Tina is standing at the bottom of a hill. Matt is standing on the hill so that when Tina's line of sight is


perpendicular to her body, she is looking at Matt's shoes.


a. If Tina's eyes are 5 feet from the ground and 14. 5 feet from Matt's shoes, what is the angle of elevation of


the hill to the nearest degree? Explain.

The smallest number of cars that can tailgate in a straight line at a football game is 12.

This can be determined by using a logic puzzle approach. If one car is in front of four cars, that's 5 cars in total. If one car is behind four cars, that's also 5 cars. And if three cars are between two cars, that's 5 sets of two cars, or 10 cars. The total number of cars is 5 + 5 + 10 = 20.

This logic puzzle approach is not applicable for all problems. It depends on the specific problem and the information provided. For certain types of problems, a different approach or method may be more appropriate. The key is to understand the problem, determine the necessary information, and select the best approach to solve it.

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Answer:

Step-by-step explanation:

To find a quadratic equation with only two points, we need to use the general form of a quadratic equation, which is:

y = ax² + bx + c

where "a", "b", and "c" are constants and "x" and "y" are the variables.

Let's say the two points we have are (x₁, y₁) and (x₂, y₂). We can use these points to form two equations:

y₁ = ax₁² + bx₁ + c

y₂ = ax₂² + bx₂ + c

We can then solve these equations simultaneously to find the values of "a", "b", and "c".

Here's an example:

Suppose we have two points, (2, 3) and (4, 7), and we want to find the quadratic equation that passes through these two points.

We start by plugging in the coordinates of the first point into the equation:

3 = a(2²) + b(2) + c

Next, we plug in the coordinates of the second point:

7 = a(4²) + b(4) + c

We now have two equations with three variables. To solve for the variables, we can use either substitution or elimination. Here, we will use substitution:

From the first equation, we can solve for "c":

c = 3 - 4a - 2b

We can now substitute this expression for "c" into the second equation:

7 = a(4²) + b(4) + 3 - 4a - 2b

Simplifying the equation, we get:

7 = 16a - 2b + 3

4a - b = 2

We can now use the first equation to solve for "b" in terms of "a" and substitute this expression into the second equation:

b = (3 - 4a - c) / 2

b = (3 - 4a - (3 - 4a - 2b)) / 2

b = -2a + 3

Substituting this expression for "b" into the equation 4a - b = 2, we get:

4a - (-2a + 3) = 2

6a = -1

a = -1/6

Now that we know "a", we can use one of the earlier equations to solve for "c":

3 = (-1/6)(2²) + b(2) + c

c = 3 + (1/3) - b

Substituting this expression for "c" into the equation c = 3 - 4a - 2b, we get:

3 + (1/3) - b = 3 + (2/3) + 2(-2a + 3)

b = -2a + (8/3) = 1/3

Therefore, the quadratic equation that passes through the points (2, 3) and (4, 7) is:

y = (-1/6)x² + (1/3)x + 1

Note that this is a quadratic equation in standard form. If we want to write it in vertex form, we can complete the square.

The length and width of the plot that will maximize the area is 60 meters by 120 meters. The largest area that can be enclosed is 7200 square meters.

To maximize the area of the rectangular plot, Farmer Ed must use 9000 meters of fencing to enclose three sides of the plot. The fourth side, which borders the river, does not need to be fenced. To find the length and width of the plot that will maximize the area, the 9000 meters of fencing must be divided into two sides of equal length, with the remaining fencing used for the third side. This results in a length of 120 meters and a width of 60 meters, which maximizes the area of the plot. The largest area that can be enclosed is 7200 square meters.

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The function is :\(y(x) = 10 + (7/8) e^8x + (97/72) e^9x + 3 e^x\)

To find y as a function of x, we need to solve the differential equation:

\(y′′′ − 17y′′ + 72y′ = 168e^x\)

Step 1: Find the characteristic equation

\(r^3 - 17r^2 + 72r = 0\)

Factor out r:

\(r(r^2 - 17r + 72) = 0\)

Factor the quadratic:

r(r - 8)(r - 9) = 0

So the roots are:

r₁ = 0, r₂ = 8, r₃ = 9

Step 2: Find the general solution

The general solution will be of the form:

\(y(x) = C1 + C2e^8x + C3e^9x + y_p(x)\)

where y_p(x) is a particular solution to the non-homogeneous equation.

Step 3: Find the particular solution

We can use the method of undetermined coefficients to find a particular solution. Since the right-hand side is an exponential function, we can guess that the particular solution is also an exponential function:

\(y_p(x) = A e^x\)

\(y_p′(x) = A e^x\)

\(y_p′′(x) = A e^x\)

\(y_p′′′(x) = A e^x\)

Substituting into the differential equation:

\(A e^x - 17A e^x + 72A e^x = 168 e^x\)

Simplifying:

\(56A e^x = 168 e^x\)

A = 3

So the particular solution is:

\(y_p(x) = 3 e^x\)

Step 4: Find the constants using initial conditions

y(0) = C₁ + C₂ + C₃ + 3 = 16

y′(0) = 8C₂ + 9C₃ + 3 = 23

\(y′′(0) = 8^2 C2 + 9^2 C3 = 24\)

Solving for the constants, we get:

C₁ = 10, C₂ = 7/8, C₃ = 97/72

Step 5: Write the final solution

Substituting the constants and the particular solution into the general solution, we get:

\(y(x) = 10 + (7/8) e^8x + (97/72) e^9x + 3 e^x\)

So the function y(x) is:

\(y(x) = 10 + (7/8) e^8x + (97/72) e^9x + 3 e^x\)

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Step-by-step explanation:

please mark me as brainlest

The probability of being dealt a card that helps improve your hand is approximately 0.7083 or 70.83%.

To determine the probability of being dealt a card that helps you, we need to consider two factors: the number of favorable cards remaining in the deck and the total number of cards remaining.

At the beginning of the game, a standard deck of 52 cards is used. After the initial deal, you and the dealer have each received two cards, so there are 52 - 4 = 48 cards remaining in the deck.

To determine the favorable cards that can improve your hand, we need to consider the possible outcomes. In this case, you have a four and a jack, and you're looking to improve your hand. Let's analyze the possibilities:

To improve your hand to a better total without exceeding 21, you would want to draw a card between 5 and 9, inclusive. In this case, you know that the dealer's one visible card is a five, so there are three more fives left in the deck. Additionally, there are four cards each of 6, 7, 8, and 9, making a total of 16 favorable cards for improving your hand.

Lastly, if you're hoping to improve your hand to a total of 20, you would need an ace, 2, 3, or 6. As we assumed earlier, there are three aces remaining, and there are four cards each of 2, 3, and 6. So there are a total of 3 + 4 + 4 + 4 = 15 favorable cards for achieving a total of 20.

In total, the number of favorable cards for improving your hand is 3 (blackjack) + 16 (improve to a better total) + 15 (improve to a total of 20) = 34.

To calculate the probability, we divide the number of favorable cards by the total number of cards remaining in the deck:

Probability = Number of Favorable Cards / Total Number of Cards Remaining

Probability = 34 / 48

Simplifying the fraction, the probability is 17 / 24 or approximately 0.7083 (rounded to four decimal places).

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Answer:

Step-by-step explanation:

a2 = -2

a + d  =  -2

a + 6d = 7

-5d  =  -9

d  =  9/5

a = -19/5

an  =  a + ( n-1 ) d

an =  -19/5 + ( n-1 ) 9/5

Answer:

C.) they both got two more questions correctly than Chris did.

Step-by-step explanation:

brian got 14 more right than chris

Chris got 2 less than damien

brian and damien both got 18/20 correct, chris got 16 right

Using for loop we can get the syntax in order to calculate this arithmetic series n is A = 0.9999.

An arithmetic collection is the sum of the phrases in an mathematics collection with a specific range of phrases. Following is a easy system for locating the sum: Formula 1: If S n represents the sum of an mathematics collection with phrases.

This system calls for the values of the primary and ultimate phrases and the range of phrases. Finite Sequence- Finite sequences have countable phrases and do now no longer cross as much as infinity. An instance of a finite mathematics collection is 2, 4, 6, 8. Infinite Sequence- Infinite arithmetic collection is the collection wherein phrases cross as much as infinity.

using while

A=0; n=1; while n<=10000     A=A+(1/(n*(n+1)));     n=n+1; end A

output

A =      0.9999

using  for loop

A=0; for n=1:10000     A=A+(1/(n*(n+1))); end A

output

A = 0.9999.

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Correct Question:

The following series can be written with a shorthand form of sigma notation (A). Use while syntax to calculate this arithmetic series:

1/(1x2) + 1/(2x3) + 1/(3x4) + ... + 1/(n x (n+1)) ......1

A = ∈ 1/(n x (n+1)); n=[1:10000]

Answer:

radius: 8.4   Area: 221.5

Step-by-step explanation:

16.8 divided by 2 = 8.4

The Formula for finding the area of circle= R₂ (radius squared: radius times itself) Multiplied by pi. pi = 3.14

8.4₂ multiplied by 3.14

Answer:

x=11

Step-by-step explanation:

(3x-21) + (x-4)=19

3x-21+x-4=19

4x-25=19

4x=44

x=11

We have: ∫(dx)/(sqrt(9 - x^2)) = -arcsin(x/3) + C. In Example 6, we have the integral:

∫(dx)/(sqrt(9 - x^2))

To use the trigonometric substitution x = a sinθ, we need to rewrite the integral in terms of θ. We can use the identity:

sin^2θ + cos^2θ = 1

to get:

cos^2θ = 1 - sin^2θ

Multiplying both sides by a^2, we get:

a^2cos^2θ = a^2 - a^2sin^2θ

Substituting x = a sinθ, we have:

x^2 = a^2sin^2θ

a^2 - x^2 = a^2cos^2θ

Substituting these into the original integral, we get:

∫(dx)/(sqrt(9 - x^2)) = ∫(dx)/(sqrt(9 - a^2sin^2θ)) = ∫(cosθ)/(sqrt(9 - a^2cos^2θ)) dθ

Now, we can use the substitution u = a cosθ, which gives:

du/dθ = -a sinθ

dθ = -(1/a)du/sinθ

Substituting this into the integral, we get:

∫(cosθ)/(sqrt(9 - a^2cos^2θ)) dθ = -∫(1/sqrt(9 - u^2)) du

This is the same integral as in Example 6, and we can evaluate it using the arcsine substitution. Thus, we get:

-∫(1/sqrt(9 - u^2)) du = -arcsin(u/3) + C

Substituting back u = a cosθ, we get:

-∫(1/sqrt(9 - a^2cos^2θ)) dθ = -arcsin(a cosθ/3) + C

Using the identity:

sin(arcsin(x)) = x

we can simplify the expression:

-arcsin(a cosθ/3) = -arcsin(x/3)

Thus, we have:

∫(dx)/(sqrt(9 - x^2)) = -arcsin(x/3) + C

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Answer:

AEC and DEB are two pairs of supplementary angles.

Answer:

D

Step-by-step explanation:

The probability that at least one child among the 3 will get influenza is 0.07.

The area of mathematics known as probability deals with numerical descriptions of how likely it is for an event to happen or for a claim to be true. A number between 0 and 1 is the probability of an event, where, broadly speaking, 0 denotes the event's impossibility and 1 denotes its certainty.

Given,

3 children in a village ages 3, 5, and 7 are vaccinated with the qiv vaccine

Probability that at least one child among the 3 will get influenza

P(at least one child among the 3 will get influenza)

=  1 - P(no one have influenza)

= 1 - (1- 0.0.378)(1-0.017)²

= 1 - 0.93

= 0.07

The probability that at least one child among the 3 will get influenza is 0.07.

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Answer:

162 cents

Step-by-step explanation:

to find the rate you want take the inches which is 21 and divide by the cost which is 63 and find the rate is 3 which means each inch costs 3 cents and so with that rate you can find the cost of 54 inchest of wire so you take your 54 inches and multiply it by 3 which is the cents per inch and find that 54 inches of wire would cost you 162 cents

A causal LTI system S whose input x[n] and output y[n] are related by the difference equation S can be represented as S = S2(S1(x[n])).

A system S can be considered a cascade connection of two causal LTI systems, S1 and S2, that the input-output relationship of S is equivalent to the input-output relationship of S1 followed by S2.

The given difference equation for S:

2y[n] - y[n-1] + y[n-3] = x[n] - 5x[n-4]

Now, this equation in terms of the output of S1 (y1[n]) and the input of S2 (x2[n]):

2y1[n] - y1[n-1] + y1[n-3] = y2[n] - 5y2[n-4]

The output of S1 (y1[n]) corresponds to the input of S2 (x2[n]) and the output of S2 (y2[n]) corresponds to the output of the overall system (y[n]). this equation with the input-output relationship of S1 and S2:

S1: 2y1[n] = x1[n] - 5x1[n-4]

S2: y2[n] = 1/2y2[n-1] - 1/2y2[n-3] + x2[n]

Comparing the equations:

x1[n] - 5x1[n-4] = y2[n] - 5y2[n-4]

The given system S can indeed be considered a cascade connection of two causal LTI systems S1 and S2 with the specified input-output relationships.

S1:System S1 has an input x1[n] and output y1[n] related by 2y1[n] = x1[n] - 5x1[n-4].

S2:System S2 has an input x2[n] and output y2[n] related by y2[n] = 1/2y2[n-1] - 1/2y2[n-3] + x2[n].

And the overall system S can be represented as S = S2(S1(x[n])).

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Answer:

Sam

Step-by-step explanation:

Just put them into a fraction for Sam its 15/5 and the ratio (basically a fraction) is 3/1 so if you simplify the 15/5, you’d get 3/1 which is the same so sam’s ratio is corrrect

Answer:

When we say that an allele is "fixed," it means that a particular allele has reached a frequency of 100% in a population.

Step-by-step explanation:

Alleles are different forms of a gene that occupy the same position on homologous chromosomes. In a population, different alleles can exist for a specific gene. However, through various evolutionary processes such as natural selection, genetic drift, or gene flow, one allele may become predominant and eventually fixate within the population.

The fixation of an allele can occur through different mechanisms. For example, if a beneficial allele provides a selective advantage to individuals carrying it, it is more likely to increase in frequency and eventually become fixed in the population. On the other hand, genetic drift, which is the random change in allele frequencies due to chance events, can also lead to the fixation of an allele, especially in small populations.

Once an allele is fixed in a population, it means that all future generations will inherit that allele, and no alternative alleles will be present at that particular gene locus.

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Answer:

B

Step-by-step explanation:

I like to put inequalities in the form y ≤ or ≥ mx + b. It makes it easier to tell which line is which on the graph. So x + y ≥ 5 can be put as y ≥ -x + 5. Next, -3x + 2y ≤ -2 can be put as 2y ≤ 3x -2. This should be further simplified by dividing everything by 2, isolating y, to y ≤ \(\frac{3}{2}\)x - 1. Now since b is the y-intercept, we know that the first inequality (x + y ≥ 5 or y ≥ -x + 5) is represented by the blue line. Since in the new form we put that inequality in (y ≥ -x + 5), the shaded part must be above the line. This means that graph A is not the answer. Next, we know that the second inequality is represented by the red line. We also know that the shaded part must be below the line, which means that it cannot be graph C. It is cannot be graph D because the shaded parts must satisfy BOTH (hence, AND. If it said OR it would be different) inequalities. This leaves B as the only correct answer.

Answer:

djal3924+3820&282039392#283830818385

Answer:

y=0.07x + 12.95

Step-by-step explanation:

The given series is equivalent to the Bessel function \(\(J_0(x)\\)).

The given series is:

\(\[ \sum_{n=1}^{\infty} \frac{(n!)^2 x^n}{(2n)!} \]\)

Let's analyze this series step by step:

1. The term \(\((n!)^2\)\) in the numerator represents the square of the factorial of n. It is the product of all positive integers from 1 to n, squared.

2. The term \(\(x^n\)\) represents the variable x raised to the power of n.

3. The term \(\((2n)!\)\) in the denominator represents the factorial of 2n. It is the product of all positive integers from 1 to 2n.

The series starts from n = 1 and goes to infinity.

The given series represents a special function called the Bessel function. Specifically, it represents the Bessel function of the first kind and of order 0, denoted as \(\(J_0(x)\)\). The Bessel functions are important in various areas of physics and engineering, particularly in solving problems involving wave propagation, heat conduction, and oscillatory phenomena.

So, the given series is equivalent to the Bessel function \(\(J_0(x)\)\).

Complete Question:

\(\[ \sum_{n=1}^{\infty} \frac{(n!)^2 x^n}{(2n)!} \]\)

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Step-by-step explanation:

there is a (V.O.A) which is vertically opposite angle

and in the given part he himself said that there is two = angles

so there for the two third angles are equal

so we can say they are similar

Based on the points obtained from Ali and Habib in the game we can say that:

a. The difference between their scores is: 38.812

b. The estimate is not enough to find the winner

To solve this problem we must perform the following algebraic operations with the given information

Information about the problem:

To calculate the difference between their scores we must perform the following subtraction:

Difference between their scores = Al's points - Habib's points

Difference between their scores = 43.901 - 5.089

Difference between their scores =38.812

Due to the difference between their scores is less than 40,000 points, it is not yet possible to define a winner.

We can say that they are the set of numbers and symbols that are related by the different mathematical operation signs such as addition, subtraction, multiplication, division among others.

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The statements that accurately compare the domain and range of the functions are: The domain of all three functions is all real numbers, The range of f(x) and h(x) is all real numbers, but the range of g(x) is all real numbers except 0.

The domain of a function refers to all the possible input values that make the function defined. For the quadratic function f(x) = 3x², there are no values of x that make the function undefined, so the domain of f(x) is all real numbers. To find the range of f(x), we can calculate the vertex of the parabola, which is (0,0). The range of f(x) is all real numbers greater than or equal to zero.

For the quadratic function f(x) = 3x², we can determine the vertex using the formula:

h = -b/2a

In this case, a = 3 and b = 0, so:

h = -0/2(3) = 0/6 = 0

The x-coordinate of the vertex is 0.

To find the y-coordinate, we evaluate the function at the vertex:

f(0) = 3(0)² = 0

So the vertex of the parabola is (0,0).

Since the coefficient of x² is positive, the parabola opens upwards, and the minimum value of the function occurs at the vertex. Therefore, the range of f(x) is all real numbers greater than or equal to zero.

For the rational function g(x) = 1/3x, the function is undefined when the denominator is equal to zero. Thus, we solve for the value(s) of x that make the denominator zero:

3x = 0

Dividing both sides by 3, we get:

x = 0

Therefore, the domain of g(x) is all real numbers except zero. The range of g(x) is all real numbers except zero, since the function cannot equal zero.

For the linear function h(x) = 3x, there are no values of x that make the function undefined. Thus, the domain of h(x) is all real numbers. The range of h(x) is also all real numbers, since the function is a straight line that passes through the origin and extend infinitely in both directions.

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The inverse Laplace transform of F(s) is:

\(f(t) = e^(-t)\)

To find the inverse Laplace transform of each function, we need to express them in terms of known Laplace transforms. Here are the solutions for each function:

a)

\(F(s) = \large{\frac{2e^{-0.5s}}{s^2-6s+9}}\)

To find the inverse Laplace transform, we first need to factor the denominator of F(s). The denominator factors as (s - 3)². Therefore, we can rewrite F(s) as:

\(F(s) = \large{\frac{2e^{-0.5s}}{(s-3)^2}}\)

Now, we know that the Laplace transform of eᵃᵗ is 1/(s - a). Therefore, the inverse Laplace transform of

\(e^(-0.5s) \: is \: e^(0.5t).\)

Applying this, we get:

\(f(t) = 2e^(0.5t) * t \\

b) F(s) = \large{\frac{s-1}{s^2-3s+2}}\)

We can factor the denominator of F(s) as (s - 1)(s - 2). Now, we rewrite F(s) as:

\(F(s) = \large{\frac{s-1}{(s-1)(s-2)}}\)

Simplifying, we have:

\(F(s) = \large{\frac{1}{s-2}}\)

The Laplace transform of 1 is 1/s. Therefore, the inverse Laplace transform of F(s) is:

\(f(t) = e^(2t) \\

c) F(s) = \large{\frac{s-1}{s^2+s-2}}

\)

We factor the denominator of F(s) as (s - 1)(s + 2). The expression becomes:

\(F(s) = \large{\frac{s-1}{(s-1)(s+2)}}\)

Canceling out the (s - 1) terms, we have:

\(F(s) = \large{\frac{1}{s+2}}\)

The Laplace transform of 1 is 1/s. Therefore, the inverse Laplace transform of F(s) is:

\(f(t) = e^(-2t) \\

d) F(s) = \large{\frac{e^{-s}(s-1)}{s^2+s-2}}\)

We can factor the denominator of F(s) as (s - 1)(s + 2). Now, we rewrite F(s) as:

\(F(s) = \large{\frac{e^{-s}(s-1)}{(s-1)(s+2)}}\)

Canceling out the (s - 1) terms, we have:

\(F(s) = \large{\frac{e^{-s}}{s+2}}\)

The Laplace transform of

\(e^(-s) \: is \: 1/(s + 1).\)

Therefore, the inverse Laplace transform of F(s) is:

\(f(t) = e^(-t)\)

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Answer:

d. All of the above are true

Step-by-step explanation:

All of the following statements,

a. the stronger the linear relationship between two variables, the closer the correlation coefficient will be to 1.

b. if the correlation coefficient between the x and y variables is negative, the sign on the regression slope will also be negative.

C. if the correlation coefficient between the dependent and independent variable is determined to be significant, the regression model for y given x will also be significant.

are true