At the 6th grade school dance, there are 132 boys, 89 girls, and 14 adults. What is the ratio of adults to boys at the school dance?​

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Answer:

  B)  K′ = (–2,2), L′ = (1,2), M′ = (1,–1), N′ = (–2,–1)

Step-by-step explanation:

Translation 2 units right adds 2 to the x-coordinate.

Translation 4 units upward adds 4 to the y-coordinate.

The translation can be represented by the relation ...

  (x, y) ⇒ (x +2, y +4)

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You can choose the correct answer by looking at the translation of K.

  K(-4, -2) ⇒ K'(-4+2, -2+4) = K'(-2, 2) . . . . . matches choice B

Answer:

The equation of the line is y = -2/3x - 29/3

Step-by-step explanation:

The slope of these points (-7,-5) and (-1,-9) is m = -2/3

Once you plug that into the y = mx + b equation, you can see that the y-intercept is -29/3.

Put all of that into the y = mx + b equation and you'll get -->  y = -2/3x - 29/3

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Answer:

  21.  D

  22.  C

Step-by-step explanation:

21. The expansion of the given expression is ...

  [tex]\displaystyle -\frac{1}{2}\left(-\frac{3}{2}x+6x+1\right)-3x=\frac{3}{4}x-3x-\frac{1}{2}-3x\\\\=\left(\frac{3}{4}-3-3\right)x-\frac{1}{2}=\boxed{-5\frac{1}{4}x-\frac{1}{2}}[/tex]

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22. The least likely team to make the championship game is the one with the lowest probability.

  3/8 < 1/2 < 2/3 < 4/5

The Bulldogs are least likely to play in the championship game.

Answer:

586 cm^3 and 486 in^2

Step-by-step explanation:

4) The volume of the triangular prims is (1/2)*(a*c*h) = 0.5*(8*9*16)=586 cm^3

5) Wrapping paper needed is equal to the surface area of the cube, 6s^2=486 in^2

Answer:

D ) none of these .. i think this help you

Answer:

hope this help you

sorry if it is mistake

Answer:

x=0,-3

Step-by-step explanation:

x(x+3)(x+3)=0

Using the zero product property

x=0  x+3=0 x+3 =0

x=0  x=-3  x=-3

Let

ATQ

[tex]\\ \sf \longmapsto Area=Length\times Width[/tex]

[tex]\\ \sf \longmapsto x(x+3)=28[/tex]

[tex]\\ \sf \longmapsto x^2+3x=28[/tex]

[tex]\\ \sf \longmapsto x^2+3x-28=0[/tex]

[tex]\\ \sf \longmapsto x^2+7x-4x-28=0[/tex]

[tex]\\ \sf \longmapsto x(x+7)-4(x+7)=0[/tex]

[tex]\\ \sf \longmapsto (x-4)(x+7)=0[/tex]

[tex]\\ \sf \longmapsto x=4\:or\:x=-7[/tex]

[tex]\\ \sf \longmapsto Width=4units[/tex]

[tex]\\ \sf \longmapsto Length=4+3=7units[/tex]

Answer:

A 6 Packages of buns and 5 packages of wieners.

Step-by-step explanation:

Because if you multiply 6x10 you get 60 and if you multiply 5x12 you get 60 exactly 1 wieners per bun

Given:

The function is:

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

To find:

The result of the operation [tex]-f(x)=-(x^2-x+1)[/tex].

Solution:

If [tex]g(x)=-f(x)[/tex], then the graph of f(x) is reflected across the x-axis to get the graph of g(x).

We have,

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

The given operation is:

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

So, it will result in a reflection across the x-axis.

Therefore, the correct option is A.

Answer:

A) reflection across the x-axis.

Step-by-step explanation: I took the test

Step-by-step explanation:

Many factors would be used to assess the effectiveness of a human rights campaign, including the following:

Social Influence. Direct Interpersonal Reach. Participant Observation. Reputation. Volume of Search & Interest. Website Traffic.

National Research.

Answer:

Pvalue

Step-by-step explanation:

The Pvalue measure which takes up a value in the range 0 - 1 is a statistical measure used on hypothesis testing to measure the likelihood of obtaining result atleast as extreme as the outcome of the statistical hypothesis test, The Pvalue is used in hypothesis testing to make a case for the alternative hypothesis, which involves being compared with the α - value.

Lower p values favors the adoption of alternative depending on how extreme th α-value is. The Pvalue is dependent on the value if the test statistic.

Answer:

a. L{t} = 1/s² b. L{1} = 1/s

Step-by-step explanation:

Here is the complete question

The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0

Solution

a. L{t}

L{t} = ∫₀⁰⁰[tex]e^{-st}t[/tex]

Integrating by parts  ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt = [tex]e^{-st}[/tex] and v = [tex]\frac{e^{-st}}{-s}[/tex] and du/dt = dt/dt = 1

So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w

So,  ∫₀⁰⁰[tex]e^{-st}t[/tex] =  [[tex]\frac{te^{-st}}{-s}[/tex]]₀⁰⁰ -  ∫₀⁰⁰ [tex]\frac{e^{-st}}{-s}[/tex]

∫₀⁰⁰[tex]e^{-st}t[/tex] =  [[tex]\frac{te^{-st}}{-s}[/tex]]₀⁰⁰ -  ∫₀⁰⁰ [tex]\frac{e^{-st}}{-s}[/tex]

= -1/s(∞exp(-∞s) - 0 × exp(-0s)) + [tex]\frac{1}{s}[/tex] [[tex]\frac{e^{-st} }{-s}[/tex]]₀⁰⁰

= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]

= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]

= -1/s[(0 - 0] - 1/s²[0 - 1]

= -1/s[(0] - 1/s²[- 1]

= 0 + 1/s²

= 1/s²

L{t} = 1/s²

b. L{1}

L{1} = ∫₀⁰⁰[tex]e^{-st}1[/tex]

= [[tex]\frac{e^{-st} }{-s}[/tex]]₀⁰⁰

= -1/s[exp(-∞s) - exp(-0s)]

= -1/s[exp(-∞) - exp(-0)]

= -1/s[0 - 1]

= -1/s(-1)

= 1/s

L{1} = 1/s

Answer:

Hello,

Just using the theorem of Thalès,

Step-by-step explanation:

Let say h the hight of the building

[tex]\dfrac{h}{3} =\dfrac{42.75}{1.23}\\\\h=104.268296...\approx{104(m)}[/tex]

Answer:

H0: µd = 0 (claim)

H1: µd ≠ 0

This is a two-tail t-test for µd

Step-by-step explanation:

This is a paired (dependent) sample test, with its hypothesis is written as :

H0: µd = 0

H1: µd ≠ 0

From the equality sign used in the hypothesis declaration, a not equal to ≠ sign in the alternative hypothesis is used for a two tailed t test

The data isn't attached, however bce the test statistic cannot be obtained. However, the test statistic formular for a paired sample is given as :

T = dbar / (Sd/√n)

dbar = mean of the difference ; Sd = standard deviation of the difference.

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Answer:

  (e) f′(x)=2xg(x)+x²g′(x)

Step-by-step explanation:

The product rule applies.

  (uv)' = u'v +uv'

__

Here, we have u=x² and v=g(x). Then u'=2x and v'=g'(x).

  f(x) = x²·g(x)

  f'(x) = 2x·g(x) +x²·g'(x)

Answer:

8/52

Step-by-step explanation:

The first thing to do is write it out;

How many aces are in a deck and how many sixes?

There are 4 of each so, 4+4 = 8 therefore our beginning ratio will be;

8/52 cards are going to be an ace or a six.

Answer:

4

Step-by-step explanation:

There are 4 reflectional symmetry

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Answer:

  B B C C A A

Step-by-step explanation:

If we number the equations 1 to 6 left to right, then we have ...

Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

[tex]A(t) = A(0)e^{-kt}[/tex]

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, [tex]A(10) = 0.7278A(0)[/tex]. We use this to find k.

[tex]A(t) = A(0)e^{-kt}[/tex]

[tex]0.7278A(0) = A(0)e^{-10k}[/tex]

[tex]e^{-10k} = 0.7278[/tex]

[tex]\ln{e^{-10k}} = \ln{0.7278}[/tex]

[tex]-10k = \ln{0.7278}[/tex]

[tex]k = -\frac{\ln{0.7278}}{10}[/tex]

[tex]k = 0.03177289938 [/tex]

Then

[tex]A(t) = A(0)e^{-0.03177289938t}[/tex]

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

[tex]A(5) = A(0)e^{-0.03177289938*5}[/tex]

[tex]A(5) = 0.8531[/tex]

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Answer:

1

Step-by-step explanation:

the maximum of f(X)=sin(X) is 1

Answer:

Step-by-step explanation:

(1 + 2i) / (1 + i)                            Rationalize the denominator.

(1 + 2i)(1+i) / (1 + i)(1-i)                  Remove the brarckets

(1 + i + 2i - 2) / (1 - i + i  - i^2)     Combine

-1 + 3i / (2)                                i^2 = - 1 in the denominator

Answers:

The boxplot is shown below.

=========================================

Explanation:

What your teacher wants is the five number summary.

This consists of:

Given in that exact order.

The given data set is  { 8, 19, 11, 20, 2, 14, 17, 9, 15}

This sorts to  {2, 8, 9, 11, 14, 15, 17, 19, 20}

From this sorted set, we see that 2 is the smallest item. So this is the min value. This is data point 1.

The max is the largest item, which in this case is 20, so this value goes in the box for data point 5.

---------------------

Count out the number of values in the sorted set. You should count out n = 9 items.

Because n is odd, this means the median is in slot n/2 = 9/2 = 4.5 = 5

The value in the 5th slot is 14 which is the median (data point 3).

-----------------------

Once you determine the median, break the sorted set up like so

L = {2, 8, 9, 11}

U = {15, 17, 19, 20}

L is the lower set of values smaller than the median

U is the upper set of values larger than the median

The median itself is not part of set L and not part of set U either. It's ignored entirely from this point on.

From here, we find the middle values of L and U

You should find that the middle value of L is (8+9)/2 = 17/2 = 8.5 which is the value of Q1 (data point 2)

And also, the middle value of set U is (17+19)/2 = 36/2 = 18 which is the value of Q3 (data point 4)

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To wrap everything up, we have this five number summary

These will determine the features of the boxplot as shown below.

In this case, there are no outliers.

Answer:

(B) -4

Step-by-step explanation:

so, do the multiplications and see :

(a - 1/a)² = a² - 2×a/a + 1/a² = a² - 2 + 1/a²

(a + 1/a)² = a² + 2×a/a + 1/a² = a² + 2 + 1/a²

now we need to subtract the second from the first :

(a² - 2 + 1/a²) - (a² + 2 + 1/a²) =

= a² - 2 + 1/a² - a² - 2 - 1/a² = -4

and that's it !

First, I'll make f(x) = sin(px) + cos(px) because this expression shows up quite a lot, and such a substitution makes life a bit easier for us.

Let's apply the first derivative of this f(x) function.

[tex]f(x) = \sin(px)+\cos(px)\\\\f'(x) = \frac{d}{dx}[f(x)]\\\\f'(x) = \frac{d}{dx}[\sin(px)+\cos(px)]\\\\f'(x) = \frac{d}{dx}[\sin(px)]+\frac{d}{dx}[\cos(px)]\\\\f'(x) = p\cos(px)-p\sin(px)\\\\ f'(x) = p(\cos(px)-\sin(px))\\\\[/tex]

Now apply the derivative to that to get the second derivative

[tex]f''(x) = \frac{d}{dx}[f'(x)]\\\\f''(x) = \frac{d}{dx}[p(\cos(px)-\sin(px))]\\\\ f''(x) = p*\left(\frac{d}{dx}[\cos(px)]-\frac{d}{dx}[\sin(px)]\right)\\\\ f''(x) = p*\left(-p\sin(px)-p\cos(px)\right)\\\\ f''(x) = -p^2*\left(\sin(px)+\cos(px)\right)\\\\ f''(x) = -p^2*f(x)\\\\[/tex]

We can see that f '' (x) is just a scalar multiple of f(x). That multiple of course being -p^2.

Keep in mind that we haven't actually found dy/dx yet, or its second derivative counterpart either.

-----------------------------------

Let's compute dy/dx. We'll use f(x) as defined earlier.

[tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\y = \ln\left(f(x)\right)\\\\\frac{dy}{dx} = \frac{d}{dx}\left[y\right]\\\\\frac{dy}{dx} = \frac{d}{dx}\left[\ln\left(f(x)\right)\right]\\\\\frac{dy}{dx} = \frac{1}{f(x)}*\frac{d}{dx}\left[f(x)\right]\\\\\frac{dy}{dx} = \frac{f'(x)}{f(x)}\\\\[/tex]

Use the chain rule here.

There's no need to plug in the expressions f(x) or f ' (x) as you'll see in the last section below.

Now use the quotient rule to find the second derivative of y

[tex]\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{f'(x)}{f(x)}\right]\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-f'(x)*f'(x)}{(f(x))^2}\\\\\frac{d^2y}{dx^2} = \frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2}\\\\[/tex]

If you need a refresher on the quotient rule, then

[tex]\frac{d}{dx}\left[\frac{P}{Q}\right] = \frac{P'*Q - P*Q'}{Q^2}\\\\[/tex]

where P and Q are functions of x.

-----------------------------------

This then means

[tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} + \left(\frac{f'(x)}{f(x)}\right)^2 + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2}{(f(x))^2} +\frac{(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)-(f'(x))^2+(f'(x))^2}{(f(x))^2} + p^2\\\\\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\[/tex]

Note the cancellation of -(f ' (x))^2 with (f ' (x))^2

------------------------------------

Let's then replace f '' (x) with -p^2*f(x)

This allows us to form  ( f(x) )^2 in the numerator to cancel out with the denominator.

[tex]\frac{f''(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*f(x)*f(x)}{(f(x))^2} + p^2\\\\\frac{-p^2*(f(x))^2}{(f(x))^2} + p^2\\\\-p^2 + p^2\\\\0\\\\[/tex]

So this concludes the proof that [tex]\frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + p^2 = 0\\\\[/tex] when [tex]y = \ln\left(\sin(px)+\cos(px)\right)\\\\[/tex]

Side note: This is an example of showing that the given y function is a solution to the given second order linear differential equation.

The focus here is the use of "Compounding interest rate" and these entails addition of interest to the principal sum of the deposit.

The below calculation is to derive maturity value when annual rate of 3.1% is applied.

Principal = $5,400

Annual rate = 3.1% semi-annually for 1 years

A = P(1+r/m)^n*t where n=1, t=2

A = 5,400*(1 + 0.031/2)^1*2

A = 5,400*(1.0155)^2

A = 5,400*1.03124025

A = 5568.69735

A = $5,568.70.

In conclusion, the accrued value she will get after one years for this account is $5,568.70,

- The below calculation is to derive maturity value when the amount compounds continuously at an annual rate of 4%

Principal = $5,400

Annual rate = 4% continuously

A = P.e^rt where n=1

A = 5,400 * e^(0.04*1)

A = 5,400 * 1.04081077419

A = 5620.378180626

A = 5620.378180626

A = $5,620.39.

In conclusion, the accrued value she will get after one years for this account is $5,620.39.

Referring to how much would Tyra's balance be from Account 2 over 3.7 years. It is calculated as follows:

Annual rate = 4% continuously

A = P.e^rt where n=3.7

A = 5,400 * e^(0.04*3.7)

A = 5,400 * e^0.148

A = 5,400 * 1.15951289636

A = 6261.369640344

A = $6,261.37

Therefore, the accrued value she will get after 3.7 years for this account is $6,261.37

Learn more about Annual rate here

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