A rectangle has side lengths of x+1 and 1. What is an expression for the area of the rectangle?
These are the choices...
3
x + 1
x + 2
2x + 3
Thank You!!!

Answer:

Step-by-step explanation:

Using the quadratic equation, you can find the solution to 2x2+15=-11x.

\(\sf{2x^2+15=-11x}\)

First, you have to add by 11x from both sides.

\(\Longrightarrow: \sf{2x^2+15+11x=-11x+11x}\)

Solve.

2x²+11x+15=0

Use the quadratic formula.

Quadratic formula:

\(\Longrightarrow: \sf{AX^2+BX+C=0}\\\\\\\Longrightarrow: \sf{x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}\)

\(\sf{x_{1,\:2}=\dfrac{-11\pm \sqrt{11^2-4\cdot \:2\cdot \:15}}{2\cdot \:2}}\)

Solve.

Use the order of operations.

PEMDAS

\(\sf{\sqrt{11^2-4\cdot \:2\cdot \:15}}\)

Multiply the numbers from left to right.

4*2*15=120

\(:\Longrightarrow\sf{\sqrt{11^2-120}\)

Do exponents.

11²=11*11=121

\(\Longrightarrow: \sf{\sqrt{121-120}\)

Subtract the numbers from left to right.

121-120=1

\(\sf{\sqrt{1}}=1\)

\(\sf{x_{1,\:2}=\dfrac{-11\pm \:1}{2\cdot \:2}}\)

\(\Longrightarrow: \sf{x_1=\dfrac{-11+1}{2\cdot \:2},\:x_2=\dfrac{-11-1}{2\cdot \:2}}\)

Solve.

\(\sf{\dfrac{-11+1}{2\cdot \:2}}=\dfrac{-10}{2*2}=\dfrac{-10}{4}=-\dfrac{10}{4}\)

\(\sf{\dfrac{-10\div2}{4\div2}=\dfrac{-5}{2}=-\dfrac{5}{2} }\)

Divide is another options.

-5/2=-2.5

\(\sf{\dfrac{-11-1}{2\cdot \:2}}\)

Solve.

\(\Longrightarrow: \sf{\dfrac{-11-1}{2\cdot \:2}}=\dfrac{-12}{2*2}=\dfrac{-12\div4}{4\div4}=\dfrac{-3}{1}=-3\)

Solutions:

\(\Longrightarrow: \boxed{\sf{-3, \ -2.5}}\)

I hope this helps. Let me know if you have any questions.

Answer:

10%

Step-by-step explanation:

450 is 10% of 4500

Answer:

1/3 of 3 hours should be 20 minutes

Answer: 1 hour

Step-by-step explanation:

(1/3)(3) = 1 hour

Answer:

2/5 or 40% chance

Step-by-step explanation:

Answer:

look at the chart it has all of them <3

There are 84,878 oranges in the stack.

We can solve it by using arithmetic series.

The first level of the stack has a rectangular base of 55 oranges by 88 oranges, which means there are 55 x 88 = 4840 oranges in the first level.

Each orange above the first level rests in a pocket formed by four oranges below, so the second level has 54 oranges by 87 oranges (one less on each side), which means there are 54 x 87 = 4698 oranges in the second level.

Similarly, the third level has 53 oranges by 86 oranges, which means there are 53 x 86 = 4558 oranges in the third level.

We can continue this pattern until we reach the top level, which has a single orange.

Therefore, the total number of oranges in the stack is:

4840 + 4698 + 4558 + ... + 1

This is an arithmetic series with a first term (a) of 4840, a common difference (d) of -142, and a number of terms (n) of 34 (since there are 34 levels in the stack).

Using the formula for the sum of an arithmetic series:

S = (n/2)(a + l)

where S is the sum, a is the first term, l is the last term, and n is the number of terms.

We can find the last term (l) using the formula for the nth term of an arithmetic series:

l = a + (n - 1)d

Substituting the values we have:

l = 4840 + (34 - 1)(-142) = 4840 - 4686 = 154

So the sum of the oranges in the stack is:

S = (34/2)(4840 + 154) = 17 x 4994 = 84878

Therefore, there are 84,878 oranges in the stack.

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The probability that a class of 50 has an average score that is less than 77 is approximately 11.90%.

To solve this problem, we need to use the central limit theorem, which states that the distribution of sample means will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
In this case, we have a population of marks on a biology final test that is normally distributed with a mean of 78 and a standard deviation of 6. We want to know the probability that a class of 50 has an average score that is less than 77.
Using the central limit theorem, we can calculate the standard error of the mean as follows:
standard error of the mean = standard deviation / square root of sample size
standard error of the mean = 6 / √50
standard error of the mean = 0.8485
Next, we need to calculate the z-score for a sample mean of 77:
z-score = (sample mean - population mean) / standard error of the mean
z-score = (77 - 78) / 0.8485
z-score = -1.18
We can use a standard normal distribution table or calculator to find the probability associated with a z-score of -1.18. The probability of a sample mean less than 77 is approximately 0.1190 or 11.90%.
Therefore, the probability that a class of 50 has an average score that is less than 77 is approximately 11.90%.

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Answer:

radius=12

Step-by-step explanation:

Hope it is helpful...

Answer:

12

Step-by-step explanation:

a radius is half the diameter

Answer:

10.50 Miles per hour

Step-by-step explanation:

42/4 = 10.50 easy

Answer:

2.5d

hope this helps

have a good day :)

Step-by-step explanation:

Answer:

A and D

Step-by-step explanation:

so sorry im in school

x+y=4

 -y.  -y

x=4-y

4-y-y=2

4-2y=2

-4      -4

-2y= -2

/-2.   /-2

y=1

x+1=4

  -1 -1

x=3

so A and D is correct

Answer:

3

Explanation:

Based on the given conditions, formulate: 3

Cross out the common factor: 3

get the result:3

Answer: 3

Answer:

2$

Step-by-step explanation:

10/5=2

since the price is decreasing, you will substract a precentage.

you could find the 40% of 499 like this

after having this substract it from the original price

the new price will be $299.4

Answer: =280.86 ft²

Step-by-step explanation:

A=\(\pi r^{2}\)/2             d=26.75    radius is half r=13.375

divide by 2 becuase it's a semicircle not full circle

=(3.14)(13.375)²/2

=280.86 ft²

Answer:

Pythagorean triple

Step-by-step explanation:

This is usually expressed as a² + b² = c² Integer triples which satisfy this equation are Pythagorean triples. The most well known examples are (3,4,5) and (5,12,13).

d. Expected life in years of the depreciable asset. The acceptable way to express the useful life of a depreciable asset is in terms of the expected life in years.

The useful life refers to the period of time over which the asset is expected to contribute to the revenue-generating activities of a business.

While options a, b, and c may be relevant factors for certain specific assets (such as units produced, hours of productivity, or miles driven), they do not encompass the overall concept of useful life. Useful life is a broader measure that takes into account the anticipated duration of productive use, regardless of specific output or activity metrics.

Expressing the useful life in years provides a common standard for comparison and allows for consistency in depreciation calculations and financial reporting. It is a practical and widely accepted approach to estimating the lifespan of a depreciable asset.

It is worth noting that the estimated useful life in years may vary depending on the nature of the asset and industry practices. It is typically determined based on factors such as technological advancements, physical wear and tear, economic obsolescence, and the intended purpose of the asset.

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Answer:

Below

Step-by-step explanation:

y varies directly with  x  means

y = k x

now with y = -93 and x = -3

   -93  =  k (-3)      shows k =  30

so   y = 30 x

now with x = -2

    y = 30 (-2) = -60

The optimal solution is x1 = 4, x2 = 0, x3 = 1, w = 0, and the minimum value of the objective function is 0.

To solve this linear programming problem using the Simplex method, we first need to convert it into standard form by introducing slack variables.

Our problem can be rewritten as follows:

Minimize w = 9x1 + 6x2

Subject to:

x1 + 2x2 + x3 = 5

2x1 + 2x2 + x4 = 8

x1 + 2x2 + 2x3 = 6

where x1, x2, x3, and x4 are all non-negative variables.

Next, we set up the initial simplex tableau:

Basic Variables x1 x2 x3 x4 RHS

x3 1 2 1 0 5

x4 2 2 0 1 8

x5 1 2 2 0 6

z -9 -6 0 0 0

The last row represents the coefficients of the objective function. The negative values in the z-row indicate that we are minimizing the objective function.

To find the pivot column, we look for the most negative coefficient in the z-row. In this case, the most negative coefficient is -9, which corresponds to x1. Therefore, x1 is our entering variable.

To find the pivot row, we calculate the ratios of the RHS values to the coefficients of the entering variable in each row. The smallest positive ratio corresponds to the pivot row. In this case, the ratios are:

Row 1: 5/1 = 5

Row 2: 8/2 = 4

Row 3: 6/1 = 6

The smallest positive ratio is 4, which corresponds to row 2. Therefore, x4 is our exiting variable.

To perform the pivot operation, we divide row 2 by 2 to make the coefficient of x1 equal to 1:

Basic Variables x1 x2 x3 x4 RHS

x3 0 1 1 -1 1

x1 1 1 0 1/2 4

x5 0 1 2 -1 2

z 0 -3 9 9/2 -18

We repeat the process until all coefficients in the z-row are non-negative. In this case, we can stop here because all coefficients in the z-row are non-negative.

Therefore, the optimal solution is x1 = 4, x2 = 0, x3 = 1, w = 0, and the minimum value of the objective function is 0.

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A good way to ensure that data issues are properly addressed is by implementing robust data quality assurance and validation processes. These processes involve various steps and techniques to identify, prevent, and resolve data issues effectively. By following these best practices, you can minimize the risk of invalidating testing results due to date-related problems.

One essential step is to establish clear data quality standards and guidelines. This includes defining data formats, data integrity rules, and validation criteria specific to date-related fields. By setting these standards, you provide a framework for ensuring consistent and accurate data across the testing process.

Another important aspect is to perform comprehensive data profiling and analysis. This involves examining the data to identify anomalies, inconsistencies, or inaccuracies related to dates. By using data profiling tools or writing custom scripts, you can detect issues such as missing values, incorrect formats, or outliers within date fields. This analysis enables you to have an overview of the data quality and pinpoint potential problems.

To address data issues effectively, it is crucial to establish data cleansing procedures. This involves correcting errors, resolving inconsistencies, and standardizing date formats. For instance, you can use data cleaning techniques like imputation to fill in missing dates, transforming data into a consistent format (e.g., YYYY-MM-DD), or removing duplicate entries. By applying these cleansing techniques, you improve the accuracy and reliability of the data used in testing.

Furthermore, implementing data validation checks is essential to identify and flag potential issues during the testing process. This can involve writing automated tests or validation scripts that verify the integrity and correctness of date-related data. By performing validation checks at different stages of the testing process, you can detect anomalies promptly and take appropriate actions to address them.

Lastly, fostering a culture of data quality awareness and accountability is crucial. Promoting education and training on data handling best practices, emphasizing the importance of accurate dates, and encouraging open communication about data issues among team members can significantly contribute to ensuring data issues are adequately addressed.

In summary, to ensure data issues are properly addressed, it is important to establish data quality standards, perform data profiling and analysis, implement data cleansing procedures, incorporate validation checks, and foster a culture of data quality awareness. By following these practices, you can mitigate the risk of invalidating testing results due to date-related problems and enhance the overall reliability and integrity of your data.

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Answer: 1st one

Step-by-step explanation:

To find the limit of (1 - sin(θ))/(1 + cos(6θ)) as θ approaches π/2, we can apply a trigonometric identity:

1 - sin(θ) = cos(θ)

1 + cos(6θ) = 2cos^2(3θ)

Substituting these expressions into the original limit, we have:

lim (1 - sin(θ))/(1 + cos(6θ)) = lim cos(θ)/(2cos^2(3θ))

Now, as θ approaches π/2, the denominator 2cos^2(3θ) approaches 0, so we can apply L'Hospital's rule to the limit:

lim cos(θ)/(2cos^2(3θ)) = lim -sin(θ)/(12cos(θ)sin(3θ))

Applying L'Hospital's rule again, we take the derivative of the numerator and denominator separately with respect to θ:

= lim -cos(θ)/(12cos(θ)cos(3θ) + 3sin(θ)sin(3θ))

= lim -1/(12cos(3θ) + 3tan(θ))

Substituting θ = π/2, we have:

lim -1/(12cos(3π/2) + 3tan(π/2)) = lim -1/0

Since the denominator approaches 0 and the numerator is negative, the limit approaches negative infinity.

Therefore, the limit of (1 - sin(θ))/(1 + cos(6θ)) as θ approaches π/2 is negative infinity.

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Answer: F divide 168 by 28 and divide 108 by the answer

Step-by-step explanation:

From the question, we are informed that an astronaut weighs 168 lbs on earth and 28 lbs on the moon. We are further told that Kristine weighs 108 lbs on earth.

To find out the weight of Kristine on the moon, we divide 168 by 28 and divide 108 by the answer. This will be:

= 168/28 = 6

We then divide 108 by 6. This will be:

= 108/6

= 18

Kristine weight on the moon is 18lbs

we have the expression

the answer is

Answer:

a t-chart to represent you proving a statement. statements are to the left of the t-chart and reasons are to the right.

Step-by-step explanation:

this is used in geometry to prove a given statement.

Answer:

10

Step-by-step explain

Step-by-step explanation:

V=pi×r^2×h

36pi =pi × (3)^2 × h

36pi ÷ pi×9 = h

4cm^2 = h

By using the concept of Permutation and Combination, Counting Principle  Phyllis can make 168 ways to select students.

Permutation and Combination are about making arrangements and making selections on the basis of certain formulas

The re-arranging of any given sorted/ordered set is called a permutation. The “permutation” word actually tells us that change the linear order of the ordered set.

Selecting items from a large collection without taking any care about their order is called combinations.

We need to select one freshman out of 12 freshmen. So selecting r things from n things we can do selection in nr ways, so we can do a selection of freshmen in 12c1 ways

Similarly, we need to select one sophomore out of 14 sophomores, so we can do a selection of sophomores in 14c1 ways

We know that ncr is equivalent to \(\frac{n!}{r!(n-r)!}\)

So, using this concept total number of ways to select a student

= ((12c1)×(14c1))

=12×14

=168 ways

Hence Phyllis can select students in 168 ways

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1. The inequality when the distance between 2 and 5 times a number is no less than 8 is 1. |5x−2|≥8

2. The correct option is A. Trey is not necessarily correct. He will need to ensure that the compass width remains the same for each arc drawn from P and R.

Inequalities are created through the connection of two expressions. It should be noted that the expressions in an inequality aren't always equal. Inequalities implies that the expressions are not equal. They are denoted by the symbols ≥  <  >   ≤.

The distance between 2 and 5 times a number is no less than 8.

Let the number be x

This will be |5x−2|≥8. The correct option is 1.

For the second question, it should be noted that Trey is not necessarily correct. He will need to ensure that the compass width remains the same for each arc drawn from P and R. The correct option is A.

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