find the area and circumference of the circle below.round your answers to the nearest hundredth

The area of the square base = x².

we have:l = w = x ... (2) ... And, h = V/lw = V/x² ... (3) ...

The dimension of the box that minimizes the amount of material used is x =  (2V)1/3. A(x) = x² + 4V/x, A'(x) = 2x - 4V/x², x =  (2V)1/3

The given volume of the box is 62500 cm³. We wish to find the dimensions of the box that minimize the amount of material used.

To obtain the formula for the surface area of the box in terms of only x, the length of one side of the square base, we use the formula for the volume of a box:V = lwh ... (1) ... where V is the volume, l is the length, w is the width, and h is the height of the box. Here, the base of the box is a square with side length x.

Hence, the area of the square base = x². Therefore, we have:l = w = x ... (2) ... And, h = V/lw = V/x² ... (3) ... We can substitute (2) and (3) in (1) to get the formula for V in terms of x as follows:V = x² V/x² A(x) = A(x) = x² + 4xhA(x) = x² + 4x(V/x²) = x² + 4V/x

Now, to find the derivative A'(x) of A(x), we differentiate A(x) with respect to x:A'(x) = 2x - 4V/x²  A'(x) = 0 when x =  (2V)1/3. Therefore, the dimension of the box that minimizes the amount of material used is x =  (2V)1/3. A(x) = x² + 4V/x, A'(x) = 2x - 4V/x², x =  (2V)1/3

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Answer:

1. n=6

2. n=19

3. n=15

4. n=20

5. n=40

6. n=16

7. n=19

8. n=6

9. n=2.4

10. n=30

Step-by-step explanation:

remember the priorities :

1. brackets

2. exponents

3. multiplications and divisions

4. additions and subtractions

inside every category you go usually from left to right, but you can use the commutative property where applicable.

4 + 5 - 3 = 4 + 5 - 3 = 4 + 5 - 3 = 4 - 3 + 5 = 6

8×3 - 5 = 24 - 5 = 19

15/3 + 10 = 5 + 10 = 15

8 + 3 - 1×2 = 8 + 3 - 2 = 9

14×3 - 2 = 42 - 2 = 40

(3 + 5)×2 = 8×2 = 16

10 + 7 + 2×1 = 10 + 7 + 2 = 19

16/8 + 4 = 2 + 4 = 6

7 + 5/5 = 7 + 1 = 8

20/4×6 = 5×6 = 30

The given expression ((y-b)²/(y-b+1)+(y-b)/(y-b+1) after being reduced to a polynomial, can be represented as y-b.

In order to reduce the given equation to a polynomial, we are required to simplify and combine like terms. First, we can simplify the expression in the numerator by expanding the square:

((y-b)²/(y-b+1) = (y-b)(y-b)/(y-b+1) = (y-b)²/(y-b+1)

Now, we can combine the two terms in the equation by finding a common denominator:

(y-b)²/(y-b+1) + (y-b)/(y-b+1) = [(y-b)² + (y-b)]/(y-b+1)

Next, we can combine the terms in the numerator by factoring out (y-b):

[(y-b)² + (y-b)]/(y-b+1) = (y-b)(y-b+1)/(y-b+1)

Finally, we can cancel out the common factor of (y-b+1) in the numerator and denominator to get the polynomial:

(y-b)(y-b+1)/(y-b+1) = y-b

Therefore, the equation ((y-b)²)/(y-b+1)+(y-b)/(y-b+1)  after being simplified, is equivalent to the polynomial y-b.

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We pIug this function into the equation above, we get: g(x+1) = 0.

Assume that g(x) is a function that decreases by 25% for every one increase in x.

This means that if we increase x by one, the vaIue of g(x) wiII be 75% of what it was befοre.

In other words, if g(x) is the function's vaIue at x, then g(x+1) is 0.75 times g. (x).

As a resuIt, we can express g(x+1) in terms of g(x) as foIIows:

g(x+1) = 0.75 * g(x) (x)

Let's see which οption meets this requirement.

A) g(x) = 0.75: When we pIug this function into the equation above, we get:

g(x+1) = 0.75 * 0.75 = 0.5625 * g(x) (x)

This means that increasing x by one increases g(x) by 25%. As a resuIt, this optiοn does not meet the condition.

B) g(x) = 75: When we pIug this function into the equation above, we get:

g(x+1) = 0.75 * 75 = 56.25

This means that increasing x by οne decreases g(x) by more than 25%.

As a resuIt, this option does not meet the condition.

C) g(x) = 0.25: When we pIug this function into the equatiοn above, we get:

g(x+1) = 0.75 * 0.25 = 0.1875 * g(x) (x)

This means that increasing x by one decreases g(x) by more than 25%. As a resuIt, this optiοn does not meet the cοndition.

D) g(x) = 25*: When we pIug this function into the equation above, we get:

g(x+1) = 0.

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Cοmplete question:

For which function does g decrease by 25% every time x increases by 1?

A) g(x) = 0.75

B) g(x) = 75

C) g(x) = 0.25

D) g(x) = 25*

The standard deviation of a normal distribution of exam scores is 8. A score that is 12 points above the mean would have a z-score of 1.5, and a score that is 20 points below the mean would have a z-score of -2.5.

The z-score can be calculated by dividing the difference between a data value and the mean of the data set by the standard deviation of the data set.

The z-score of a score that is 12 points above the mean in a normal distribution of exam scores with a standard deviation of 8.

z = (x−μ)/σ = (x−μ)/σ = (12−0)/8 = 1.5

The z-score of a score that is 12 points above the mean in a normal distribution of exam scores with a standard deviation of 8 is 1.5.

The z-score of a score that is 20 points below the mean in a normal distribution of exam scores with a standard deviation of 8.

z = {x-μ}/{σ}  = {-20-0}/{8}  = −2.5

The z-score of a score that is 20 points below the mean in a normal distribution of exam scores with a standard deviation of 8 is -2.5.

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Answer:

$56 divided by 7 hours is $8 per hour sh get payed.

Step-by-step explanation:

Here is one way :

  at the x-axis intercepts , the value of 'y' = 0  (obviously)

   0  = x^2 -64

    x^2 = 64

     x = ± 8

Answer:

X+4

Step-by-step explanation:

Area = l *b

x^2 + 13x + 36 = (X+9) * b

x^2 + 9x + 4x + 36 = (X+9) * b

X(X+9) + 4(X+9) = (X+9) * b

(X+4) (X+9) = (X+9) * b

b = (X+4)

The percent error from the real and estimated values is 16%.

The difference between a precise value and an approximation to it is the approximation error in a data value. Either an absolute mistake or a relative error can be used to describe this error. It is possible for measurement mistake or computing machine precision to produce an approximation error. The percent error is the distinction between the estimated value and the real value in relation to the actual value. In other terms, the relative error is multiplied by 100 to calculate the percent error.

The estimated value= 170

the real value= 147

Percent error ={ (170-147)/147}*100= 0.1564*100

Percent error = 15.64% ≈ 16%

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According to the VAT rate, the total cost of the item including VAT is Birr 10,925 assuming a VAT rate of 15%.

What is VAT rate ?

VAT (Value Added Tax) rate is a percentage that is added to the price of goods or services. VAT is usually charged on top of the base price of a product or service, and the total price is calculated as (base price) + (base price x VAT rate). The VAT rate is typically set by the government and may be subject to change over time.

To find the total cost of the item including VAT, we need to know the VAT rate.

Let's assume the VAT rate is 15%. To calculate the VAT amount, we can multiply the price before VAT by the VAT rate:

VAT amount = 0.15 x 9,500 = Birr 1,425

The total cost of the item including VAT is the sum of the price before VAT and the VAT amount:

Total cost = 9,500 + 1,425 = Birr 10,925

Therefore, the total cost of the item including VAT is Birr 10,925 assuming a VAT rate of 15%.

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Answer:

sales tax 6.40

total cost 86.40

Step-by-step explanation:

First we need to find the sales tax.

80 * 8%

80 *.08 = 6.40

Add this to the original price.

80+6.40

86.40

The total cost is 86.40

The possible expression that can be a rule for the number sequence is:[tex]$a_n = 4n + 1$[/tex],

As an illustration, the expression x + y is one where x and y are words with an addition operator in between. There are two types of expressions in mathematics: numerical expressions, which only comprise numbers, and algebraic expressions, which also include variables.

One possible expression that can be a rule for the number sequence is:

[tex]$a_n = 4n + 1$[/tex], where n is the position of the term in the sequence.

Using this expression, we can find the values of the first few terms as follows:

[tex]$a_1 = 4(1) + 1 = 5$[/tex]

[tex]$a_2 = 4(2) + 1 = 9$[/tex]

[tex]$a_3 = 4(3) + 1 = 13$[/tex]

[tex]$a_4 = 4(4) + 1 = 17$[/tex]

[tex]$a_5 = 4(5) + 1 = 21$[/tex]

Thus, possible expression that can be a rule for the number sequence is:[tex]$a_n = 4n + 1$[/tex],

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Answer:

5n, where n is equal to 0, 1, 2, 3, 4

Step-by-step explanation:

5, 9, 13, 17, 21,

Answer: 2

Step-by-step explanation:

If there are three times as many students when there are 6 students you do 6 divided by three to give you two.

Answer:

yes it is right now you can write it

If the question is simply asking for the product of 5 and 5/6, then writing the reciprocal is not necessary. To multiply fractions, we can simply multiply the numerators and the denominators separately. In this case, we would get:

5 x 5/6 = (5 x 5)/(1 x 6) = 25/6

However, if the question specifically asks for the result to be expressed as a mixed number or a whole number, then we would need to simplify the fraction by converting it to a mixed number or a whole number. In some cases, it may be helpful to write the reciprocal in order to simplify the fraction more easily.

For example, if the question asks us to express 5 x 5/6 as a mixed number or a whole number, we could first write the reciprocal of 5/6 as 6/5, and then multiply it with 5 to get:

5 x 5/6 = 5 x 6/5 = 30/5

We can simplify this fraction by dividing both the numerator and the denominator by 5:

30/5 = 6

Therefore, if the context or the desired result requires the expression 5 x 5/6 to be simplified further, it may be helpful to write the reciprocal of 5/6 to make the simplification process easier. However, in general, it is not necessary to write the reciprocal to multiply fractions.

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Answer:

D= 7(2g-H)/U

Step-by-step explanation:

This question is related to formal languages and automata theory, specifically to the topic of context-free grammar and derivations.

a) bbabbab:

S -> bS (Apply S -> bS)

-> bbS (Apply S -> bS)

-> bbbA (Apply S -> aA)

-> bbbaS (Apply A -> aS)

-> bbbaaS (Apply S -> aA)

-> bbbaaa (Apply A -> a)

Therefore, the derivation of bbabbab is S -> bS -> bbS -> bbbA -> bbbaS -> bbbaaS -> bbbaaa -> bbabbab.

b) bbbaab:

S -> bS (Apply S -> bS)

-> bbS (Apply S -> bS)

-> bbbA (Apply S -> aA)

-> bbbaS (Apply A -> aS)

-> bbbaaS (Apply S -> aA)

-> bbbaaa (Apply A -> a)

-> bbbaab (Apply A -> b)

Therefore, the derivation of bbbaab is S -> bS -> bbS -> bbbA -> bbbaS -> bbbaaS -> bbbaaa -> bbbaab.

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The average power output of a bike rider at a speed of 20 miles per hour is 556 W

Given, Riding a bike burns 100 Calories per mile. Speed of riding a bike is 20 miles per hour. Power is a measure of the rate at which work is done over time.

Therefore, the formula for average power is given by,

P = W/t

Where, P is power, W is work and t is time.

Now, We know that riding a bike burns 100 Calories per mile, but we have to find out the work done.

So, work done = 100 Calories × distance traveled= 100 × 1.609 × 1000 = 160900 J

Time taken = distance/speed= 1.609/20 = 0.08045 hours

Putting values in the formula of power, we get,

P = W/t= 160900/0.08045= 1999999.9 J/hour

Converting hour to seconds,1 hour = 3600 seconds1 J/s = 1 watt1 J/hour = 1 / 3600 watt

Therefore, power = 1999999.9 / 3600= 555.5556 watts (approx)

Therefore, the average power output of a bike rider at a speed of 20 miles per hour is 556 W (approx). Answer: 556

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The οptiοn in the questiοn that is equivalent tο 3/8 is 2/5.

The term "equivalent" in math refers tο twο meanings, numbers, οr quantities that are the same. The equivalence οf twο such quantities shall be denοted by a bar οver an equivalent symbοl οr Equivalent Sign. It alsο means a lοgical equivalence between twο values οr a set οf quantities. Equivalence is similar but mοre universal than equality. If twο sets οf equatiοns have the same sοlutiοns, we might cοnsider them similar, but they are nοt identical.

We can simplify 3/8 and 2/5 sο that they have a cοmmοn denοminatοr:

3/8 = 3*(5/5)/(85/5) = 15/40

2/5 = 2(8/8)/(5*8/8) = 16/40

Since 15/40 and 16/40 have the same denοminatοr, we can cοmpare their numeratοrs tο see which is larger:

15/40 < 16/40

Since 16/40 is larger, we can cοnclude that 2/5 is greater than 3/8.

Therefοre, the οptiοn that is equivalent tο 3/8 is 2/5.

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Answer:

144°

Step-by-step explanation:

x = (10 - 2)(180)/10 = 144

Answer:

224

Step-by-step explanation:

392 ÷ 7 = 56

56 *4 = 224

hope this helps

Answer:

  (a) {0 ≤ x ≤ 60; 0 ≤ y ≤ 90; x + y ≤ 120}

  (b) p = 50x +75y; 30 tons of loam and 90 tons of peat will maximize profit

Step-by-step explanation:

You want a graph of constraints and the solution to the problem of maximizing profit from sale of 60 tons of loam and 90 tons of peat at a profit of $50/t and $75/t, respectively, if at most 120 tons can be sold.

Let x and y represent tons of loam and peat, respectively. Then the constraints can be written as the inequalities ...

We want to maximize profit, so maximize ...

  p = 50x +75y

The attached graph shows the graphs of the inequalities. The restrictions on the variables confine the applicable graph to the first quadrant.

The black line shows the objective function. The value of the profit function is maximized when the line is as far from the origin as possible. The boundary point of the constraint solution space that maximizes profit is (x, y) = (30, 90).

Sale of 30 tons of loam and 90 tons of peat soil will maximize profit.

__

Additional comment

Since selling peat is the most profitable, profit will be maximized when we sell as much peat as possible. Sale of only 90 tons is possible, so profit will be maximized when we sell that many, then make up the rest of the 120 ton maximum by selling loam. We don't need a graph or a list of constraints to figure that out.

The correct relative frequency for X is:

X: $2 $3 $4 $7

P(X): 0.38 0.15 0.28 0.19

First we need to determine the probability of each value of X occurring based on the given data.

We can use the following formula to calculate the probability of a particular value of X:

P(X = x) = f(x)/n

Where

Let's calculate the probabilities for each value of X:

P(X = $2) = 57/150 = 0.38

P(X = $3) = 23/150 = 0.15

P(X = $4) = 42/150 = 0.28

P(X = $7) = 28/150 = 0.19

Therefore, the correct relative frequency for X is:

X: $2 $3 $4 $7

P(X): 0.38 0.15 0.28 0.19

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To find the next two terms of the sequence, we need to first find the common difference between consecutive terms:

52 - 44 = 8

64 - 52 = 12

80 - 64 = 16

We notice that the common difference is increasing by 4 for each term. Therefore, the next two terms of the sequence are:

80 + 20 = 100

100 + 24 = 124

To determine the nth term of the quadratic sequence, we can use the formula:

an = a1 + (n-1)d + bn^2

where a1 is the first term, d is the common difference, b is the coefficient of n^2, and n is the term number.

Using the first four terms of the sequence, we can form a system of equations:

44 = a1 + b

52 = a1 + d + b

64 = a1 + 2d + b

80 = a1 + 3d + b

Solving for a1 and b, we get:

a1 = 20

b = 24

Substituting these values into the formula for an, we get:

an = 20 + (n-1)4 + 24n^2

an = 24n^2 + 4n - 4

To find the 30th term of the sequence, we simply substitute n = 30 into the formula we just derived:

a30 = 24(30)^2 + 4(30) - 4

a30 = 21,596

To prove that the quadratic sequence will always have even terms, we notice that the first term is even (44 = 2 x 22), and the common difference is even (8 = 2 x 4). Therefore, every term of the sequence can be expressed as an even number plus an even multiple of n^2, which is always even. Hence, the quadratic sequence will always have even terms.

Step-by-step explanation:

Sequence is 44;52;64;80;.....44;52;64;80;.....

General formula is Tn=2n2+2n+40

Answer:

A $0.05n< $0.45

Step-by-step explanation:

It is given that she has "less than" rather than "less than or equal to" 45 cents. Therefore, your answer will contain the "less than" sign, which is signified as <.

Therefore, being the only answer with <, your answer will be A.

A nickel is $0.05, therefore:

$0.05 of n is less than $0.45, or $0.05n< $0.45

A) $0.05n < $0.45 is your answer.

~

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The solution is {y} = (2311, 2557, 2015) and hence the original message is CAT.

Suppose Alice uses the RSA methods as follows. She starts with a message consisting of several letters, and assigns a = 1,b=2, -2=26. She then encrypts each letter separately. For example, if her message is cat, she calculates 3 = mod n 19 = 2 mod n and 209 = mod n. Then she sends the encrypted message {G}=1,2,3 to Bob.The solution to the given question is as follows:Eva knows that n = 8881 = 79 × 112, where p = 79 and q = 112 are primes.Using the extended Euclidean algorithm and the fact that 13*5081 = 1 (mod 8880), one can compute the inverse of e modulo 8880 as d ≡ 5081 ≡ -3805 (mod 8880).To compute the message, Alice encrypts each letter of the message separately using the encryption function y ≡ x13 (mod 8881).Eva receives {G}=1,2,3 from Alice, and hence she wants to decipher it by computing 13 ≡ y ≡ G13 (mod 8881), where y is the message.To compute y, we have 1^{13} ≡ 1 (mod 8881), 2^{13} ≡ 8192 ≡ - 6889 + 2 (mod 8881) since 8192 = 2 × 4096 and 4096 = 8881 - 4785, hence, 2^{13} ≡ - 4785 × 2 + 2 (mod 8881), i.e., 2^{13} ≡ 2311 (mod 8881), and 3^{13} ≡ 3 × 3^{12} ≡ 3 × (3^{4})^{3} ≡ 3 × 6561^{3} ≡ 3 × 4299 ≡ 12897 ≡ 2557 (mod 8881).Therefore, the decrypted message is {y} = (2311, 2557, 2015) and hence the original message is CAT.Hence, the solution is {y} = (2311, 2557, 2015) and hence the original message is CAT.

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The dimensions of the triangle with maximum area are a base of 11 cm and a height of 11 cm.

We are given that the sum of the base and height of a triangle is 22 cm. Let's call the base of the triangle "b" and the height "h". Then we have the equation:

b + h = 22

We want to find the dimensions of the triangle that will give us the maximum area.

The formula for the area of a triangle is A = (1/2)bh.

We can use the equation b + h = 22 to solve for one of the variables in terms of the other.

we can solve for h:

h = 22 - b

Substituting this into the formula for the area, we get:

A = (1/2)b(22 - b)

The equation 0 = 11b - (1/2)b² can be rearranged to:

(1/2)b² - 11b = 0

We can then use the quadratic formula to solve for "b"

b = 11 ± 11

So the value of b that gives us the maximum area is b = 11 cm. Substituting this back into the equation h = 22 - b, we get:

h = 22 - 11 = 11 cm

Triangle has a base of 11 cm and a height of 11 cm.

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The five fold health hazards prevalent in Nepal are respiratory diseases due to high altitude, Germ cell mutagenicity, Carcinogenicity, Reproductive toxicity and sanitation related issues which causes diarrhea, cholera like diseases.

Nepal is a small country in hilly region of Himalayas between China and India. It has not developed much in terms of health care facilities and advanced treatment methods due to which people migrate to India for treatments in several extreme diseases. The low life expectancy is because of poor employment which causes financial insecurity and hence people are either malnourished, or die of lack of proper treatment.

The Nepal Health Sector Program – Implementation Plan (NHSP-IP) has been developed in order to meet the Health Sector Strategy which has its agenda to improvise the needs of the people in near future.

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The proportion of students with a z-score between -1.20 and -0.30 is about 0.304. So for every 100 students.

What is the z-score?

Z-score indicates how much a given value differs from the standard deviation. The Z-score, or standard score, is the number of standard deviations a given data point lies above or below the mean. Standard deviation is essentially a reflection of the amount of variability within a given data set.

Assuming that the grading follows a standard normal distribution (i.e., with a mean of 0 and a standard deviation of 1), we can use the z-score formula to find the proportions of students that fall within each grade range:

For an A grade: z-score > 1.65

For a B grade: 0.67 < z-score ≤ 1.65

For a C grade: -0.40 < z-score ≤ 0.67

For a D grade: -1.60 < z-score ≤ -0.40

To convert from the standard normal distribution to the distribution with the course mean and standard deviation, we can use the formula:

z-score = (x - mean) / standard deviation

We don't know the actual mean and standard deviation for the course, so we'll use the given range for the D grade (-1.60 < x < -0.4σ) to estimate the standard deviation. Since the D grade range is 1.2 standard deviations wide (from -1.60 to -0.4σ), we can solve for the standard deviation:

1.2σ = -1.60

σ = -1.60 / 1.2

σ = 1.33

(Note that we have a negative value for σ, which is not possible for a standard deviation. This is because we are using the estimated value of σ to convert from the standard normal distribution to the course distribution, which may not perfectly match the actual distribution of grades.)

Now we can use the z-score formula with the estimated mean and standard deviation to find the proportions of students in each grade range:

For an A grade: z-score > 1.65

z-score = (x - mean) / standard deviation

z-score = (1.65 - 0) / 1.33

z-score = 1.24

From standard normal distribution tables or a calculator, we can find that the proportion of students with a z-score greater than 1.24 is about 0.107. So for every 100 students, we can expect about:

A grades: 100 * 0.107 = 10.7 or approximately 11

For a B grade: 0.67 < z-score ≤ 1.65

z-score = (0.67 - 0) / 1.33

z-score = 0.50

From standard normal distribution tables or a calculator, we can find that the proportion of students with a z-score between 0.50 and 1.24 is about 0.205. So for every 100 students, we can expect about:

B grades: 100 * 0.205 = 20.5 or approximately 21

For a C grade: -0.40 < z-score ≤ 0.67

z-score = (-0.40 - 0) / 1.33

z-score = -0.30

From standard normal distribution tables or a calculator, we can find that the proportion of students with a z-score between -0.30 and 0.50 is about 0.307. So for every 100 students, we can expect about:

C grades: 100 * 0.307 = 30.7 or approximately 31

For a D grade: -1.60 < z-score ≤ -0.40

z-score = (-1.60 - 0) / 1.33

z-score = -1.20

Hence, From standard normal distribution tables or a calculator, we can find that the proportion of students with a z-score between -1.20 and -0.30 is about 0.304. So for every 100 students,

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The shipping company will need 2,688 one cubic centimeter blocks to fill the box.

The volume of the box can be calculated as:

Volume = length x width x height

Volume = 14 cm x 12 cm x 16 cm

Volume = 2,688 cubic cm

Since each block is also one cubic cm in volume, we can simply divide the total volume of the box by the volume of one block to find the number of blocks needed to fill the box:

Number of blocks = Volume of box / Volume of one block

Number of blocks = 2,688 cubic cm / 1 cubic cm

Number of blocks = 2,688

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