Is f(x)=5x and g(x)=x/5 inverse of each other?

Answer:

x ≈ 399.43

Step-by-step explanation:

Using the rule of logarithms

\(log_{b}\) x = n ⇒ x = \(b^{n}\)

Given

\(log_{e}\) (x + 4) = 6 , then

x + 4 = \(e^{6}\) ( subtract 4 from both sides )

x = \(e^{6}\) - 4 ≈ 399.43 ( to the nearest hundredth )

Answer:

Step-by-step explanation:

3 2/5

Answer:

3 2/5

Step-by-step explanation:

Answer:

A) -3/-11

Step-by-step explanation:

-3/-11-3/11

-3/-11

Reduce the fraction with -1.

3/11

Hope this help :)

Answer:

the computer was 280, and the printer was 6 times that. do the math an you have ur answer

The cost of a printer is $240 and the cost of a computer is $1440.

Given that, the cost of both computer and printer is $1680.

In mathematics, an equation is a formula that expresses the equality of two expressions, by connecting them with the equals sign =.

Let the cost of printer be x.

Richard paid six times as much for his computer as he did for his printer.

Then, the cost of computer will be 6x

Now, 6x+x=1680

⇒ 7x=1680

⇒ x=1680/7

⇒ x=$240

So, 6x=$1440

Therefore, the cost of a printer is $240 and the cost of a computer is $1440.

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Answer:

(4,5)

Step-by-step explanation:

Answer:

rational number.................

The relative change is approximately -14.9%, indicating a decrease of approximately 14.9% in per person consumption of beef from 2000 to 2020.

To find the absolute and relative change as a percentage in the U.S. per person consumption of beef from 2000 to 2020, we can follow these steps:

Calculating the absolute change:

Absolute change = Final value - Initial value

= 57.7 pounds - 67.8 pounds

= -10.1 pounds

Step 2: Calculate the relative change:

Relative change = (Absolute change / Initial value) * 100

= (-10.1 pounds / 67.8 pounds) * 100

≈ -14.9%

The absolute change is -10.1 pounds, indicating a decrease in per person consumption of beef.

The relative change is approximately -14.9%, indicating a decrease of approximately 14.9% in per person consumption of beef from 2000 to 2020.

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-1.96 and 1.96 are the essential z values for a two-tailed test with a significance level of 0.05.

A null hypothesis is a statistical hypothesis that presupposes that there is no link between two variables in a population or that there is no meaningful difference between two populations. It serves as the beginning point of a statistical hypothesis test and is typically marked as H0. A null hypothesis serves as a reference point against which the alternative hypothesis may be compared. The alternative hypothesis is accepted if there is sufficient evidence from the data to reject the null hypothesis. The null hypothesis is not rejected if there is insufficient evidence in the data to support it.

Using a significance threshold of 0.05, we may use a conventional normal distribution table or calculator to get the essential z value. The value that corresponds to the 0.025 percentile is the crucial z value (or the 0.975 percentile, depending on the direction of the alternative hypothesis).

The z value for the 0.025 percentile, as determined by a typical normal distribution table, is -1.96.

Hence, -1.96 and 1.96 are the essential z values for a two-tailed test with a significance level of 0.05.

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Answer: The answer is 3.

Answer:

3

Step By step:

Answer:

96.25%

Step-by-step explanation:

The empirical rule regarding mean and standard deviation is that

Therefore between mean and +2 standard deviation or between mean and -2 standard deviation you will find  approximately 95/2 = 47.5% of the values

Between mean and ± 3 standard deviations you will find approximately 97.5/2 = 48.75% of the values

Therefore between 2 standard deviations above and 3 standard deviations below you will find approximately 47.5 + 48.75 = 96.25% of values

Answer:

2.5 each

Step-by-step explanation:

dividing by 2......

Answer:

2.5 yards of ribbon.

Step-by-step explanation:

5 divided by 2= 2.5

Hope this helps :)

Answer:

y = 2x + 6

Step-by-step explanation:

We are given that a line runs through the points (-4,-2) and (-2,2).

We want to write the equation of this line in slope-intercept form.

Slope-intercept form is written as y=mx+b, where m is the slope and b is the value of y at the y intercept, hence the name.

First, we need to find the slope of the line.

The slope can be found using the formula \(\frac{y_2-y_1}{x_2-x_1}\), where \((x_1, y_1)\) and \((x_2, y_2)\) are points.

Even though we have two points, let's label the values of the points to avoid confusion & mistakes.

\(x_1=-4\\y_1=-2\\x_2=-2\\y_2=2\)

Now let's find the slope (remember: the formula has subtraction):

m=\(\frac{y_2-y_1}{x_2-x_1}\)

m=\(\frac{2--2}{-2--4}\)

This can be simplified:

m=\(\frac{2+2}{-2+4}\)

Add the numbers together.

m=\(\frac{4}{2}\)

Divide.

m = 2

The slope of the line is 2.

We can substitute this into the equation for m.

Here is the equation of the line so far:

y = 2x + b

We now need to find b.

As the equation passes through both (-4, -2) and (-2, 2), we can use either one of them to help solve for b.

Taking (-4, -2) for instance:

Substitute -4 as x and -2 as y in the equation we have.

-2 = 2(-4) + b

Multiply.

-2 = -8 + b

Add 8 to both sides.

6 = b

Substitute 6 as b into the equation.

y = 2x + 6

Topic: finding the equation of the line (slope-intercept form)

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Answer:

10:4 and 20:8 are equivalent to 15:6.

Step-by-step explanation:

15/6 = 2.5

10/4 = 2.5 yes

7/2 = 3.5 no

20/8 = 2.5 yes

Answer:

17.9

Step-by-step explanation:

Answer: r= 5/2 or r= 2 1/2

Step-by-step explanation:

The firm should produce good  x₁ = 4 and x₂ = 8 to minimize cost of C = x₁² + x₂² .

What is Lagrange multipliers ?

Lagrange multipliers is a method used in optimization problems to find the maximum or minimum value of a function subject to one or more constraints. The method involves introducing a Lagrange multiplier, which is a scalar variable, to the objective function and the constraint(s). This forms the Lagrangian function, which is then optimized with respect to the original variables and the Lagrange multiplier.

To minimize the cost C = x₁² + x₂²

We can use the method of Lagrange multipliers to solve this problem. Let's define the Lagrangian function L as:

L(x₁, x₂, λ) = x₁²+ x₂² + λ(400 - 5x₁²- 8x₁x₂ - 15x₂²)

Taking partial derivatives of L with respect to x₁, x₂, and λ, we get:

∂L/∂x₁ = 2x₁ - 10λx₁ - 8λx₂ = 0

∂L/∂x₂ = 2x₂ - 30λx₂ - 8λx₁ = 0

∂L/∂λ = 400 - 5x₁² - 8x₁x₂ - 15x₂² = 0

Solving these equations simultaneously, we get:

x₁ = 4

x₂ = 8

λ = 1/20

Hence, the firm should produce a combination of x₁ = 4 and x₂ = 8 to minimize the cost of C = x₁² + x₂² while satisfying the production quota of 400.

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Research Report:

Title: The Number of Coins Carried by Teachers

Introduction:

This research report aims to investigate the number of coins carried by teachers. The study seeks to understand the reasons behind carrying coins and whether there are any patterns or correlations between the number of coins and certain factors such as age, gender, and occupation.

The data was collected through a survey distributed among teachers from various educational institutions. The findings of this study provide insights into teachers' habits and preferences when it comes to carrying coins.

Results and Analysis:

A total of 300 teachers participated in the survey. The data revealed that the majority of teachers (60%) carry less than 5 coins, while 25% carry between 5 and 10 coins. Only a small percentage (15%) reported carrying more than 10 coins.

Further analysis based on demographic factors indicated that age and occupation had a significant influence on the number of coins carried. Older teachers were more likely to carry fewer coins, with 70% of teachers above the age of 50 carrying less than 5 coins.

Additionally, primary school teachers tended to carry more coins compared to secondary school teachers.

Discussion and Interpretation:

The findings suggest that the number of coins carried by teachers is influenced by various factors.

Teachers may carry coins for a range of reasons, such as purchasing small items, providing change for students, or utilizing vending machines.

The lower number of coins carried by older teachers could be attributed to a shift towards digital payment methods or a preference for carrying minimal cash.

The discrepancy between primary and secondary school teachers could be due to differences in daily activities and responsibilities.

This research provides valuable insights into the habits and preferences of teachers regarding the number of coins they carry.

Understanding these patterns can assist in designing more efficient payment systems within educational institutions and potentially guide the development of tailored financial solutions for teachers.

Further research could explore the reasons behind carrying coins in more depth and investigate how the digitalization of payments affects teachers' behavior in different educational contexts.

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Answer:

sorryajaajqiqoqoqiqsjsjwuwuqqu

Answer:

Profitability ratios, leverage ratios, and liquidity ratios.

Step-by-step explanation:

Answer:

There is an accident ahead, which often causes traffic to slow.

Step-by-step explanation:

Looking at the question closely, we understand that the traffic jam occurred at a time that is not generally regarded as a rush hour.

A traffic jam due to road repair is not really an option because roads are closed during repairs. Shop malls announce their special sales days ahead of time so that many people will hear about it.

However, it is likely that an accident occurred. An road accident may suddenly occur on an otherwise less busy road. This often leads to an unusual delay in traffic irrespective of the hour of the day when it occurs.

A) x -14

Hope this helps!! :)

Answer:

a) To solve the differential equation y''-2y'-3y= e^4x, we first find the characteristic equation:

r^2 - 2r - 3 = 0

Factoring, we get:

(r - 3)(r + 1) = 0

So the roots are r = 3 and r = -1.

The general solution to the homogeneous equation y'' - 2y' - 3y = 0 is:

y_h = c1e^3x + c2e^(-x)

To find the particular solution, we use the method of undetermined coefficients. Since e^4x is a solution to the homogeneous equation, we try a particular solution of the form:

y_p = Ae^4x

Taking the first and second derivatives of y_p, we get:

y_p' = 4Ae^4x

y_p'' = 16Ae^4x

Substituting these into the original differential equation, we get:

16Ae^4x - 8Ae^4x - 3Ae^4x = e^4x

Simplifying, we get:

5Ae^4x = e^4x

So:

A = 1/5

Therefore, the particular solution is:

y_p = (1/5)*e^4x

The general solution to the non-homogeneous equation is:

y = y_h + y_p

y = c1e^3x + c2e^(-x) + (1/5)*e^4x

b) To solve the differential equation y'' + y' - 2y = 3xe^x, we first find the characteristic equation:

r^2 + r - 2 = 0

Factoring, we get:

(r + 2)(r - 1) = 0

So the roots are r = -2 and r = 1.

The general solution to the homogeneous equation y'' + y' - 2y = 0 is:

y_h = c1e^(-2x) + c2e^x

To find the particular solution, we use the method of undetermined coefficients. Since 3xe^x is a solution to the homogeneous equation, we try a particular solution of the form:

y_p = (Ax + B)e^x

Taking the first and second derivatives of y_p, we get:

y_p' = Ae^x + (Ax + B)e^x

y_p'' = 2Ae^x + (Ax + B)e^x

Substituting these into the original differential equation, we get:

2Ae^x + (Ax + B)e^x + Ae^x + (Ax + B)e^x - 2(Ax + B)e^x = 3xe^x

Simplifying, we get:

3Ae^x = 3xe^x

So:

A = 1

Therefore, the particular solution is:

y_p = (x + B)e^x

Taking the derivative of y_p, we get:

y_p' = (x + 2 + B)e^x

Substituting back into the original differential equation, we get:

(x + 2 + B)e^x + (x + B)e^x - 2(x + B)e^x = 3xe^x

Simplifying, we get:

-xe^x - Be^x = 0

So:

B = -x

Therefore, the particular solution is:

y_p = xe^x

The general solution to the non-homogeneous equation is:

y = y_h + y_p

y = c1e^(-2x) + c2e^x + xe^x

c) To solve the differential equation y" - 9y' + 20y = x^2*e^4x, we first find the characteristic equation:

r^2 - 9r + 20 = 0

Factoring, we get:

(r - 5)(r - 4) = 0

So the roots are r = 5 and r = 4.

The general solution to the homogeneous equation y" - 9y' + 20y = 0 is:

y_h = c1e^4x + c2e^5x

To find the particular solution, we use the method of undetermined coefficients. Since x^2*e^4x is a solution to the homogeneous equation, we try a particular solution of the form:

y_p = (Ax^2 + Bx + C)e^4x

Taking the first and second derivatives of y_p, we get:

y_p' = (2Ax + B)e^4x + 4Axe^4x

y_p'' = 2Ae^4x +

Answer:

  a. 16.9 m

  b. 4.082 seconds

  c. no

  d. 2.041 seconds

  e. 21.41 m

Step-by-step explanation:

A graphing calculator can help you answer these questions quickly and easily. The graph is attached.

__

Evaluate the function at t=3.

  h(t) = -4.9t^2 +20t +1

  h(3) = (-4.9×3 +20)×3 +1 = 5.3×3 +2 = 16.9

The baton will be at a height of 16.9 meters after 3 seconds.

__

The axis of symmetry of quadratic ax^2+bx+c is located at x=-b/(2a). The point of release and the later point at the same height will be symmetrically located about the axis of symmetry. The peak height will be on the axis of symmetry of the graph. That value of time is ...

  tpeak = -20/(2(-4.9)) = 20/9.8 ≈ 2.041 . . . seconds

The baton will be at the release height again after 2×2.041 = 4.082 seconds.

__

As we saw in part (b), the maximum height will be when t=2.041 seconds. That height is found by evaluating the function for that value of t.

 h(2.041) = (-4.9(2.041) +20)(2.041) +1 = 21.41 . . . meters

The baton will not reach a height of 25 meters.

__

We found this answer in part (b): 2.041 seconds to reach the maximum height.

__

We found this answer in part (c): 21.41 meters is the maximum height.

_____

Additional comment

The time at maximum height is 20/9.8 seconds = 100/49 = 2 2/49 seconds. In general, the answer values shown here are rounded to 4 significant figures. The graphing calculator rounds to 3 decimal places.

The dairy farmer should use 1330 pounds of the 70% protein supplement and 570 pounds of the 20% protein ratio to make a 1900-pound high-grade 55% protein ration.

To determine the amounts of the 70% protein supplement and the 20% protein ratio needed to make a 1900-pound high-grade 55% protein ration, we can set up a system of equations.

Let's assume x represents the pounds of the 70% protein supplement and y represents the pounds of the 20% protein ratio.

The total weight equation is given by:

x + y = 1900

The protein content equation is given by:

(0.70x + 0.20y) / 1900 = 0.55

Simplifying the second equation, we get:

0.70x + 0.20y = 0.55 × 1900

0.70x + 0.20y = 1045

Now, we can solve this system of equations to find the values of x and y.

Rearrange the first equation to solve for x:

x = 1900 - y

Substitute this expression for x in the second equation:

0.70(1900 - y) + 0.20y = 1045

1330 - 0.70y + 0.20y = 1045

1330 - 0.50y = 1045

-0.50y = 1045 - 1330

-0.50y = -285

y = -285 / -0.50

y = 570

Now substitute this value of y back into the first equation to find x:

x + 570 = 1900

x = 1900 - 570

x = 1330

Therefore, the dairy farmer should use 1330 pounds of the 70% protein supplement and 570 pounds of the 20% protein ratio to make a 1900-pound high-grade 55% protein ratio.

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Answer:

\([x + 5][2x+ 5]\)

Step-by-step explanation:

Given

\(2x^2 + 15x + 25\)

Required

Factorize

Expand the x term

\(2x^2 + 5x + 10x+ 25\)

Group into 2

\([2x^2 + 5x] + [10x+ 25]\)

Take the GCF of each group

\(x[2x + 5] + 5[2x+ 5]\)

Factor out 2x + 5

\([x + 5][2x+ 5]\)

x = 301/56

Algebra is the branch of mathematics that deals with numbers and values which are represented with letters and symbols.

Sometimes, we do not want to mention a particular number, we can represent the number by a letter or a suitable symbol. This approach is algebraic.

For example, d + d = 2d

This is an example of an algebraic expression

Given the algebraic expression,

\(3\frac{1}{8} \div (x - 4 \frac{7}{28} ) = \frac{17}{18} + 1 \frac{5}{6} \)

Step 1:

Simplify the expression by converting the mixed numbers into improper fractions:

25/8 ÷ ( x - 119/28) = 17/18 + 11/6

Step 2: Simplify the right side of the equation

25/8 ÷( x - 119/28) = 17/18 + 33/18

25/8 ÷( x - 119/28) = 50/18

Step 3: Multiply both sides of the equation by (x - 119/28)

25/8 = 50/18 × (x - 119/28)

Step 4: Multiply both sides of the equation by (18/50)

25/8 × (18/50) = (x - 119/28)

Step 5: Simplify the left side of the equation

9/8 = (x - 119/28)

Step 6: Add 119/28 to both sides of the equation

9/8 + 119/28 = x

Step 7: Calculate the value of x:

x = 9/8 + 119/28

= 63/56 + 238/56

= 301/56

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