1. Find the following limits. 2x2 - 8 a) lim * 4x2 + 2 b) lim -0 c) lim 2x +5x+3 2x+3

Answer:

Step-by-step explanation:

this is sorta a thought process to figure this out.   so think it thur with me.  b/c they are saying that the line is the same length with the lines thur it... see them?  so then that means that two sides of the right triangle are of a certain length so that means we know the base and length of the sides of a box, if we were to add the other right triangle to it.. see that?

so we can say that  6*7.5 = box   and we'll have two total boxes, see that?  the top portion of the parallelogram and the bottom are each a box .. but cut down the middle and put as two opposite parts of a triangle.

so

2*(6*7.5) = 90 [tex]in^{2}[/tex]

Answer:

Step-by-step explanation:

the n number of value of x

[tex] \displaystyle x_{1},x _{2} \dots x_{n}[/tex]

let it be

[tex] \displaystyle x_{1} = x _{2} = x_{3}{\dots }= x_{n} = a[/tex]

now, the A.M of x is

[tex] \rm \displaystyle \: A.M = \frac{ x_{1} + x_{2} + \dots \dots \: + x_{n} }{n} [/tex]

since every value equal to a

substitute:

[tex] \rm \displaystyle \: A.M = \frac{ a + a + \dots \dots \: + a}{n} [/tex]

[tex] \rm \displaystyle \: A.M = \frac{ na}{n} [/tex]

reduce fraction:

[tex] \rm \displaystyle \: A.M = a[/tex]

the G.M of x is

[tex] \rm\displaystyle \: G.M =( x_{1} \times x _{2} {\dots }\times x_{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

since every value equal to a

substitute:

[tex] \rm\displaystyle \: G.M =( a \times a{\dots }\times a{)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M =( {a}^{n} {)}^{ {1}^{}/ {n}^{} } [/tex]

recall law of exponent:

[tex] \rm\displaystyle \: G.M = a[/tex]

the H.M of x is

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ x_{1}} + \frac{1}{ x_{2} } {\dots } \: { \dots}\frac{1}{x _{n} } } [/tex]

since every value equal to a

substitute:

[tex] \displaystyle \: H.M = \frac{n}{ \frac{1}{ a} + \frac{1}{ a } {\dots } \: { \dots}\frac{1}{a } } [/tex]

[tex] \displaystyle \: H.M = \frac{n}{ \dfrac{n}{a} } [/tex]

simplify complex fraction:

[tex] \displaystyle \: H.M = n \times \frac{a}{n} [/tex]

[tex] \displaystyle \: H.M = a \: [/tex]

so

[tex] \displaystyle \: A.M = G.M = H.M = a[/tex]

hence,

[tex]\text{Proven}[/tex]

Answer:

By considering the material distribution and recoverable percentages, we can calculate the amount of recoverable materials for the urban area's recycling program.

To determine the amount of recoverable materials in an urban area's recycling program, we need to consider the material distribution and the percentage of each component that is recoverable. Based on the provided data, let's calculate the recoverable weight for each component:

Food: 46% of municipal solid waste (MSW) weighs 5129 tons, so the recoverable weight is 46% of 5129 tons.

Paper and cardboard: 11% of MSW weighs 1226 tons, and assuming a certain percentage of paper and cardboard is recoverable, we multiply this percentage by 1226 tons to obtain the recoverable weight.

Plastics: 9% of MSW weighs 1003 tons, and the recoverable weight is determined similarly to paper and cardboard.

Glass: 7% of MSW weighs 780 tons, and the recoverable weight is calculated based on the recoverable percentage.

Metals: 5% of MSW weighs 558 tons, and the recoverable weight is determined using the recoverable percentage.

Clothing/textiles, ashes/dust, and unclassified materials: The recoverable weight for each component is calculated based on the respective percentages.

By summing up the recoverable weights of all components, we can determine the total amount of recoverable materials for the urban area's recycling program.

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Answer:

158

Step-by-step explanation:

The given question is incomplete. The complete question is:

Omar rented a truck for one day. There was a base fee of $17.95, and there was an additional charge of 98 cents for each mile driven. Omar had to pay $23 when he returned the truck. For how many mile did he drive the truck?

Answer: Omar drove the truck for 5.15 miles

Step-by-step explanation:

Base fee = 17.95 $

Additional charge per mile = 98 cents = 0.98 $     ( 100cents = 1$)

Now Omar payed = 22 $

Let the miles he travelled = x

Now , [tex]17.95+0.98\times x=23[/tex]

Solvimg for x :

[tex]x=5.15miles[/tex]

Thus Omar drove the truck for 5.15 miles

Answer:

2(x+2) = x+4

Step-by-step explanation:

The pair that shows equivalent expressions is:

2(x+2) = x+4

This is because when we distribute the 2 to the terms inside the parentheses, we get:

2x + 4 = x + 4

By subtracting x from both sides of the equation, we get:

2x - x + 4 = 4

Simplifying further, we have:

x + 4 = 4

Therefore, the expression 2(x+2) is equivalent to x+4.

Hope this helps!

Answer:

23%, 23/100, 0.23

Step-by-step explanation:

23%

= 23/100

= 0.23

Answer:

13.8

Step-by-step explanation:

Hello There!

Basic concept: Geometry

We can solve for x using trigonometric ratios

Here are the Trigonometric Ratios

Remember SOHCAHTOA

Sin = Opposite over Hypotenuse

Cos = Adjacent over Hypotenuse

Tan = Opposite over Adjacent

For the angle that has a measure of 37 degrees

We are given its adjacent side length (11) and we need to find the hypotenuse

Basic concept: Alegbra 1

Adjacent and Hypotenuse corresponds with Cos so we are going to use cosine to create an equation and solve for x

Remember cos = adjacent over hypotenuse

So

[tex]cos37=\frac{11}{x}[/tex]

now we solve for x

what we want to do is get rid of the 11, to do so we divide each side by 11

[tex]\frac{11}{cos37} =x\\cos37=.79863551\\\frac{11}{0.79863551} =13.773749224[/tex]

we're left with x = 13.773749224

finally we want to round to the nearest tenth

The answer would be 13.8

Answer:

4x^2 - 4x - 19

Step-by-step explanation:

Given the following functions

f(x) = 2x^2 - 4x - 3

g(x) = 2x^2 - 16

f(x)+ g(x) =2x^2 - 4x - 3+ 2x^2 - 16

f(x)+ g(x) = 2x^2+2x^2 -4x - 3 - 16

f(x)+ g(x) =4x^2 - 4x - 19

Hence the required sum is 4x^2 - 4x - 19

The simple interest of $1500.00 for 4 months at 6 3/4% annual interest is $101.25. Thus, the correct answer is B) $101.25.

To calculate simple interest, we use the formula: Interest = Principal x Rate x Time. In this case, the principal (P) is $1500.00, the rate (R) is 6 3/4% (or 0.0675 as a decimal), and the time (T) is 4 months.

First, we need to convert the rate from a percentage to a decimal by dividing it by 100: 6 3/4% = 6.75/100 = 0.0675.

Next, we plug these values into the formula: Interest = $1500.00 x 0.0675 x 4 = $101.25.

Therefore, the simple interest on $1500.00 for 4 months at 6 3/4% annual interest is $101.25. Thus, the correct answer is B) $101.25.

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a) The revenue function in terms of ticket price is R(x) = 63000x - 3000x². b) The price that maximizes revenue from ticket sales is $10.50. c) The ticket price that is so high that no revenue is generated is $21.

(a) To model the revenue in terms of ticket price, we can use the given information. The revenue is calculated by multiplying the number of attendees by the ticket price.

Let x represent the price of a ticket and R represent the revenue. We know that the average attendance at recent games has been 33,000. With every dollar decrease in the ticket price, attendance increases by 3000. So, the attendance can be represented as 33,000 + 3000(10 - x) = 33,000 + 30000 - 3000x.

Therefore, the revenue function R(x) can be expressed as:

R(x) = x * (33,000 + 30000 - 3000x)

= 63000x - 3000x²

Hence, the revenue function in terms of ticket price is R(x) = 63000x - 3000x².

(b) To find the price that maximizes revenue from ticket sales, we need to find the maximum point of the revenue function. This can be determined by finding the vertex of the parabolic function, which represents the maximum point.

The revenue function is R(x) = 63000x - 3000x², which is in the form of a quadratic equation. The maximum point of the quadratic function occurs at x = -b / (2a), where a = -3000 and b = 63000.

Plugging in the values, we have:

x = -63000 / (2 * (-3000))

= -63000 / (-6000)

= 10.5

Therefore, the price that maximizes revenue from ticket sales is $10.50.

(c) To find the ticket price that generates no revenue, we need to consider the scenario where the revenue is equal to zero. From the revenue function R(x) = 63000x - 3000x², we can set R(x) = 0 and solve for x:

63000x - 3000x² = 0

Factoring out x, we have:

x(63000 - 3000x) = 0

This equation is satisfied when either x = 0 or 63000 - 3000x = 0.

For x = 0, it means that the ticket price is zero, which is not possible.

Solving 63000 - 3000x = 0 for x, we get:

3000x = 63000

x = 21

Therefore, the ticket price that is so high that no revenue is generated is $21.

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Inglis please. Thank you.

For function f:(a,b) → R (continuous), and c ∈ (a,b), then

(i) If derivative is negative before c and positive after c, then f has an absolute minimum at c.

(ii) If derivative is positive before c and negative after c, then f has an absolute maximum at c.

Part (i) : If derivative of a function f(x) is negative for values of x between a and c, and positive for values of x between c and b, then the function has an absolute minimum at c.

This means that at point c, function reaches its lowest-value compared to all other points in the interval (a, b). The negative derivative before c indicates a decreasing trend, while the positive derivative after c indicates an increasing trend.

The change from decreasing to increasing at c suggests a minimum point. By the continuity of the function, we can conclude that the minimum value is achieved at c.

Part (ii) : Conversely, if  derivative of a function f(x) is positive for values of x between a and c, and negative for values of x between c and b, then the function has an absolute maximum at c.

This means that at point c, the function reaches its highest-value compared to all other points in the interval (a, b). The positive derivative before c indicates an increasing trend, while the negative derivative after c indicates a decreasing trend.

The change from increasing to decreasing at c suggests a maximum point. By the continuity of the function, we can conclude that the maximum value is achieved at c.

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Prove the theorems below: Let f:(a,b) → R be continuous. Let c ∈ (a,b) and suppose f is differentiable on (a, c) and (c, b).

(i) if f'(x) < 0 for x ∈ (a, c) and f'(x) > 0 for x ∈ (c, b), then f has an absolute minimum at c.

(ii) if f'(x) > 0 for x € (a, c) and f'(x) < 0 for x ∈ (c, b), then f has an absolute maximum at c.

The answer to question is given below.

(i) To find the mean, E(N), use the following formula:

mean = E(N) = ∑[n · P(N = n)] for all values of n.

The distribution given above is geometric: P(N = n) = p^ n (1-p)

where n = 1,2,3,...

Therefore, E(N) = ∑[n · P(N = n)] = ∑ [n · p^ n (1-p)] for n = 1,2,3,...

Since this sum is infinite, we have to truncate the summation and compute the mean for a finite number of terms. We can use 40 since there are 40 independent observations. Therefore, we have:

E(N) ≈ ∑[n · P(N = n)] for n = 1 to 40

= 1 · p(1-p) + 2 · p^2(1-p) + 3 · p^3(1-p) + ... + 40 · p^40(1-p)

= (1-p) ∑[n · p^n] for n = 1 to 40

= (1-p) [p + 2p^2 + 3p^3 + ... + 40p^40]

= (1-p) p ∑[n · p^(n-1)] for n = 1 to 40

= (1-p) p [1 + 2p + 3p^2 + ... + 40p^39]

= p(1-p) [1p^(1-1) + 2p^(2-1) + 3p^(3-1) + ... + 40p^(40-1)]

= p(1-p) ∑[n · p^(n-1)] for n = 1 to 40

= p(1-p) ∑[(n-1+1) · p^(n-1)] for n = 1 to 40

= p(1-p) [∑[(n-1) · p^(n-1)] + ∑[1 · p^(n-1)]] for n = 1 to 40

= p(1-p) [∑[n · p^(n-1)] - ∑[p^(n-1)]] + p(1-p) ∑[p^(n-1)] for n = 1 to 40

= p(1-p) [d/dp ∑[p^n]] - p(1-p) (1/(1-p)) + p(1-p) (1/(1-p))

= p(1-p) [d/dp (1/(1-p)) ∑[(1-p)p^n]] + 1

= p(1-p) [d/dp (1/(1-p)) (1-p)/(1-p)^(40+1)] + 1

= p(1-p) [d/dp (1-p)^(-40)) + 1

= p(1-p) (40(1-p)^(-41)) + 1

= 40p/(1-p) - 40

(ii) Since this is a geometric distribution, the maximum likelihood estimator (MLE) for the unknown parameter p is given by:

MLE: ê = x/n

where x is the number of successes (species of insect caught in a trap) and n is the sample size (40). To find an equation that determines the MLE ê, differentiate the log-likelihood and equate it to zero to find the maximum of the likelihood function. The log-likelihood for a geometric distribution is given by:

L = ∑ [log(P(N = n))] for n = 1 to 40

= ∑ [log(p^n (1-p))] for n = 1 to 40

= ∑ [n · log(p) + log(1-p)] for n = 1 to 40

= (40 · log(p) + ∑ [n · log(p)] + ∑ [log(1-p)])

Use the formula MLE = x/n to replace p by x/n. This gives:

L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])

Differentiate L with respect to x/n and equate to zero to obtain the MLE ê:0 = dL/d(x/n) = (40/x) - (n/x) - ∑ [1/(1 - x/n)]

Solving this equation will give ê in terms of x and n.

(iii) The Fisher information, I(p), is defined as:

I(p) = -E[d^2/dp^2 L] = E[d/dp (d/dp L)]

where L is the log-likelihood function.

From part (ii), we have:

L = (40 · log(x/n) + ∑ [n · log(x/n)] + ∑ [log(1-x/n)])

Therefore,

∂L/∂p = (40/x) - (n/x) and∂^2L/∂p^2 = -40/x^2.The Fisher information is therefore:

I(p) = -E[d^2/dp^2 L] = E[40/x^2] = 40E[x/n]^(-2) = 40/p^2.

Using the asymptotic normality of the MLE, the 95% confidence interval for p is approximately given by:

p ± 1.96 · sqrt(Var(p))

where Var(p) = 1/I(p) = p^2/40.

Using Ô = 0.75, we have ê = x/n = (N1 + N2 + ... + N40)/40 = (28 + 30 + ... + 22)/40 = 0.625.

Therefore, p ± 1.96 · sqrt(Var(p))= 0.625 ± 1.96 · sqrt(0.625^2/40)= (0.478, 0.772).

(iv) When N1 = 100, the maximum likelihood estimate, ê(1), can be found iteratively as follows:

ê(1) = 0.70MLE = ê(1) = x/n

where x is the number of species of insect caught in a trap and n = 40. Therefore, ê(1) can be computed from the data. For example, if x = 25, then ê(1) = 25/40 = 0.625. To obtain ê(2), we need to solve the equation obtained in part (ii) for n = 100:0 = (40/x) - (100/x) - ∑ [1/(1 - x/100)]

We can use Newton's method to solve this equation numerically. Let ê(2) be the root obtained after one iteration of Newton's method. Then, we have:

ê(2) = ê(1) - f(ê(1))/f'(ê(1))where f(p) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] and f'(p) = -∑ [x/100(x - 100)^2].

For example, if x = 25 and ê(1) = 0.625, then:

ê(2) = 0.625 - f(0.625)/f'(0.625)= 0.625 - (-0.1155)/(0.1869)= 0.625 + 0.6171= 1.242.

This value is not valid since the MLE must lie between 0 and 1. Therefore, we need to use a different starting value of ê. Let ê(1) = 0.80. Then, we have:

f(0.80) = (40/x) - (100/x) - ∑ [1/(1 - x/100)] = 0.0142f'(0.80) = -∑ [x/100(x - 100)^2] = -0.1026

Using Newton's method, we have:

ê(2) = 0.80 - f(0.80)/f'(0.80)= 0.80 - (0.0142)/(-0.1026)= 0.9367

ê(3) can be obtained in a similar manner by solving the equation obtained in part (ii) for n = 100 using ê(2) as the starting value.

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The probability that they shoot free throws in alphabetical order depends on the number of players participating. If there are only two players, the probability would be 1/2 or 0.5.

Since each player has a different name, there are only two possible orders in which they can shoot free throws: alphabetical order or reverse alphabetical order. Out of these two possibilities, only one is the desired outcome (alphabetical order). Therefore, the probability of shooting free throws in alphabetical order is 1 out of 2, which can be expressed as 1/2 or 0.5

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Answer:

It’s a: The area of the floor would be a rational number

Step-by-step explanation:

Answer:

s = 4.5 cm

l= 3.75 cm

Step-by-step explanation:

Mathematically, when two shapes are similar, the ratio of their corresponding sides are equal

According to this rule, we have it that;

4/6 = 3/s

4 * s = 3 * 6

4s = 18

s = 18/4

s = 4.5 cm

Similarly;

4/5 = 3/l

4 * l = 5 * 3

4l = 15

l = 15/4

l = 3.75 cm

The reflection principle is used to find the number of paths for a simple random walk from a starting point S0 to a destination point S15, without hitting a specific line.

In this scenario, we want to find the number of paths for a random walk from S0 = 2 to S15 = 5, without crossing the line y = 2. We can use the reflection principle to simplify the problem.

The reflection principle states that if a path hits a specific line and goes below it, we can reflect the portion of the path below the line to create a new path above the line. This new path is symmetric to the original path.

In our case, the line y = 2 acts as the reflecting line. We reflect the portion of the paths that hit the line y = 2 above the line. By doing so, we transform the problem into finding the number of paths from S0 = 2 to S15 = 5, without crossing or touching the line y = 2.

Using the principles of combinatorics and counting, we can calculate the number of valid paths without hitting the line y = 2. This involves considering the number of steps taken in the positive and negative y-directions, while ensuring that the path remains above the line y = 2. The specific calculations and details would require a more extensive analysis of the random walk and its possible movements.

By applying the reflection principle and counting the valid paths, we can determine the number of paths for the given scenario.

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The appropriate classification for the hypothesis test in this scenario is a two-tailed test.

A two-tailed test is used when we are interested in determining whether the mean running time has changed as a result of the introduced change in the production method, without specifying a specific direction of the change. We want to consider the possibility of both an increase and a decrease in the mean running time.

The null hypothesis (H0) for a two-tailed test states that there is no change in the mean running time. In this case, it would be stated as: H0: μ = μ0, where μ represents the population mean running time and μ0 represents the hypothesized mean running time under the previous production method.

The alternative hypothesis (Ha) for a two-tailed test states that there is a change in the mean running time, either an increase or a decrease. In this case, it would be stated as: Ha: μ ≠ μ0, indicating that the mean running time is not equal to the hypothesized mean running time.

The claim being tested is that the mean running time has changed as a result of the introduced change in the production method. This claim is not specific to whether the mean running time has increased or decreased, but rather focuses on the existence of a change.

By classifying the hypothesis test as two-tailed, we are considering the possibility of both an increase and a decrease in the mean running time. This allows for a more comprehensive evaluation of the effects of the introduced change in the production method.

In summary, the hypothesis test is classified as a two-tailed test. The null hypothesis states that there is no change in the mean running time, while the alternative hypothesis states that there is a change, whether it is an increase or a decrease. The claim being tested is that the mean running time has changed as a result of the introduced change in the production method.

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Answer:

decrease by 10％ each year

Answer:

SHe has walked 2 more miles more than the other person

Step-by-step explanation:

Probability of selecting two defective cylinders ≈ 0.1071. Probability of selecting at least one defective cylinder ≈ 0.6429

To calculate the probability of selecting two defective cylinders when two cylinders are chosen at random without replacement, we need to consider the total number of cylinders and the number of defective cylinders. Given: Total number of cylinders: 8, Number of defective cylinders: 3. Probability of selecting two defective cylinders: To calculate this probability, we first need to determine the total number of ways to choose two cylinders out of the eight available. This can be calculated using the combination formula (nCr). Total ways to choose two cylinders out of eight: C(8, 2) = 8! / (2! * (8-2)!) = 28.

Next, we need to determine the number of ways to choose two defective cylinders out of the three available. Number of ways to choose two defective cylinders out of three: C(3, 2) = 3! / (2! * (3-2)!) = 3. Therefore, the probability of selecting two defective cylinders is: P(Two defective cylinders) = Number of ways to choose two defective cylinders / Total ways to choose two cylinders = 3/28 ≈ 0.1071 (rounded to four decimal places). Probability of selecting at least one defective cylinder: To calculate this probability, we can consider the complementary event, which is the probability of selecting no defective cylinders. Then, we subtract this probability from 1 to obtain the probability of selecting at least one defective cylinder.

Number of ways to choose two non-defective cylinders out of five remaining non-defective cylinders: C(5, 2) = 5! / (2! * (5-2)!) = 10. Total ways to choose two cylinders out of eight: C(8, 2) = 28 (as calculated earlier). Number of ways to choose at least one defective cylinder = Total ways to choose two cylinders - Number of ways to choose two non-defective cylinders= 28 - 10 = 18. Therefore, the probability of selecting at least one defective cylinder is: P(At least one defective cylinder) = Number of ways to choose at least one defective cylinder / Total ways to choose two cylinders= 18/28 ≈ 0.6429 (rounded to four decimal places).

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The correct answer is All employees in Lucia's company.

Lucia, a company director, randomly selected 30 employees to discover the proportion of employees who eat eggs in the morning.

Out of 30 employees, 22 of them consume eggs in the morning.

The population being studied in this scenario is D) All employees in Lucia's company.

What is the population in statistics?

The population, in statistical terms, refers to the entire set of data collected or available to researchers, which can be people, objects, measurements, or events, among other things.

It refers to a collection of individuals, objects, or events with at least one common feature of interest.

The population being investigated in a statistical study is the complete group of individuals, items, or objects that the researcher is interested in studying and drawing inferences from.

It is significant since it enables researchers to collect and analyze data to establish associations or inferences between groups, predict future results, and construct models.

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The matrix A for the linear transformation T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4} is:
[4 0 0]  
[0 4 0]
[0 0 4]
[0 0 0]
[0 0 0]

The given linear transformation is T: P2 → P4 and T(p) = 4x2p.

We are to find the matrix A for T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4}.

Consider the linear transformation of each element of the first basis B.

We have; T(1) = 4x2(1) = 4x2(1) + 0x3 + 0x4 = 4x2T(x) = 4x2(x) = 0x2 + 4x3 + 0x4 = 0x2 + 4x3T(x^2) = 4x2(x^2) = 0x2 + 0x3 + 4x4 = 0x2 + 0x3 + 4x4

Thus, the matrix of T relative to B is: [4 0 0] [0 4 0] [0 0 4]

Next, we will find the coordinates of each element of the basis B' under the basis B.

Using the relations;x3 = x3x^3 = x2.x   [x3]B = [0 0 1]T(x^3) = 4x2(x^3) = 0x2 + 0x3 + 0x4 = 0x2 + 0x3 + 0x4

Thus, the coordinate vector of x3 relative to B is [0 0 1].

Using the relation; x4 = x4 - x3x^4 = x^4 - x2.x   [x4]B = [0 -1 0]T(x^4) = 4x2(x^4) = 0x2 + 0x3 + 4x4 = 0x2 + 0x3 + 4(x3 + x2.x) = 0x2 + 0x3 + 4x3 + 0x2 = 0x2 + 4x3

Thus, the coordinate vector of x4 relative to B is [0 -1 0].

Thus, the matrix of T relative to B' is [4 0 0 0 0] [0 4 0 0 0] [0 0 4 0 0] [0 0 0 0 0] [0 0 0 0 0]

Therefore, the matrix A for T relative to the bases B = {1, x, x2} and B' = {1, x, x2, x3, x4} is: [4 0 0] [0 4 0] [0 0 4] [0 0 0] [0 0 0].

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Answer:

Aceleracion = 0.8 m/s²

Step-by-step explanation:

Dados los siguientes datos;

Velocidad inicial = 12 m/s

Velocidad final = 20 m/s

Tiempo, t = 10 segundos

Para encontrar la aceleración;

Aceleración se puede definir como la tasa de cambio de la velocidad de un objeto con respecto al tiempo.

Esto simplemente significa que la aceleración viene dada por la resta de la velocidad inicial de la velocidad final a lo largo del tiempo.

Por lo tanto, si restamos la velocidad inicial de la velocidad final y la dividimos por el tiempo, podemos calcular la aceleración de un objeto. Matemáticamente, la aceleración viene dada por la fórmula;

[tex] Aceleracion = \frac{final \; velocidad - inicial \; velocidad}{tiempo}[/tex]

Sustituyendo en la fórmula, tenemos;

[tex] Aceleracion = \frac{20 - 12}{10}[/tex]

[tex] Aceleracion = \frac{8}{10}[/tex]

Aceleracion = 0.8 m/s²