The ages of members of a stamp collecting group are normally distributed with a mean of 55 years and a standard deviation of 4 years.
There are 104 members in the group.
About how many members are expected to be between 51 years old and 59 years old?

The implications of Constructivist approach in teaching mathematics is that it helps the students to develop a better way to comprehend the content of what they are being taught.

A constructive approach of teaching mathematics believes that students construct knowledge rather than just passively take in information.

A constructive approach of teaching also believes that students integrate new knowledge with existing knowledge to create a deeper understanding of the mathematics being taught to them.

Therefore, constructive approach of teaching can help the students to comprehend the content of what they are being taught.

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Cameron and Rebecca are organizing a lunch for their friends. Within the dishes to be served, they will prepare a tomato, carrot and broccoli salad. The number of tomatoes and carrots are 10-partners and the total amount of vegetables is 12. If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners. What is the number of tomatoes, carrots and broccoli bought by Cameron and Rebecca?

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

Number of tomatoes = x

Number of carrots = 10 - x

Number of broccoli = y

System of Equations:

x + 10 - x + y = 12

y = 12 - 10

y = 2

x - 2 + y + 3 = 10 (If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners)

x + 1 + y = 10

x + y = 9

y = 9 - x

Substitution of y:

x + 10 - x + y = 12

x + 10 - x + 9 - x = 12

-x + 19 = 12

-x = 12 - 19

x = 7 ⇒ 10 - x = 3

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

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The area of triangle UVW to the nearest square inch is approximately 2 square inches.

The area of triangle UVW can use the Heron's formula:

Area = √(s(s-a)(s-b)(s-c))

s is the semiperimeter of the triangle and a, b and c are the lengths of its sides.

First, we can calculate the semiperimeter s:

s = (u + v + w)/2

s = (5.5 + 4.4 + 9.6)/2

s = 9.75

The length of each of the three sides of the triangle:

a = v = 4.4 inches

b = u = 5.5 inches

c = w = 9.6 inches

Then, we can plug these values into the Heron's formula and simplify:

Area = √(s(s-a)(s-b)(s-c))

Area = √(9.75(9.75-4.4)(9.75-5.5)(9.75-9.6))

Area = √(9.75(5.35)(4.25)(0.15))

Area ≈ √(3.296)

Area ≈ 1.81

Rounding this to the nearest square inch, we get:

Area ≈ 2 square inches

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To find the volume of the figure, we use the formula: (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm). The correct answer is (B)

To find the volume of a joined shape, you have to solve for the volume of each component of the shape and sum them together.

In this case, we have 2 shapes joined and both are cuboids.

Volume of a cuboid = length x breath x height

Volume₁ = 8 x 7 x 2 = 2 x 8 x 7

Volume₂ = 8 x 1 x 6 = 1 x 6 x 8

Total volume = (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm)

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Jodi will need 10 half-cup measurements of flour to make 2 batches of cookies.

To double the recipe, Jodi needs to multiply the amounts of each ingredient by 2.

The original recipe calls for 2½ cups of flour. Doubling this amount gives:

2½ cups * 2 = 5 cups of flour

We may divide the total amount of flour required (5 cups) by the quantity in each 12-cup measurement to get the number of 1-1/2 cups Jodi needs.

5 cups / ½ cup = 10

Therefore, Jodi will need 10 half-cup measurements of flour to make 2 batches of cookies.

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To solve for f, we can rearrange the equation:

1975f = t1758 + Q2

First, we can isolate f by dividing both sides of the equation by 1975:

f = (t1758 + Q2) / 1975

So the final answer is:

f = (t1758 + Q2) / 1975

Answer:

Let x be the amount invested at 9% and y be the amount invested at 6%

.According to the problem:

x + y = 11,335 ----(1) (the total amount  invested is $11,335)0.09x - 0.06y = 865.35 ----(2) (the interest earned from the amount invested at 9% exceeds the interest earned from the amount invested at 6% by $865.35)

We can use these two equations to solve for x and y.

Multiplying equation (1) by 0.06, we get

:0.06x + 0.06y = 680.10 ----(3)

Subtracting equation (3) from equation (2), we get:

0.03x = 185.25

x = 6,175

Substituting x = 6,175 into equation (1), we get:

y = 5,160

Therefore, $6,175 is invested at 9% and $5,160 is invested at 6%.

Step-by-step explanation:

i) The capacities of different vessels which completely fill P vessel with exact number of fillings are:

(ii) The capacities of different vessels which completely fill Q vessel with exact number of fillings are:

One vessel of 18 I

Two vessels of 9 I each

Three vessels of 6 I each

Six vessels of 3 I each

Nine vessels of 2 I each

Eighteen vessels of 1 I each

(iii) The common capacities of vessels which completely fill P as well as Q vessels with exact number of fillings are:

Two vessels of 6 I each

Three vessels of 4 I each

Six vessels of 2 I each

Twelve vessels of 1 I each

(iv) The greatest capacity of vessel that completely fills P as well as Q vessels with exact number of fillings is 6 I.

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[tex]x=\textit{lbs of aluminum at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{lbs of aluminum at 54\%}\\\\ ~~~~~~ 54\%~of~y\implies \cfrac{54}{100}(y)\implies 0.54 (y) \\\\\\ \textit{48 lbs of aluminum at 52\%}\\\\ ~~~~~~ 52\%~of~48\implies \cfrac{52}{100}(48)\implies 0.52 (48)\implies 24.96 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{array}{lcccl} &\stackrel{lbs}{quantity}&\stackrel{\textit{\% of lbs that is}}{\textit{AL only}}&\stackrel{\textit{lbs of}}{\textit{AL only}}\\ \cline{2-4}&\\ \textit{1st alloy}&x&0.3&0.3x\\ \textit{2nd alloy}&y&0.54&0.54y\\ \cline{2-4}&\\ mixture&48&0.52&24.96 \end{array}~\hfill \begin{cases} x + y = 48\\\\ 0.3x+0.54y=24.96 \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{using the 1st equation}}{x+y=48\implies y=48-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{0.3x~~ + ~~0.54(48-x)~~ = ~~24.96}\implies 0.3x+25.92-0.54x=24.96 \\\\\\ 0.30x-0.54x=-0.96\implies -0.24x=-0.96\implies x=\cfrac{-0.96}{-0.24} \\\\\\ \boxed{x=4}\hspace{9em}y=48-x\implies \boxed{y=44}[/tex]

Answer:3

Step-by-step explanation:

subtract 3 from 3 and repeat on the other side of the equal sign. So 6 minus 3 equals 3.

The distribution of the weight of one dozen eggs is approximately normally distributed with a mean of 68 grams and a standard deviation of 4.2 grams.

The distribution of the weight of one dozen eggs is also approximately Normally distributed. The mean of the distribution is the same as the mean of the individual eggs, which is 68 grams.

The standard deviation of the distribution is the square root of the sum of the variances of the individual eggs and the sample size, which is:

= 2.2 * ✓12

= 4.2 grams.

Therefore, the distribution of the weight of one dozen eggs is approximately normally distributed with a mean of 68 grams and a standard deviation of 4.2 grams.

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a) Approximately 778.7 grams of thorium 234 will be left after 0.5 years.

b) Approximately 2.1 years will have passed when 115 grams of thorium 234 are left.

(a) To find the amount of thorium 234 left after 0.5 years, we can use the equation:

[tex]A_t = A_0 e^{(-10.498t)}[/tex]

where A₀ is the initial amount, which is 1000 grams, t is the time in years, and At is the amount after t years.

Substituting t = 0.5, we get:

[tex]At = 1000 e^{(-10.498\times0.5)} = 778.7 \ grams[/tex]

Therefore, approximately 778.7 grams of thorium 234 will be left after 0.5 years.

(b) To find the time at which 115 grams of thorium 234 will be left, we can use the same equation and solve for t:

[tex]A_t = A_0 e^{(-10.498t)[/tex]

Dividing both sides by A₀ and taking the natural logarithm of both sides, we get:

ln(At/A₀) = -10.498t

Substituting At = 115 grams and A₀ = 1000 grams, we get:

ln(115/1000) = -10.498t

Solving for t, we get:

t = (ln(115/1000)) / (-10.498) ≈ 2.1 years

Therefore, approximately 2.1 years will have passed when 115 grams of thorium 234 are left.

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Hüseyin concluded: "I was able to map quadrilateral EFGH onto quadrilateral EJGK using a sequence of rigid transformations and dilations, so the figures are similar."  The error that he make is option A: Huseyn  used a vertical stretch, so the quadrilaterals aren't similar.

Hüseyin's transformations is one that keeps the EFGH's shape and proportions, so EJGK is similar to EFGH. Similar figures have congruent angles and proportional sides.

The error is seen when he uses vertical stretch, that is violating similarity condition of quadrilaterals EFGH and EJGK. A vertical stretch would alter the proportions of the sides and angles in quadrilateral EFGH, making it non-similar to that of  EJGK.

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The reduced row echelon form of the matrix is:

[tex]\left[\begin{array}{ccc}-1144/259&\\-321/259\\25/74\end{array}\right][/tex]

Given the matrix below

[tex]\left[\begin{array}{ccc}7&-1&-3 |-63\\0&-6&-2 |-20\\0&1&1 | 6\end{array}\right][/tex]

Divide the first row by 7:

[tex]\left[\begin{array}{ccc}1&-1/7&-3/7 |-9\\0&-6&-2 |-20\\0&1&1 | 6\end{array}\right][/tex]

Add 1/7 times the first row to the second row:

[tex]\left[\begin{array}{ccc}1&-1/7&-3/7 |-9\\0&-37/7&-17/7 |-107/7\\0&1&1 | 6\end{array}\right][/tex]

Add 3/7 times the second row to the first row:

[tex]\left[\begin{array}{ccc}1&0&-16/37 |-296/37\\0&-37/7&-17/7 |-107/7\\0&1&1 | 6\end{array}\right][/tex]

Add 17/37 times the second row to the third row:

[tex]\left[\begin{array}{ccc}1&0&-16/37 |-296/37\\0&-37/7&-17/7 |-107/7\\0&0&20/37 | 25/37\end{array}\right][/tex]

Multiply the third row by 37/20:

[tex]\left[\begin{array}{ccc}1&0&-16/37 |-296/37\\0&-37/7&-17/7 |-107/7\\0&0&1 | 25/74\end{array}\right][/tex]

Add 3/7 times the third row to the second row:

[tex]\left[\begin{array}{ccc}1&0&-16/37 |-296/37\\0&-37/7&0 |-321/259\\0&0&1 | 25/74\end{array}\right][/tex]

Add 16/37 times the third row to the first row:

[tex]\left[\begin{array}{ccc}1&0&0 |-1144/259\\0&-37/7&0 |-321/259\\0&0&1 | 25/74\end{array}\right][/tex]

Multiply the second row by -7/37:

[tex]\left[\begin{array}{ccc}1&0&0 |-1144/259\\0&1&0 |321/259\\0&0&1 | 25/74\end{array}\right][/tex]

This is the reduced row echelon form of the given matrix.

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C. The total profit, if all shares for both stocks are sold at the close of business on day 7, is $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. There is no difference in profit or loss compared to part C.

Part 2

A. Diversity can be evaluated based on the portfolio in question

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

Part 1:

A. TechCo (TCO) and HealthCo (HCO) have their shares priced at $100 and $50 respectively.

Name of Stock Symbol Quantity of Shares Purchased Purchase Price per Share Total Purchase Value

TechCo TCO 75 $100 $7,500

HealthCo HCO 150 $50 $7,500

B. Here is an illustration of the potential ending values:

Name of Stock Closing Price Day 1 Closing Price Day 2 Closing Price Day 3 Closing Price Day 4 Closing Price Day 5 Closing Price Day 6 Closing Price Day 7

TechCo $101 $99 $100 $98 $102 $101 $103

HealthCo $51 $50 $52 $49 $51 $52 $54

C. If we sell all shares on day 7, we can calculate the profit/loss and the brokerage fee:

TechCo: 75 shares * $103/share = $7,725

HealthCo: 150 shares * $54/share = $8,100

Total = $7,725 + $8,100 = $15,825

Brokerage fee = $15,825 * 1.5% = $237.38

Profit = $15,825 - $15,000 (initial investment) - $237.38 (brokerage) = $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. Since we've already assumed selling on the best day, there is no difference in profit or loss compared to part C.

Part 2:

A. Diversity can be evaluated based on the portfolio in question. A portfolio that encompasses investments in a variety of sectors, asset classes, and regions would exhibit a high level of diversity.

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

An investment portfolio that primarily consists of stocks or other assets with high risks would be considered as an aggressive portfolio. A portfolio that primarily consists of bonds or low-risk assets is considered to be a conservative one.

A balance between the two would be a moderate portfolio.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

For example, a young professional with a high-risk tolerance might carry an aggressive portfolio to maximize growth, while a retired individual might have a conservative portfolio to preserve their capital.

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Answer:

Company policies help employees to carry out their daily operations. - Yes, part of internal control accounting.

Bank reconciliation needs to be prepared for verification. - Yes, part of internal control accounting.

The organization's financial data in the computer should be in password-protected files. - Yes, part of internal control accounting.

Step-by-step explanation:

Part A: The equation of the circle in standard form is x² + (y - 7)² = 2².

Part B: Center: (h, k) = (0, 7), Radius: 2

Part C: Please check the attachment.

In this problem we find the definition of the equation of a circle in general form, whose standard form must be found:

(x - h)² + (y - k)² = r²

Where:

This can be done by completing the square. First, write the equation of the circle:

x² + y² = 14 · y - 45

Second, complete the square by algebra properties:

x² + y² - 14 · y + 45 = 0

x² + (y² - 14 · y + 49) = 4

x² + (y - 7)² = 2² (PART A)

Third, write the coordinates of the center and the radius of the circle:

Center: (h, k) = (0, 7), Radius: 2 (PART B)

Finally, we proceed to graph the resulting expression.

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The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

We have to given that;

Function is,

⇒ f (x) = 6√x

Now, We can find the inverse of the function is,

⇒ f (x) = 6√x

⇒ y = 6√x

Solve for x;

⇒ y/6 = √x

⇒ (y/6)² = x

⇒ x = y² / 36

⇒ f ⁻¹ (x) = x² / 36

Thus, The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

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Answer: £234

Step-by-step explanation:

First, we need to subtract the deposit (£350) from the total cost of the trip (£1754):

£1754 - £350 = £1404

Then we need to divide the remaining amount of money (£1404) by the number of months Janice will be on holiday (6 months):

£1404 ÷ 6 months = £234 each month

Hope this helps!

The end behavior of the quadratic function is f(x) = (x + 5)² - 2

We have the table of values which represents a quadratic function f(x).

A quadratic equation is represented as;

f(x) = a(x - h)² + k

Where, Vertex = (h, k)

If we plot the graph according to the given table, the vertex will be,

(h, k) = (-5, -2)

Substitute, (h, k) = (-5, -2) in f(x) = a(x - h)² + k

we have, f(x) = a(x + 5)² - 2

Also, the graph, we have the point (0, 23)

This means that

a(0 + 5)² - 2 = 23

25a = 25

a = 1

Substitute a = 1 in f(x) = a(x + 5)² - 2

f(x) = (x + 5)² - 2

Hence, the equation is polynomial f(x) = (x + 5)² - 2

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Answer: the awnser is b

Step-by-step explanation:

Answer:

[tex]573.9 \text{ in}^3[/tex]

Step-by-step explanation:

First, we can find the volume of the rectangular prism using the formula:

[tex]V_\square = l \cdot w \cdot d[/tex]

where [tex]l[/tex] is the prism's length, [tex]w[/tex] is its width, and [tex]d[/tex] is its depth.

Plugging the given dimensions into the formula:

[tex]V_\square = 8 \cdot 12 \cdot 6[/tex]

[tex]V _\square= 96 \cdot 6[/tex]

[tex]\boxed{V_\square = 576 \text{ in}^3}[/tex]

Next, we can find the volume of the half-sphere using the formula:

[tex]V_\circ = \dfrac{2}{3} \pi r^3[/tex]

where [tex]r[/tex] (or [tex]d/2[/tex]) is the half-sphere's radius.

Plugging the given diameter value into the formula:

[tex]V_\circ = \dfrac{2}{3} \pi (2/2)^3[/tex]

[tex]V_\circ = \dfrac{2}{3} \pi (1)^3[/tex]

[tex]\boxed{V_\circ=\dfrac{2}{3}\pi \text{ in}^3}[/tex]

Finally, we can find the volume of the composite figure by subtracting the volume of the half-sphere from the volume of the rectangular prism.

[tex]V = V_\square - V_\circ[/tex]

[tex]V = 576 \text{ in}^3 - \dfrac{2}{3}\pi \text{ in}^3[/tex]

We can evaluate this using a calculator.

[tex]\boxed{V\approx 573.9 \text{ in}^3}[/tex]

The probability of randomly choosing two boys is 0.174

If the selection is done at random, then the probability of choosing a boy is equal to the quotient between the number of boys and the total number of people.

There are 12 boys and 16 + 12 = 28 students in total, then the probability is:

P = 12/28

Now, for the second selection the probability is computed in the same way, but now there are 11 boys and 27 students (one was already choosen), then:

P = 11/27

the joint probability (for the two selections) is given by the product between the individual probabilities, we will get:

P = (12/28)*(11/27)

P = 0.174

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The minimum number of tickets the agent must sell to have a 95% chance of selling at least one prize-winning ticket is 368 tickets

To unravel this issue, we are able to utilize the concept of likelihood. Let's begin with discovering the likelihood of not offering any prize-winning tickets with one ticket:

- The likelihood of not winning with one ticket is 499/500 (since there's as it were one winning ticket out of 500).

- The likelihood of not winning with n tickets is (499/500)[tex]^n[/tex](since the probabilities are autonomous and we duplicate them together).

Presently, we need to discover the least number of tickets the specialist must offer to have a 95% chance of offering at the slightest one prize-winning ticket. In other words, we need to discover the least esteem of n such that:

- The likelihood of not winning with n tickets is less than or rises to 5% (or 0.05).

- This can be identical to (499/500)[tex]^n[/tex]≤ 0.05.

We will illuminate for n utilizing logarithms:

- Take the normal logarithm of both sides:

ln[(499/500)[tex]^n[/tex]] ≤ ln(0.05).

- Utilize the logarithmic property of types to bring down the example:

n ln(499/500) ≤ ln(0.05).

- Fathom for n by partitioning both sides by ln(499/500) and taking the ceiling work (since n must be an entire number):

n = ceil[tex][ln(0.05) / ln(499/500)].[/tex]

Employing a calculator, we get 

n = ceil[tex][ln(0.05) / ln(499/500)][/tex]≈ 368

In this manner, the specialist must offer at the slightest 368 tickets to have a 95% chance of offering at the slightest one prize-winning ticket. 

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right isosceles triangle.

Right Scalene Triangle

Angles

Acute angle:  An angle whose measure is less than 90 degrees.

Right angle:  An angle whose measure IS 90 degrees.  As a consequence, the two rays that form the angle are perpendicular.

Obtuse angle:  An angle whose measure is greater than 90 degrees.

Triangles (defined by side lengths)

Equilateral Triangle:  A triangle with all three sides congruent (all three sides have the same length)

Isosceles Triangle:  A triangle that contains two congruent sides (two sides with the same length).

Scalene Triangle: A triangle with three different length sides.

Triangles (defined by angle measures)

Acute Triangle:  A triangle whose angles are all acute angles

Right Triangle:  A triangle that contains one right angle.

Obtuse Triangle:  A triangle that contains one obtuse angle.

By angle

Our triangle has an angle that appears to be a right angle (the two sides look perpendicular).  This would be a Right Triangle (when classified by angles).

By side length

All three sides of this triangle are different in length.  This would be a Scalene Triangle (when classified by side length).

Combining the attributes:

This is both a Right Triangle and a Scalene Triangle.  Together, it is a Right Scalene Triangle.

Right Isosceles

To be a Right Isosceles Triangle, the triangle needs to be both a Right Triangle, and an Isosceles Triangle.

A Right triangle contains one right angle, and an Isosceles Triangle contains a pair of sides that are the same length.

While this triangle does contain a right angle, all three sides appear to be different length (scalene).

This is not a Right Isosceles Triangle.

Acute Isosceles

To be an Acute Isosceles Triangle, the triangle needs to be both an Acute Triangle, and an Isosceles Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and an Isosceles Triangle contains a pair of sides that are the same length.

However, this triangle contains a right angle (equal to 90 degrees, not less than), so it cannot be an Acute Triangle, and this triangle does not contain a pair of sides that are the same length, so it cannot be an Isosceles Triangle.

This is not an Acute Isosceles Triangle.

Acute Scalene

To be an Acute Scalene Triangle, the triangle needs to be both an Acute Triangle, and a Scalene Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and a Scalene Triangle has sides that are all different length.

While this triangle does have sides that are all different length, it contains a right angle, whose measure IS 90 degrees (not less than 90 degrees).  So it cannot be an Acute Triangle

This is not an Acute Scalene Triangle.

Answer:

Set your calculator to Degree mode.

10² = 6.78² + 4² - 2(6.78)(4)(cos x)

100 = 61.9684 - 54.24(cos x)

38.0316 = -54.24(cos x)

cos x = -.701173

[tex]x = {cos}^{ - 1} ( - .701173)[/tex]

x = 134.52°

After simplification,

(2x + 1)/2(x^2)

Given,

(x/4)+(1/8)/(x^2/4)

Simplifying,

Firstly take LCM of 4 and 8,

LCM of 4 and 8 will be 8.

Then,

(2x +1) /(8) / x^2 /4

Further,

Take reciprocal of the denominator,

(2x +1) /(8) (4/x^2)

(2x + 1)/2(x^2)

Hence after simplification the expression becomes (2x + 1)/2(x^2).

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Answer:

Answer is -2/21r + (-19)

Step-by-step explanation:

(-1/7r - 9 -2/3r) - (-5/7r + 10)

= -1/7r - 9 - 2/3r + 5/7r - 10  (distributing the negative sign)

= (-1/7r + 5/7r) + (-2/3r) - 9 - 10  (grouping the like terms)

= 4/7r - 2/3r - 19  (combining the like terms)

= (12r - 14r)/21r - 38r/21r  (getting a common denominator)

= -2r/21r - 38r/21r

= (-2r - 38r)/21r

= -40r/21r

= -40/21

Therefore, the simplified expression is B. -2/21r + (-19).

398.223 redondeado a la decena más cercana es 400.

Redondear un número a la decena más cercana implica determinar a qué decena se acerca más. En este caso, estamos redondeando el número 398.223.

Para hacer esto, observamos el dígito en la posición de las decenas, que es el 9 en 398.223. El dígito a la derecha de la posición de las decenas es el lugar de las centésimas, que en este caso es 8.

Para redondear a la decena más cercana, seguimos la regla general de redondeo: si el dígito en el siguiente lugar de valor más bajo (en este caso, el lugar de las centésimas) es 5 o mayor, redondeamos hacia arriba el dígito en el lugar de valor objetivo (en este caso, las decenas). Si el dígito es menor que 5, redondeamos hacia abajo.

En nuestro caso, el dígito en el lugar de las centésimas (8) es mayor que 5, por lo que redondeamos hacia arriba el dígito en el lugar de las decenas (9) al número siguiente más alto. Por lo tanto, 398.223 redondeado a la decena más cercana se convierte en 400.

En resumen, redondear 398.223 a la decena más cercana nos da 400 porque el dígito en el siguiente lugar de valor más bajo (8) es mayor que 5.

Más información sobre el redondeo se puede encontrar aquí: https://brainly.com/question/10851954

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