% certain that the true population parameter (in this case, the proportion of adult Californians who support marijuana legalization) resides. It cannot be extrapolated to adults in other states or to all Americans as a whole because it is specific to the demographic of adult inhabitants of California who were polled.

Accordingly, Option A is the right response, and we are 95% certain that between 50.6% and 55.4% of adult inhabitants of California agree that marijuana should be legal.

Answer:

Prove that for any x0∈X and any r>0, the open ball Br(xo) is open. My attempt: Let y∈Br(x0). By definition, d(y,x0)<r

Step-by-step explanation:

Prove that for any x0∈X and any r>0, the open ball Br(xo) is open. My attempt: Let y∈Br(x0). By definition, d(y,x0)<r

Answer:This statement is not necessarily true

Step-by-step explanation:In general, the rate at which thermal energy is transferred depends on several factors, including the temperature difference between the material and its surroundings, the type of material, and its thermal conductivity. Some materials, such as air and glass, are poor conductors of heat and are therefore considered insulators. However, other materials, such as metal, are good conductors of heat and are not considered insulators.

To solve the given system of equations using elimination, we can add the two equations together to eliminate one of the variables. Doing this, we get 3x + 9y + (-10x - 6y) = 33 + (-14), which simplifies to -7x + 3y = 19. We can then solve for y by dividing both sides of the equation by 3, giving us y = 19/3 = 6.33. We can substitute this value for y in either of the original equations to solve for x. For example, if we substitute 6.33 for y in the first equation, we get 3x + 9 * 6.33 = 33, which simplifies to 3x = 3.67. Dividing both sides of the equation by 3, we get x = 1.22. Therefore, the solution of the system is (1.22, 6.33), which is approximately (1, 6). The answer choice that is closest to this solution is (d) (-1, 4). Thus, the solution of the system is (-1, 4).

Answer:

Initial balance 500, compounds 4 times at an interest rate of 15%

Step-by-step explanation:

Initial balance 500, compounds 4 times at an interest rate of 15%

Answer:

1. A(t) = 65t + 700

2.B(t) = 40t + 900

3. High School A and High School B will both have 1,220 students in the 8th year

Step-by-step explanation:

1. The equation for the number of students in High School A represents a linear function.

⭐ What is a linear function?

Let's write the equation for High School A in the [tex]y = mx+b[/tex] format, known as slope-intercept form:

∴ High School A: [tex]y = 65x + 700[/tex]

2. Let's write the equation for High School B in the [tex]y = mx + b[/tex] format, known as slope-intercept form:

∴ High School B: [tex]y = 40x + 900[/tex]

3. To find at what year High School A and High School B will have the same number of students, we need to solve a system of linear equations.

⭐What is a system of linear equations?

For this problem, let's set both linear equations equal to each other to see at what point will the high school populations be the same.

[tex]A(t) = B(t)[/tex]

[tex]65t + 700 = 40t + 900[/tex]

[tex]25t = 200[/tex]

[tex]t = 8[/tex]

Now we know that in the 8th year, High School A will have the same population as High School B.

We need to find what the population will be in year 8.

Thus, substitute the value of t into one of the functions and solve.

I am choosing to substitute t into A(t), but you can also do B(t).

[tex]A(8) = 65(8) + 700[/tex]

[tex]A(8) = 520 + 700[/tex]

[tex]A(8) = 1,220[/tex]

⚠️!!! CAUTION !!! ⚠️

Some people may stop at this point and write that in the 8th year, both high schools will have a population of 1,220 students.

However, you should also substitute 8 into the other function you didn't substitute it into to make sure that 8 is correct.

[tex]B(8) = 40(8) + 900[/tex]

[tex]B(8) = 320 + 900[/tex]

[tex]B(8) = 1,220[/tex]

∴ In the 8th year, both high schools will have a population of 1,220 students.

No, A is not defective.

The rank–nullity theorem is a theorem in linear algebra, which asserts that the dimension of the domain of a linear map is the sum of its rank (the dimension of its image) and its nullity (the dimension of its kernel).

Given here  A be a 4 x 4 matrix and let λ be an eigenvalue of multiplicity 3 and rank 1, then by rank nullity theorem we have

    Rank(A-λI) + Nullity((A-λI) =4

⇒dimN(N-λI) = Nullity(A-λI)

⇒dimN(N-λI) = 4 - Rank(A-λI)

⇒dimN(N-λI)  = 4-1

⇒dimN(N-λI) = 3

Therefore 3 linearly independent eigenvectors belong to eigenvalue λ. Since λ is an eigenvalue of multiplicity 3, this ensures that there is a different eigenvalue different from λ. Along with 3 linearly independent eigenvectors in the eigenspace of λ, this gives us 4 linearly independent eigenvectors in the matrix A. Now we An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. Thus A is nondefective

Hence, matrix A is not defective.

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Using the points A(1,5), B(1,-4), C(-3,5), D(-3,-4) the following are matched:

Distance between points B and C = √97.

Distance between points A and B = 9.

Midpoint between points  B and C = (-1, 1/2).

Midpoint between C and D = (-3, 1/2).

Slope of the line through B and D = 0.

The distance between two points of a line is given as:

Distance = [tex]\sqrt{c - a)^2 + (d - b)^2}[/tex]

We have,

A(1,5), B(1,-4), C(-3,5) and D(-3,-4).

The distance between A and B.

= [tex]\sqrt{(1 - 1)^2 + (-4 - 5)^2}[/tex]

= √81

= 9

The distance between B and C.

= [tex]\sqrt{(-3 - 1)^2 + (5 + 4)^2}[/tex]

= √(16 + 81)

= √97

The midpoint (x, y) between B and C.

x = (1 - 3)/2 = -1

y = (-4 + 5) / 2 = 1/2

The midpoint (x, y) between C and D.

x = (-3 -3) / 2 = -6/2 = -3

y = (5 - 4) / 2 = 1/2

The slope of the line through points B and D.

= (-4 + 4) / (-3 - 1)

= 0/ -4

= 0

Thus,

The distance between A and B is 9.

The distance between B and C is √97.

The midpoint between B and C is (-1, 1/2).

The midpoint between C and D is (-3, 1/2).

The slope of the line through B and D is 0.

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Answer: Pythagorean triplets.

Step-by-step explanation:

not sure but this is the triangle that helps at all.

Answer:

Step-by-step explanation:

to find mean you add all numbers and divide by the amount of numbers, 10+4+8+15+17+6+19+16 is 95, 95 divided by 8 is 11.875

Answer:

1

Step-by-step explanation:

A)

[tex]( \sqrt{3 } - \sqrt{2} )( \sqrt{3} + \sqrt{2} )( \sqrt{5 } - \sqrt{4} )( \sqrt{5} + \sqrt{4} )....[/tex]

using difference of two squares

[tex]( \sqrt{a} + \sqrt{b} )(\sqrt{a} - \sqrt{b} ) = a - b[/tex]

we can rewrite the A) to 1×1×1×1.... so the answer would be 1

Answer:

22

There are 22 multiples that end in 22.

Answer:

  (c)  A line that goes through (0, 0) and (3.6, 7.2)

Step-by-step explanation:

You want to find the correct descriptor of the proportional relationship that includes points (x, y) = (2.1, 4.2), (3.6, 7.2), (12.2, 24.4).

The graph of the points in the table will go through points with coordinates that have the x-value of the table listed first. Points (24.4, 12.2), (7.2, 3.6) are not on the graph. (The table y-value is listed first for those ordered pairs.)

A proportional relationship graphs as a straight line through the origin, so this graph can be described as ...

  A line that goes through (0, 0) and (3.6, 7.2)

Answer:

 (c)  A line that goes through (0, 0) and (3.6, 7.2)

Step-by-step explanation:

Answer:

x=5

Step-by-step explanation:

(Not sure how I was supposed to do the model..so I ended up doing it this way..)

3x-5=10

Add 5 on both sides and then you get

3x=15

divide both sides by 3 and then you get

x=5

(Hopefully you can see my model even if it's just a little...)
For the model I drew 3 rectangles and put an x inside each one of them. for the -5, I drew 5 squares with a minus symbol/sign inside. And then I put an equal sign beside and I drew 10 squares with nothing inside but you can draw positive signs in it to show that it's positive..

So basically mines was

3 rectangles with x inside -5 squares with - inside = 10 squares

Answer:  2x-4

Step-by-step explanation:

Answer:

solution

you must multiply a difference of x and 2 which is. >{x-2}

mathematical

4 x {x-2}

The number of small rooms reserved is 4.

The number of large rooms reserved is 8.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign.

We have,

Small rooms are denoted as S.

Large rooms are denoted as L.

Now,

We will make two equations.
L = 2S _____(1)

8L + 6S = 88 ______(2)

Putting (1) in (2) we get,

8L + 6S = 88

8 x (2S) + 6S = 88

16S + 6S = 88

22S = 88

S = 88/22

S = 8/2

S = 4

Now,

Putting S = 4 in (1) we get,

L = 2 x 4

L = 8

Thus,

The number of small rooms and large rooms reserved is 4 and 8.

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Answer:

(a)  $897.72:  26 yr

     $1526.49:  10 yr

(b)  See below.

(c)  x = 705

     more, $705, longer it will take

Step-by-step explanation:

Given equation:

[tex]t=16.708 \ln \left(\dfrac{x}{x-705}\right)[/tex]

where:

[tex]\begin{aligned}x=897.72 \implies t& =16.708 \ln \left(\dfrac{897.72}{897.72-705}\right)\\& =16.708 \ln \left(4.65815691...\right)\\&=16.708(1.53861985...)\\&=25.70726...\\&=26\; \rm years\end{aligned}[/tex]

[tex]\begin{aligned}x=1526.49 \implies t& =16.708 \ln \left(\dfrac{1526.49}{1526.49-705}\right)\\& =16.708 \ln \left(1.85819669...\right)\\& =16.708(0.619606496...)\\&=10.352385...\\&=10 \; \rm years\end{aligned}[/tex]

To approximate the total amounts paid (in dollars) over the term of the mortgage, multiply the monthly payment by the term.

Please note I have provided two calculations per monthly payment:

(1) by using the exact term, and (2) using the rounded term from part (a).

[tex]\implies \$897.72 \times 25.7072605... \times 12 =\$276935.06[/tex]

[tex]\implies \$897.72 \times 26 \times 12=\$280088.64[/tex]

[tex]\implies \$1526.49 \times 10.3523853... \times 12=\$189633.75[/tex]

[tex]\implies \$1526.49 \times 10 \times 12 =\$183178.80[/tex]

To calculate the amount of interest costs (in dollars) in each case, subtract $150,000 from the total amounts paid:

[tex]\$897.72\implies 276935.06-150000=\$126935.06[/tex]

[tex]\$897.72 \implies 280088.64-150000=\$130088.64[/tex]

[tex]\$1526.49 \implies 189633.75-150000=\$39633.75[/tex]

[tex]\$1526.49\implies 183178.80-150000=\$33178.80[/tex]

The natural logarithm of a negative number cannot be taken.

Therefore, x > 705.

So the vertical asymptote for the model is:

The monthly payment must be more than $705, and the closer to this value the payment is, the longer it will take to pay off the mortgage.

Answer:

  x ≤ 40 . . . miles per hour

Step-by-step explanation:

You want to know the values of x for which d(x) = 0.05x² +x is at most 120.

  d(x) ≤ 120 . . . . . . . . the required limit

  0.05x² +x ≤ 120 . . . . . substitute the expression for d(x)

  x² +20x -2400 ≤ 0 . . . . . multiply by 20, subtract 2400

  (x -40)(x +60) ≤ 0 . . . . . . . . factor. Factors are zero for x=40, x=-60.

The product will be negative for x-values between the zeros:

  -60 ≤ x ≤ 40

We only care about positive values of x:

  0 ≤ x ≤ 40 . . . . .  miles per hour

With the help of Linear relationships The difference in the rate of change between the two containers is 1/5 gallon per minute.

what is Linear relationships?

A statistical word used to express a straight-line relationship between two variables is a linear relationship (or linear association). Graphs and mathematical equations of the form y = mx + b can both be used to represent linear relationships. In everyday life, linear relationships are quite prevalent.

Given:

Volume of water in each container.

To find:

Difference in the rate of change.

Solution:

Take any two points on container 1.

Let the points are (10, 2) and (20, 4).

m = (y2 - y1) / ( x2 - x1 )

   = ( 4 - 2 ) / ( 20 - 10 )

   = 2/ 10

   = 1/ 5

Rate of change for container 1 is .

Take any two points on container 2.

Let the points are (5, 2) and (10, 4).

m = (y2 - y1) / ( x2 - x1 )

   = ( 4 - 2 ) / ( 10 - 5)

   = 2 / 5

Rate of change for container 2 is 2/5

Difference = (2 /5 ) - ( 1/ 5 )

                  = 1 / 5  

Hence, The difference in the rate of change between the two containers is 1/5 gallon per minute.

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The students corresponding to those 200 numbers would beasked to participate in the survey.

What is corresponding ?

Having or participating in the same relationship (such as kind, degree, position, correspondence, or function) especially with regard to the same or like wholes (such as geometric figures or sets) corresponding parts of similar triangles. : related, accompanying.

Have given,

a) The administrators could number an alphabetical list of students from 1 to 2,500.Then use a random number generator from a calculator or computer to generate 200 unique random numbers from 1 to2,500.(if you don’t say “unique”, then say “ignoring repeats”)The students corresponding to those 200 numbers would beasked to participate in the survey.

b) One advantage is that stratified random sampling guarantees that each of the school-level strata will have some representation, because it is possible that a simple random sample would miss one or more of the strata completely.(NOTE: It is NOT sufficient to say that a stratified sample “makes it MORELIKELY that all school-levels are represented in the sample”… the student must state that stratifying by school level GUARANTEES or ENSURES that all school levels are represented in the sample.)

The students corresponding to those 200 numbers would beasked to participate in the survey.

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Answer:

f(g(-1)) = -8

Step-by-step explanation:

⭐ The two equations we are given are:

This problem is an example of composite functions.

⭐What are composite functions?

First, we have to compute the value for the 2nd function you see. The 2nd function is inside f(x), and it is g(-1).

Essentially, we have to find the corresponding y-value for when x = -1 in g(x)

[tex]g(-1) = 2(-1) + 5[/tex]

[tex]g(-1) = 3[/tex]

Next, take this value and substitute it as the x-value for f(x).

[tex]f(3) = 1-x^2[/tex]

Now we have to find the y-value for f(x) for when x = 3.

[tex]f(3) = 1 -3^2[/tex]

⚠️⚠️⚠️!!! CAUTION !!! ⚠️⚠️⚠️

Some people may make the mistake of computing f(3) like this:

[tex]f(3) = 1 - 3^2[/tex]

[tex]f(3) = 1+9\\f(3) = 10[/tex]

This is wrong because only 3 is being squared, not -3. Make sure to read the equations you are given carefully to avoid mistakes like this.

[tex]f(3) = 1-3^2\\f(3) = 1-9\\f(3) = -8[/tex]

∴ f(g(-1)) = -8

The management assertions for the given parts is explains below.

What is assertion?

Management assertions are statements made about specific elements of a business by members of management. The idea is mainly applied to the auditing of financial statements of a company, where the auditors rely on various business assertions.

Consider, the following.

(a) Recorded sales transactions have been recorded

          Category of management.              Classes of transactions                

           Name of assertion.          .               Occurance

(b)  There are no lines are other restrictions in accounts receivable

          Category of management.             Account balances                

          Name of assertion.                          Rights and obligations.          

(c) All sales transactions have been recorded

           Category of management.             Classes of transactions                

            Name of assertion.                        Completeness

(d) Receivables are appropriately classified as to trade and other receivables in the financial  statements and are clearly described.

       Category of management.              Presentation and disclosure

       Name of assertion.                        Classification and understandabilty

(e) sales transactions have been recorded in proper period

         Category of management.                   Classes of transactions.            

          Name of assertion.                               Cut off.  

(f) Account receivables recorded at correct amounts

         Category of management.                  Account balances      

         Name of assertion.                              Valuation and allocation

(g) sales transactions recorded at appropriate accounts

         Category of management.                 Class of transactions                

          Name of assertion.                            Classification

(h) All required disclosure about sales and receivables are at recorded amounts

     Category of management.                 Presentation and disclosure                  

     Name of assertion.                             Completeness

(k) Sales transactions have been recorded at correct amounts

        Category of management.                   Account balances                  

        Name of assertion.                         Completeness

(L) Recorded accounts receivable exist

        Category of management.                Account balances.                  

        Name of assertion.                            Existence

(m) Disclosure related to sales and receivable relate to entity

         Category of management.               Presentation and disclosure                

         Name of assertion.                           Occurance and obligation

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Answer:

  see the attachment for a graph

Step-by-step explanation:

You seem to want a graph of the inequalities ...

The equations of the boundary lines are given by replacing the inequality symbol with an equal sign:

These are graphed in the usual way. It is often convenient to find the y-intercept, then use the slope to locate other points on the line.

y = -3/2x -2

  The y-intercept is (0, -2). The line has a rise/run of -3/2, so goes down 3 units for each 2 to the right. Another point on the line is (2, -5).

y = 1/2x +6

  The y-intercept is (0, 6). The line has a rise/run of 1/2, so goes up 1 unit for each 2 to the right. Another point on the line is (2, 7).

When the "or equal to" symbol (≤ or ≥) is used, the boundary line is solid. When the "or equal to" case is not part of the solution, the boundary line is dashed.

The line with negative slope through (0, -2) is solid; the line with positive slope through (0, 6) is dashed.

All you need to determine shading is a variable with a positive coefficient, and the inequality symbol.

  y ≤ . . . . . tells you shading is below the solid line

  y > . . . . . tells you shading is above the dashed line

We notice x has a positive coefficient in the second inequality, so we could determine shading from ...

  > x . . . . . tells you shading is left of the line (where x values are less than those on the line)

Of course, you can rearrange the inequality so a variable of interest has a positive coefficient. For example, we could add 3/2x to the first inequality to get ...

  3/2x + y ≤ -2

Then, looking at the x-variable, we see ...

  x ≤ . . . . . tells you shading is to the left of the line

Answer:

See attachment.

Step-by-step explanation:

When graphing inequalities:

Treat the inequalities as equations (swap the inequality sign for an equals sign) to find two points on the line to help draw the lines.

Inequality 1

[tex]y \leq -\dfrac{3}{2}x-2[/tex]

[tex]\begin{aligned}x=0 \implies y&=-\dfrac{3}{2}(0)-2\\y&=-2\end{aligned}[/tex]

[tex]\begin{aligned}x=2 \implies y&=-\dfrac{3}{2}(2)-2\\y&=-5\end{aligned}[/tex]

Plot the points (0, -2) and (2, -5).

Draw a solid straight line through the points.

Shade under the line.

Inequality 2

[tex]y > \dfrac{1}{2}x+6[/tex]

[tex]\begin{aligned}x=0 \implies y&=\dfrac{1}{2}(0)+6\\y&=6\end{aligned}[/tex]

[tex]\begin{aligned}x=4 \implies y&=\dfrac{1}{2}(4)+6\\y&=8\end{aligned}[/tex]

Plot the points (0, 6) and (4, 8).

Draw a dashed straight line through the points.

Shade above the line.

The solution to the two inequalities is the overlap of the shaded parts.