Translate this phrase into an algebraic expression.
40 less than twice Jenny's height
Use the variable j to represent Jenny's height.

Answer:

when X=1 y = -9/2

Step-by-step explanation:

\( - 4 < \frac{9}{8} > {}^{1} = - 4 < \frac{9}{8} > = - \frac{ 36}{8} = - \frac{9}{2} \)

The traces of f(x,y) in this case are given by fixing one of the variables and letting the other vary.

When y=0, f(x,y)=0 for all values of x. When x=0, f(x,y)=undefined for all values of y, since the denominator 3x³-6 cannot be zero. When y=1, f(x,y)=-3x/3x³+6, which simplifies to -1/(x²+2). This means that as x approaches positive or negative infinity, f(x,1) approaches zero, and as x approaches zero, f(x,1) approaches negative infinity. Based on these traces, the graph that corresponds to the function f(x,y) is the graph labeled B. This is because the traces indicate that the function has a horizontal asymptote at y=0, and vertical asymptotes at x=+-√(2). The graph in option B exhibits all of these properties, with a horizontal line at y=0 and two vertical lines at x=+-√(2). The other graphs do not exhibit the correct asymptotic behavior for the given traces.

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Part (a):
To prove that if a is singular, then adj A must also be singular, we can use the fact that the determinant of a matrix and its adjugate are related by the equation:

A(adj A) = det(A)I

If A is singular, then det(A) = 0, which means that the left-hand side of the equation above is the zero matrix. Since the adjugate of A is obtained by taking the transpose of the matrix of cofactors, and since the matrix of cofactors involves computing determinants of submatrices of A, we know that if A is singular, then at least one of these submatrices will also have determinant 0. Therefore, the transpose of the matrix of cofactors will have at least one row or column of zeros, which means that adj A is also singular.

Part (b):
To show that if n ≥ 2, then det(adj A) = [det(A)]ⁿ⁻¹, we can use the fact that the product of a matrix and its adjugate is equal to the determinant of the matrix times the identity matrix, i.e.,

A(adj A) = det(A)I

Taking the determinant of both sides, we get

det(A)(det(adj A)) = [det(A)]ⁿ

Since n ≥ 2, we can divide both sides by det(A) to get

det(adj A) = [det(A)]ⁿ⁻¹

which is what we wanted to prove.

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Answer:

la C

Step-by-step explanation:

Grams to milliliters is a conversion from a unit of weight to a unit of volume.

1 gram of water is equal to 1 milliliter. This conversion is valid for any substance that has a density of 1g/mL, which means that one gram of that substance will occupy one milliliter of space.

It's important to note that the density of a substance will change with temperature and pressure. Therefore, it's important to know the density of the substance you are trying to convert, as well as the conditions of temperature and pressure at which the conversion is being made.

Additionally, this conversion is only valid for liquids and not for solids, as solids have a fixed volume and changing the weight won't change the volume.It's widely used in cooking, chemistry, and other fields where the amount of a substance is measured by weight and volume. Knowing the

conversion factor or having a conversion table handy allows to convert quickly and easily between the two measurements.

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The amount in Brianna's savings account will be equal to the amount in Shanice's checking account on day 6.

Given that Shanice goes on vacation, and she has $1,750 saved in her checking account for vacation, and each day, she withdraws $125 to pay for meals and entertainment expenses, while Brianna has $550 in her savings account when Shanice goes on vacation, and she deposits $75 from her waitressing job each day into her savings account, to determine on which day will the amount in Brianna's savings account be equal to the amount in Shanice's checking account the following calculation must be made:

Therefore, the amount in Brianna's savings account will be equal to the amount in Shanice's checking account on day 6.

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the function is described as this

Given data

The given function of ƒ(x) = -2x + 5

This is a linear equation and is compared with equation of straight line which is y = mx + c

where:

m is the slope of the equation

c is the intersect

comparing both equations

2x + 5 and mx + c

m = 2

c = 5

Hence the characteristics of the function is fully described while comparing with linear equation. This shows that the slope is 2 and the intersect is 5

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Strategies such as fast-tracking, crashing, prioritization, and resource optimization can be employed to reduce the project completion time by 5 weeks.


To reduce the completion time of the project by 5 weeks, we need to analyze the provided information and make appropriate adjustments. The initial completion time of the project is 25 weeks.

To achieve a reduction of 5 weeks, we can consider several strategies:

1. Fast-tracking: This involves overlapping or parallelizing certain project activities that were initially planned to be executed sequentially. By identifying tasks that can be performed concurrently, we can potentially save time. However, it's important to evaluate the impact on resource allocation and potential risks associated with fast-tracking.

2. Crashing: This strategy focuses on expediting critical activities by adding more resources or adopting alternative approaches to complete them faster. By compressing the schedule of critical tasks, we can reduce the overall project duration. However, this may come at an additional cost.

3. Prioritization: By reevaluating the project tasks and their priorities, we can allocate resources more efficiently. This ensures that critical activities receive higher attention and are completed earlier, resulting in an accelerated project timeline.

4. Resource optimization: Analyzing the resource allocation and identifying potential areas for optimization can lead to time savings. By ensuring that resources are utilized effectively and efficiently, we can streamline the project execution process.

It's important to note that implementing any of these strategies requires careful evaluation, considering factors such as project constraints, risks, cost implications, and stakeholder agreements. A comprehensive analysis of the project plan, resource availability, and critical path can guide the decision-making process for reducing the project completion time.

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Answer 41

Step-by-step explanation:

If you need me to show you how just comment

Answer:

73

41

always

Step-by-step explanation:

Just completed it

Answer:

10.9899097

Step-by-step explanation:

Trigonometry.

Tan = opposite/adjacent

Adjacent = Opposite / Tan(20)

The depth of the water increasing when the water is 13 feet deep is approximately 1.68 feet per minutes.

We know that the conical water tank has a radius of 12 feet and is 26 feet high.

We also know that water is flowing into the tank at a rate of 30ft³/min. In other words, our derivative of the volume with respect to time t is:

                                   \(\frac{dV}{dt} = \frac{10 ft^3}{min}\)

We want to find how fast the depth of the water is increasing when the water is 13 feet deep. So, we want to find dh/ dt.

First, remember that the volume for a cone is given by the formula:

                V = 1/3 π r² h    

We want to find dh/dt. So, let's take the derivative of both sides with respect to the time t. However, first, let's put the equation in terms of h.

We can see that we have two similar triangles. So, we can write the following proportion:

         \(\frac{r}{h} = \frac{12}{36}\)

Multiply both sides by h:

    ⇒ \(r = \frac{12}{36} h\)

So, let's substitute this in r:

\(V = \frac{1}{3} \pi \frac{12}{36} h^{2} h\)

Square:

\(V = \frac{1}{3} \pi \frac{144}{3888} (h^{2} ) h\)

Simplify:

\(V = \frac{144}{3888} \pi h^{2}\)

Now, let's take the derivative of both sides with respect to t:

\(\frac{d}{dt} (V) = \frac{d}{dt} [\frac{144}{3888} ] \pi h^{3}\)

Simplify:

\(\frac{dV}{dt} =\frac{144}{3888} \pi (3h^{2} ) \frac{dh}{dt}\)

We want to find dh/dt when the water is 13 feet deep. So, let's substitute 13 for h. Also, let's substitute 10 for dV/dt. This yields:

\(10 = \frac{144}{3888} \pi (3(13^{2} ) \frac{dh}{dt}\)

\(10 = \frac{144}{3888} \pi (507) \frac{dh}{dt}\)

\(10 = \frac{73008}{3888} \pi \frac{dh}{dt}\)

\(38880 = 73008 \pi \frac{dh}{dt}\)

\(\frac{dh}{dt} = \frac{38880}{73008} \pi\)

\(\frac{dh}{dt} = \frac{38880}{73008} X\frac{22}{7}\)

≈ 1.6737109 feet / min

The water is rising at a rate of approximately 1.68 feet per minute.

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Answer:

4/9 of an hour (26.666 minutes)

Step-by-step explanation:

4 * x * 40/60 = 1

x * 160/60 = 1

x=60/160 = 6/16 = 3/8 (this is the rate for one worker)

~~~~~~~~~~~~~~~~

6 * (3/8) * x = 1

x = 4/9

Answer:

p(t) = 200(2)^t

Step-by-step explanation:

an exponential function is in the form a(b)^x.

a is the initial value. the initial value is 200 in the problem, so a = 200.

b is the growth/decay rate. because it doubles, b = 2.

therefore, the function is p(t) = 200(2)^t

The variance of a random variable X, denoted as V(X), is a measure of the spread of the values of X around its mean. If you want to calculate V(X) and V(X-13), the variances will be different.

The variance of a random variable X, V(X), is calculated by subtracting the mean of X from each of the values of X, squaring each result, and then taking the average of these squares. This can be represented as V(X)=E[(X-μ)2].

The variance of X-13, V(X-13), is calculated by subtracting the mean of X-13 from each of the values of X-13, squaring each result, and then taking the average of these squares. This can be represented as V(X-13)=E[(X-13-μ)2].

Since the mean of X is not necessarily the same as the mean of X-13, it is clear that the two variances, V(X) and V(X-13), will be different.

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answer:

∠4, ∠6, ∠8

step-by-step explanation:

If the magnitude of the second derivative, |f''|, is small on a given 6-interval centered at a, then the slopes of the tangent lines change slowly as r increases over the interval.

The second derivative, f'', measures the rate of change of the slope of a function f. If |f''| is small, it implies that the rate of change of the slope is small. In other words, the function is relatively "smooth" or "flat" on the interval.

When the function is smooth, the tangent lines to the curve change slowly as the parameter r increases. This means that for small changes in r, the slopes of the tangent lines will not vary significantly. The graph of the function will exhibit a gradual change, without sudden or rapid fluctuations in the steepness of the curve.

Conversely, if |f''| were large on the 6-interval, it would indicate that the function has more pronounced changes in slope. In such cases, the tangent lines would change quickly as r increases, resulting in a graph with more abrupt variations in steepness.

Therefore, when the magnitude of the second derivative is small on a given 6-interval with center a, the slopes of the tangent lines change slowly as r increases over the interval.

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In this problem, we are testing whether the second-graders in a certain school district have greater math skills than the nationwide average. We are given a sample of 65 second-graders with a sample mean score of 53

a. The null hypothesis, H0, is stated as: u = 50 (The population mean score of the second-graders in the school district is equal to the nationwide average score).

The alternative hypothesis, H1, is stated as: u > 50 (The population mean score of the second-graders in the school district is greater than the nationwide average score).

b. To find the critical value and test statistic, we use the t-distribution and the formula:

t = (x - μ) / (σ / \(\sqrt{n}\))

where x is the sample mean, μ is the population mean under the null hypothesis, σ is the population standard deviation, and n is the sample size. Plugging in the given values, we can calculate the test statistic.

Using the TI-84 calculator or a statistical software, we can find the critical value corresponding to the significance level α = 0.10 for a one-tailed t-test with the appropriate degrees of freedom (df = n - 1).c.

To determine whether to reject or not reject the null hypothesis, we compare the test statistic with the critical value. If the test statistic is greater than the critical value, we reject the null hypothesis. If the test statistic is less than or equal to the critical value, we do not reject the null hypothesis.

By comparing the test statistic with the critical value, we can draw a conclusion on whether the second-graders in the school district have greater math skills than the nationwide average.

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Answer:

y + x = 7

Step-by-step explanation:

Standard form should be

y - 9 = -(x+2)

y = -x -2 +9

y + x = 7

Answer:

y + x = 7

Step-by-step explanation:

hope that helps :)

could you mark my answer as brainliest ? :)

In this problem

Looking at the graph

The domain is all real numbers greater than or equal to -3 and less than -1

therefore

The domain is the interval

The qualities of the given set (x, y) satisfying the inequality 4 < \(x^{2}\) + \(y^{2}\) < 9 are as follows: the set includes points within the annular region bounded by the circles with radii 2 and 3, but it does not include the points on the boundaries of these circles.

The given inequality represents a region between two circles in the xy-plane. The first circle has a radius of 2 (since 4 = \(2^{2}\)) and the second circle has a radius of 3 (since 9 = \(3^{2}\)).

Based on this information, we can determine the qualities of the given set as follows:

The set excludes points on the boundary of the inner circle (\(x^{2}\) + \(y^{2}\) = 4), which means it does not include the points lying exactly on the circle.

The set includes points within the area between the inner and outer circles.

The set excludes points on the boundary of the outer circle (\(x^{2}\) + \(y^{2}\) = 9), so it does not include the points lying exactly on the outer circle.

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The complete question is:

Determine The Qualities Of The Given Set. (Select All That Apply.) {(X, Y) | 4 < X^2 + Y^2 &Lt; 9} Open Connected Simply-Connecte

Answer:

Y'(-5, 5)

Step-by-step explanation:

To find the coordinates of Y' after the translation, we apply the given rule to the coordinates of point Y(-3, 4).

Using the translation rule (x, y) → (x - 2, y + 1), we can substitute the coordinates of Y(-3, 4) into the rule:

x' = x - 2 = -3 - 2 = -5

y' = y + 1 = 4 + 1 = 5

Therefore, the coordinates of Y' are (-5, 5).

The terms "cross-sectional study" and "time-series study" refer to different types of research designs. A cross-sectional study collects data from a population at a specific point in time, whereas a time-series study collects data from the same population over an extended period.

Based on this definition, it is difficult to classify a graph as either a cross-sectional or time-series study without additional context.

A graph alone does not provide enough information about the research design. It would be best to refer to the accompanying study or research report to determine the type of study represented by the graph.
Therefore, the long answer to your question is that a graph cannot be classified as a cross-sectional or time-series study without further information about the research design.

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Answer:

Step-by-step explanation:

The experimental probability of the cube landing on three is 1/10.

No, every 2-approximation algorithm for finding a minimum vertex cover is not necessarily a 2-approximation algorithm for finding a maximum independent set.

The reason for this is that the two problems, minimum vertex cover and maximum independent set, are not symmetric in their definitions and objectives.

In the minimum vertex cover problem, the goal is to find the smallest possible set of vertices that covers all edges in the graph. On the other hand, in the maximum independent set problem, the objective is to find the largest possible set of vertices such that no two vertices in the set are adjacent.

An approximation algorithm for the minimum vertex cover problem guarantees that the size of the vertex cover found by the algorithm is at most twice the size of the optimal minimum vertex cover. This means that the algorithm provides a solution that is within a factor of 2 of the optimal solution.

However, this does not imply that the same algorithm will provide a solution within a factor of 2 of the optimal maximum independent set. The reason is that the concepts of vertex cover and independent set are complementary. A vertex cover is a set of vertices that covers all edges, whereas an independent set is a set of vertices with no adjacent vertices.

Therefore, while a 2-approximation algorithm for minimum vertex cover guarantees that the size of the vertex cover is at most twice the size of the optimal solution, it does not necessarily imply that the algorithm will find a maximum independent set with a size within a factor of 2 of the optimal solution.

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If he can earn 10% interest compounded annually then he have \(\$148,643.627\) by the time he's 30 yrs old, without putting anymore in , to have a million dollars the bank when he turns 50.

In the given question,

Augie wants to become a millionaire by the time he's 50 years old.

He can earn \(10\%\) interest compounded annually.

We have to find how much money he have by the time he's 30 yrs old, without putting anymore in ,to have a million dollars the bank when he turns 50.

We we learn about compound interest.

In the compound interest we receive payment in the interest also.

The formula of Compound Amount is

\(A=P(1+\frac{r}{100})^n\)

where \(A=\) Amount After interest

\(P=\) Principal Amount

\(r=\) Interest rate

\(n=\) Time

From the given question \(A=1,000,000\), \(r=10\%\), \(n=20\)

Now putting the value in the formula

\(1,000,000=P(1+\frac{10}{100})^{20}\)

Simplifying

\(1,000,000=P(1+\frac{1}{10})^{20}\)

\(1,000,000=P(1\times\frac{10}{10}+\frac{1}{10})^{20}\)

\(1,000,000=P(\frac{10}{10}+\frac{1}{10})^{20}\)

\(1,000,000=P(\frac{10+1}{10})^{20}\)

\(1,000,000=P(\frac{11}{10})^{20}\)

\(1,000,000=P({1.1})^{20}\)

\(1,000,000=P\times6.7275\)

Divide by \(6.7275\) on both side

\(\frac{1,000,000}{6.7275}=P\times\frac{6.7275}{6.7275}\)

\(148,643.627=P\)

\(P=148,643.627\)

Hence, if he can earn 10% interest compounded annually then he have \(\$148,643.627\) by the time he's 30 yrs old, without putting anymore in , to have a million dollars the bank when he turns 50.

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