[10] solve the equation R/6=9
(A) R = 8 (B) R = 15 (C) R = 54 (D) R = 96​

The conjugate of the expression √x - 5 - 2 when x > 5 is -(√x - 3) / (√x + 2).

The conjugate of an expression is obtained by multiplying both the numerator and denominator of the expression by the conjugate of the denominator.

We have the following expression below:

√x - 5 - 2

To determine the conjugate of the expression above, we first consider the denominator:

√x - 5

The conjugate of this expression is:

√x + 5

Therefore, to obtain the conjugate of the expression given above, we multiply the numerator and denominator by √x + 5.√x - 5 - 2 * √x + 5 / √x - 5 - 2 * √x + 5

This expression simplifies to:

-(√x - 3) / (√x - 5 + 2 * √x + 5)= -(√x - 3) / (√x + 2)

Therefore, the conjugate of the expression √x - 5 - 2 when x > 5 is -(√x - 3) / (√x + 2).

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Answer:

a₂₀ = 69

Step-by-step explanation:

The nth term of an arithmetic sequence is

\(a_{n}\) = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 12 and d = 15 - 12 = 3 , then

a₂₀ = 12 + (3 × 19) = 12 + 57 = 69

A) The control limits for the x-bar chart are approximately:

Upper Control Limit (UCL) = 55.752 lb

Lower Control Limit (LCL) = 53.748 lb

B) The control limits for the 3-sigma R-chart are approximately:

Upper Control Limit (UCR) = 3.759 lb

Lower Control Limit (LCR) = 1.363 lb

To calculate the control limits for the x-bar chart and the 3-sigma R-chart, we need to use the given information about the sample size and the average range.

a) Control limits for the x-bar chart:

The control limits for the x-bar chart are typically calculated using the formula:

UCL = x(bar) + A₂ × R-bar

LCL = x(bar) - A₂ × R-bar

Where:

UCL = Upper Control Limit

LCL = Lower Control Limit

x(bar) = Overall mean

A₂ = Constant depending on the sample size (from statistical tables)

R-bar = Average range

In this case, the sample size is 7, so we need to find the value of A₂ from the statistical tables. For a sample size of 7, A₂ is approximately 0.577.

Using the given information:

x(bar) = 54.75 lb (Overall mean)

R-bar = 1.78 lb (Average range)

A₂ = 0.577

Substituting these values into the formula, we can calculate the control limits for the x-bar chart:

UCL = 54.75 + 0.577 × 1.78

UCL ≈ 55.752

LCL = 54.75 - 0.577 × 1.78

LCL ≈ 53.748

Therefore, the control limits for the x-bar chart are approximately:

Upper Control Limit (UCL) = 55.752 lb

Lower Control Limit (LCL) = 53.748 lb

b) Control limits for the 3-sigma R-chart:

The control limits for the R-chart can be calculated using the formula:

UCR = D₄ × R-bar

LCR = D₃ × R-bar

Where:

UCR = Upper Control Limit for R-chart

LCR = Lower Control Limit for R-chart

D₄, D₃ = Constants depending on the sample size (from statistical tables)

For a sample size of 7, the values of D₄ and D₃ are approximately 2.115 and 0.765, respectively.

Using the given information:

R-bar = 1.78 lb (Average range)

D₄ = 2.115

D₃ = 0.765

Substituting these values into the formula, we can calculate the control limits for the 3-sigma R-chart:

UCR = 2.115 × 1.78

UCR ≈ 3.759

LCR = 0.765 × 1.78

LCR ≈ 1.363

Therefore, the control limits for the 3-sigma R-chart are approximately:

Upper Control Limit (UCR) = 3.759 lb

Lower Control Limit (LCR) = 1.363 lb

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The complete question is :

35 sample of size 7 each were taken from a fertiliser filling machine at lawm ltd. the result were overall mean = 54.75lb average range = 1.78lb

a) For the given sample size, the control 3-sigma lmits for x bar

Upper Contol Litit (UCL-) = . (round your response to thee decimal places) Lower Control Limit (LCL)= . (round your response to three decimat places).

b) The control lirtits for the 3-sigma R-chart are:

Upper Control Limit (UCR)=  . (cound your rospanse to thee decimal places).

Lower Control Limit (LC,R)=  .(round your response fo three decimal places)

To find the y-coordinate of point P on the unit circle, given that its x-coordinate is 5/13, we can utilize the Pythagorean identity for points on the unit circle.

The Pythagorean identity states that for any point (x, y) on the unit circle, the following equation holds true:

x^2 + y^2 = 1

Since we are given the x-coordinate as 5/13, we can substitute this value into the equation and solve for y:

(5/13)^2 + y^2 = 1

25/169 + y^2 = 1

To isolate y^2, we subtract 25/169 from both sides:

y^2 = 1 - 25/169

y^2 = 169/169 - 25/169

y^2 = 144/169

Taking the square root of both sides, we find:

y = ±sqrt(144/169)

Since we are dealing with points on the unit circle, the y-coordinate represents the sine value. Therefore, the y-coordinate of point P is:

y = ±12/13

So, the y-coordinate of point P can be either 12/13 or -12/13.

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♥️ \(\large{\underline{\textcolor{red}{\mathcal{SUMIT\:\:ROY\:\:(:\:\:}}}}\)

Answer:

DAY 2

Step-by-step explanation:

To determine the points where the probability distribution function is a maximum for the given states of a particle in a three-dimensional box, we need to find the values of x, y, and z that maximize the wave function. In this case, we are given two different states: (a) n_X = 1, n_Y = 1, n_Z = 1, and (b) n_X = 2, n_Y = 2, n_Z = 1. We will find the points where the probability distribution function is a maximum for each state.

(a) For the state n_X = 1, n_Y = 1, n_Z = 1, the wave function is given by Ψ(x, y, z) = √(8/L^3) * sin(πx/L) * sin(πy/L) * sin(πz/L). To find the maximum points, we need to maximize the absolute value of this function. Since sin oscillates between -1 and 1, the maximum value of the wave function occurs at the points where sin(πx/L) = 1, sin(πy/L) = 1, and sin(πz/L) = 1. This means the maximum points are at (x, y, z) = (L, L, L).

(b) For the state n_X = 2, n_Y = 2, n_Z = 1, the wave function is given by Ψ(x, y, z) = √(8/L^3) * sin(2πx/L) * sin(2πy/L) * sin(πz/L). Similarly, we need to find the points where sin(2πx/L) = 1, sin(2πy/L) = 1, and sin(πz/L) = 1. This gives us the maximum points at (x, y, z) = (L/2, L/2, L).

In summary, for the state n_X = 1, n_Y = 1, n_Z = 1, the maximum points of the probability distribution function are at (x, y, z) = (L, L, L). For the state n_X = 2, n_Y = 2, n_Z = 1, the maximum points are at (x, y, z) = (L/2, L/2, L).

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Cyber safety is the use of information and communication technologies in a responsible and safe manner. It involves protecting information and keeping it secure, but it also entails handling information responsibly, showing consideration for others online, and following proper internet etiquette.

The Essential Eight are a collection of eight mitigation techniques: application control, application patching, configuring Microsoft Office macro settings, user application hardening, limiting administrative rights, operating system patching, multi-factor authentication, and regular backups.

Best Cybersecurity Advice for 2023:

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Ivan could work any whole number of hours greater than 9 to earn more than $63.

Let's denote the number of hours Ivan works by 'h'.

Ivan charges $8 per hour, so his earnings 'E' can be expressed as:

E = 8h - 9

We want to find the possible numbers of hours Ivan could work to earn more than $63, so we can set up the inequality:

8h - 9 > 63

Adding 9 to both sides, we get:

8h > 72

Dividing both sides by 8, we get:

h > 9

Therefore, Ivan must work more than 9 hours to earn more than $63. However, since the number of hours worked must be a positive integer, the possible numbers of hours Ivan could work are:

10, 11, 12, ...

In other words, Ivan could work any whole number of hours greater than 9 to earn more than $63.

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The equivalent expression of the statement is as follows:

\((\frac{y^{-3} X z^{5} }{z^{-4} X y^{6} }) ^{2}\)

The expression is as follows:

Quantity y raised to the negative third power times z raised to the fifth power end quantity over quantity z raised to the negative fourth power times y raised to the sixth power end quantity all raised to the negative second power

The equivalent expression of the statement is as follows:

\((\frac{y^{-3} X z^{5} }{z^{-4} X y^{6} }) ^{2}\)

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Answer:

1/9

You are being asked to "complete the square"

you need to find a value to make the quadratic a "perfect square:

the focus is -2/3 you need to "add" one half of -2/3 squared to the

equation (1/2 * -2/3)^2 .... (-2/6)^2 .... 4/36 .... 1/9

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

double 5 is 10

10+3=13

13 multiplied by 4=52

Answer:

$2,436

Step-by-step explanation:

I added the percentage together 30+7+7=42 and I multiplied 0.42 by $4,200 and then I got $1,764 which is the amount I spend on expenses each month. Last, I did 4,200-1,764 and got the answer

The probability that the first white ball drawn in powerball will not be an 18 is 0.985.

Concept used:

Probability of an event = number of favorable outcomes / total number of outcomes.

We have 69 balls, out of which 1 is an 18. Thus, there are 68 balls that are not 18.

Therefore, the probability of the first white ball drawn in powerball will not be an 18 is given by,

P(event)= number of favorable outcomes / total number of outcomes= 68/69 = 0.985 (approx)

Thus, the required probability that the first white ball drawn in powerball will not be an 18 is 0.985, rounded to three decimal places.

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ANSWER

1.08p = t

x = m/4

V = 12s^2

EXPLANATION

The proportional relationship between the quantity y and the quantity x has a constant of proportionality k and is expressed by the equation y = kx. If the equation can be rewritten in other forms as above, it is proportional.

Equation 1.

120 - x = L.

This cannot be rewritten as y = kx

Equation 2:

1.08p = t

This can be rewritten as y = kx

Equation 3

x = m/4

This can be rewritten as y = kx

Equation 4

V = 12s^2

This can be rewritten as y = kx

Hence, 1.08p = t, x = m/4 and V = 12s^2 equations represent proportional relationships.

Answer:

c

hopefully this will help you

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We know:-

\( \bigstar \boxed{ \rm Area \: of \: circle = \pi {r}^{2} }\)

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So:-

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\( \dashrightarrow\sf Area \: of \: circle = 314 \times {9}^{2}\)

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\( \dashrightarrow\sf Area \: of \: circle = 314 \times 9 \times 9 \\ \)

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\( \dashrightarrow\sf Area \: of \: circle = 314 \times 81 \\ \)

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\( \dashrightarrow\bf Area \: of \: circle = 25434 \: unit^2\\ \)

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.°. Option D is correct

Answer:

sorry dont know

Step-by-step explanation:

good luck tho

\

f(x) = 0.5x-1.5x+1

4x<-1

-1 ≤ x ≤ 3

x < 3

To double, the population must increase by a factor of 2, which would take ln(2) / ln(6) = ln(2) / ln(6) * 10 hours = 2.817 hours.

To solve for the time it took for the bacterial population to double, we need to use the exponential growth formula:

\(N(t) = N_0 * e^{(kt)\)

where N0 is the initial population, N(t) is the population at time t, k is the growth rate, and e is the base of the natural logarithm (approximately 2.71828).

Since the population is increased sixfold in 10 hours, we can set up the equation as:

\(N_0 * e^{(kt)} = 6 * N_0\)

Taking the natural logarithm of both sides gives:

kt = ln(6).

Dividing both sides by k and using the fact that the natural logarithm of 2 is approximately 0.693,

ln(2) / ln(6) = ln(2) / ln(6) * 10 hours = 2.817 hours.

So it took approximately 2.817 hours for the population to double.

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Explanation:

The y axis is vertical. Anything perpendicular to this is horizontal.

All horizontal linear equations are of the form y = k, where k is any number.

In this case, k = 5 so that the line y = 5 goes through all points with y coordinate 5. Two points on this line are (-4,5) and (1,5).

Note how y = 5 is equivalent to y = 0x+5. We see the slope is 0 and the y intercept is 5. Compare this to y = mx+b.

The interest earned increases with higher principal amounts, higher interest rates, and longer time periods.

Principal (P) | Interest Rate (r) | Time Period (t) | Interest Earned (I)

$1,000 | 2% | 1 year | $20

$5,000 | 4% | 2 years | $400

$10,000 | 3.5% | 3 years | $1,050

$2,500 | 1.5% | 6 months | $18.75

$7,000 | 2.25% | 1.5 years | $236.25

To calculate the interest earned (I), we can use the simple interest formula: I = P * r * t.

For the first row, with a principal of $1,000, an interest rate of 2%, and a time period of 1 year, the interest earned is calculated as follows: I = $1,000 * 0.02 * 1 = $20.

For the second row, with a principal of $5,000, an interest rate of 4%, and a time period of 2 years, the interest earned is calculated as follows: I = $5,000 * 0.04 * 2 = $400.

For the third row, with a principal of $10,000, an interest rate of 3.5%, and a time period of 3 years, the interest earned is calculated as follows: I = $10,000 * 0.035 * 3 = $1,050.

For the fourth row, with a principal of $2,500, an interest rate of 1.5%, and a time period of 6 months (0.5 years), the interest earned is calculated as follows: I = $2,500 * 0.015 * 0.5 = $18.75.

For the fifth row, with a principal of $7,000, an interest rate of 2.25%, and a time period of 1.5 years, the interest earned is calculated as follows: I = $7,000 * 0.0225 * 1.5 = $236.25.

These calculations show the interest earned for different savings principals, interest rates, and time periods.

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Answer: The first 3 options

Step-by-step explanation:

You have to make sure that the given variable would be able to equal -5. The  \(\geq\) sign in the second and third answer show that the variable could be equivalent to -5.

Answer:

g > -7, f ≤ -5 and g ≥ -5

Step-by-step explanation:

the < and > symbols mean less than and greater than respectively, but not including. Therefore -5 cannot be a valid solution if the range is (for example) g > -5. However, ≤ and ≥ mean less/greater than or equal to, so -5 would be a valid solution for g ≥ -5 (for example).

Answer:

c

Step-by-step explanation:

A quadratic function looks like a U, which is called a parabola.

Answer:

Last option, y = -4/3x - 2/3

Step-by-step explanation:

The equation of the line is written in the slope-intercept form,

y = mx + b

where m represents the slope

number that describes both the direction and the steepness of the line

slope = rise/run

           b represents the y-intercept  

point where crosses the y-axis

Given:

Let change 3x - 4y = 2 into y = mx + b form,

3x - 4y = 2

subtract 3x from both sides,

3x - 3x - 4y = 2 - 3x

- 4y = 2 - 3x

divided both sides by -4 to get y alone,

- 4y / -4 = 2 - 3x / - 4

  y = (3x / 4) - (2 / 4)

→ y = (3/4)x - (1/2)

Slope of perpendicular lines to another is the negative reciprocal

(ex. slope of 2 becomes -1/2)

So for y = (3/4)x - (1/2) the slope will be,

slope = -4/3

We need to find b (y-intercept),

we know it passes through point (1, - 2)

a point is (x, y) format

y = -(4/3)x + b

plug in the point (1, - 2),

-2 = -4/3(1) + b

-2 = -4/3 + b

add 4/3 to both sides to get b alone,

-2 + 4/3 = -4/3 + 4/3 + b

-2/3 = b

the perpendicular line equation is,

y = -4/3x - 2/3

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The dimensions of the poster with the smallest area are = 64cm X 56cm where Width of the poster is 64 and Height of the poster is 56.

In order to establish a condition equation that represents the printed area's width (w) in terms of its height (h), we must first use what we know about the printed area.

Ap(w,h)=wh                   [Printed area]

Ap(w,h)=wh                  [ Substitute Ap(w,h)=1536]

1536=wh

w=1536h                      [width  ''w''  in terms of height ''h'' ]

The smallest surface area must be determined. We need to find out the area as a function of the variable h using the condition equation.

W=w+16                                              [ Width  of the rectangular poster ]

H=h+24                                                 [ Height  of the rectangular poster ]

Ar(w,h)=(w+16)(h+24)                        [Area of the rectangular poster ]

Ar(w,h)=(w+16)(h+24)

Put the value of W in terms of ‘h’ that we got above.

Ar(w)=(1536h+16)(h+24)

Ar(w)=36864h+16h+1920

Ar(w)=36864/w+16w+1920           (objective function)

Determine the domain of objective function

D={w∈R:w>0}                           [Domain of the Objective area function]

Ar(w)=36864/w+16w+1920

A′r(w)=−36864w2+16              [First derivative of the Objective area function]

A′r(w)=0                                     [critical point condition]

−36864/w2+16=0

16=36864/w2

w2=36864/16

w2=2304

w=±48

w=−48                       [w=−48∉D]

w=48                         [w=48∈D]

w=48                        [Critical point of the Objective area function ]

Applying sufficient condition for extremes (Second derivative test)

A′r(w)=−36864/w2+16

A′′r(w)=73728/w3[Second derivative of the Objective area function]

A′′r(48)=0.6                     [A′′r(48)>0]

w=48                                [Minimum point of the Objective area function ]

Calculate the dimensions of the rectangular poster with minimum area

w=48 Width of the print area

h=1536/48=32           (Height of the print area)

W=48+16=64             (Width of the rectangular poster)

H=32+24=56             (Height of the rectangular poster)

The dimensions of the poster with the smallest area are: ⟹64cm×56cm

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After 35 minutes there will be 40 bacteria remaining.

The process of a constant percentage rate decrease in an amount over time is referred to as "exponential decay." The formula to calculate exponential decay is given as, \(N_t=N_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\). Here, Nt is the quantity after time t, N0 is the initial quantity, t1/2 is the half-life, and t is time.

For the first situation, Nt=360, N0=900, t=10 minutes. Therefore, substituting the given values get the value of t1/2. So,

\(\begin{aligned}360&=900\left(\frac{1}{2}\right)^{\frac{10}{t_{1/2}}} \\\frac{360}{900}&=\left(\frac{1}{2}\right)^{\frac{10}{t_{1/2}}}\\0.4&=\left(\frac{1}{2}\right)^{\frac{10}{t_{1/2}}}\\ \ln(0.4)&=\frac{10}{t_{1/2}}\ln(0.5)\\t_{1/2}&=10\times\frac{\ln(0.5)}{\ln(0.4)}\\&=7.6\end{aligned}\)

Now, for the second situation,  Nt=40.  We have to find the time at which there will be 40 bacteria remaining. Then,

\(\begin{aligned}40&=900\left(\frac{1}{2}\right)^{t/7.6}\\0.04&=\left(\frac{1}{2}\right)^{t/7.6}\\\ln(0.04)&=\frac{t}{7.6}\ln(0.5)\\t&=7.6\times\frac{\ln(0.04)}{\ln(0.5)}\\&=7.6\times4.64\\&=35.26\\&\approx35\end{aligned}\)

The answer is 35 minutes.

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It seems like you're trying to match one-to-one functions with their inverse functions.

Based on the given functions, I have identified the pairs as follows:

Function: f(x) = x - 10
Inverse Function: f^(-1)(x) = x + 10

Function: f(x) = √(2r)
Inverse Function: f^(-1)(x) = x^2 / 2

Function: f(x) = x + 17
Inverse Function: f^(-1)(x) = x - 17

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We need to use 2 cups of lemon juice and 8 cups of water as we need to mix to get 10 cups of lemonade.

In mathematics, multiplication is a method of finding the product of two or more numbers. It is one of the basic arithmetic operations, that we use in everyday life.

here, we have.

We need to use 2 cups of lemon juice and 8 cups of water to get a total of 10 cups of lemonade.

We know that the lemonade recipe is:

1 cup of lemon juice and 4 cups of water.

This gives us a total of:

1 cup + 4 cups =  5 cups of lemonade.

Then if we want 10 cups of lemonade, we just need to use the double of each one of the ingredients.

this is:

2*(1 cup) = 2 cups of lemon juice

2*(4 cups) = 8 cups of water.

We need to use 2 cups of lemon juice and 8 cups of water to get a total of 10 cups of lemonade.

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Answer:

DOMAIN : R/2

RANGE : R/5

VERIT TRANS =2

HORIZ TRANS =5

Step-by-step explanation:

Answer:

Option "3" is correct

Step-by-step explanation:

Given:

equation -3r + 5 = 28

Explanation:

Assume;

Given number = r

The product of negative three and a number

⇒ -3(r)

Plus five equals twenty-eight

⇒ -3r + 5 = 28

So,

Option "3" is correct .

The velocity, acceleration, and speed of the particle are
v(t) = 2ti + 2j + (7/t)k, a(t) = 2i - (7/t^2)k, s(t) = √((4t^4 + 4t^2 + 49) / t^2) respectivelty.

To find the velocity, acceleration, and speed of a particle with the given position function, we can differentiate the position function with respect to time. Let's go step by step:

1. Position function:
The given position function is r(t) = t^2i + 2tj + 7ln(t)k.

2. Velocity:
To find the velocity, we need to differentiate the position function with respect to time.
Velocity (v) is the derivative of position (r) with respect to time (t).

Taking the derivative of each component of the position function:
d/dt (t^2) = 2t
d/dt (2t) = 2
d/dt (7ln(t)) = 7/t

So, the velocity function (v) is:
v(t) = 2ti + 2j + (7/t)k

3. Acceleration:
To find the acceleration, we need to differentiate the velocity function with respect to time.
Acceleration (a) is the derivative of velocity (v) with respect to time (t).

Taking the derivative of each component of the velocity function:
d/dt (2t) = 2
d/dt (2) = 0
d/dt (7/t) = -7/t^2

So, the acceleration function (a) is:
a(t) = 2i + 0j - (7/t^2)k
or simply,
a(t) = 2i - (7/t^2)k

4. Speed:
The speed of a particle is the magnitude of its velocity vector.
The magnitude of a vector is found using the Pythagorean theorem.
For the velocity function v(t) = 2ti + 2j + (7/t)k, the magnitude of v(t) is:

|v(t)| = √((2t)^2 + (2)^2 + (7/t)^2)
      = √(4t^2 + 4 + 49/t^2)
      = √((4t^4 + 4t^2 + 49) / t^2)

Therefore, the speed function (s) is:
s(t) = √((4t^4 + 4t^2 + 49) / t^2)

This is the velocity, acceleration, and speed of the particle with the given position function.

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