Mrs. Young has p goats and q cows on his farm. He has 23 fewer cows than goats.

What are the missing values in the table?

PLSSSS QUICK

Yes, there is an analogous ratio between 2:1 and 20:10.

We just cancel by a common factor. So 4:2=2:1 . The simplest representation of the ratio 4 to 2 is the ratio 2 to 1. Also, since each pair of numbers has the same relationship to one another, the ratios are equivalent.

By dividing the terms of each ratio by their greatest common factor, we may simplify both ratios to explain why.

As the greatest common factor for the ratio 2:1 is 1, additional simplification is not necessary.

The greatest common factor for the ratio 20:10 is 10. When we multiply both terms by 10, we get:

20 ÷ 10 : 10 ÷ 10

= 2 : 1

As a result, both ratios have the same reduced form, 2:1, making them equal.

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The surface area of the surface generated by rotating the curve y = x² about the y-axis, and we found that the surface area is approximately 54.33 square units.

In this case, the curve we want to rotate is y = x², and we want to rotate it about the y-axis. To use the formula above, we need to express the equation of the curve in terms of x. Therefore, we need to rewrite y = x² as x = √y.

Next, we need to find the derivative of x = √y with respect to y, which is:

dx/dy = 1/2√y

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

Surface Area = 2π ∫[0,4] √y √(1+(1/2√y)²) dy

Simplifying the expression inside the square root, we get:

Surface Area = 2π ∫[0,4] √(y+(1/4)) dy

We can evaluate this integral using the power rule of integration, which gives:

Surface Area = 2π [2/3(y+(1/4))^(3/2)]₀⁴

Simplifying further, we get:

Surface Area = 2π [2/3(17/4)^(3/2)]

Surface Area ≈ 54.33 square units

Therefore, the surface area of the surface generated by rotating the curve y = x² about the y-axis is approximately 54.33 square units.

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Complete Question:

How do you find the area of the surface generated by rotating the curve about the y-axis y = x^2 , 0 ≤ x ≤ 2 ?

10% chance of contamination by a particular: It's not clear what random variable is being referred to here, but if it's the probability of contamination.

In general terms, a distribution refers to the way something is divided or spread out. In the context of statistics and probability theory, a distribution is a mathematical function that describes the likelihood of different possible outcomes or values that a variable can take.

There are various types of distributions, but some of the most commonly used ones include:

Normal distribution: also known as the Gaussian distribution, it is a continuous probability distribution that is symmetrical around the mean, with most of the data falling within one standard deviation of the mean.

Binomial distribution: this is a discrete probability distribution that describes the likelihood of a certain number of successes in a fixed number of trials.

Poisson distribution: another discrete probability distribution that describes the likelihood of a certain number of events occurring in a fixed interval of time or space.

Exponential distribution: a continuous probability distribution that describes the time between events occurring at a constant rate.

Distributions are essential in statistical analysis as they can help to understand and analyze data, make predictions, and draw conclusions about a population based on a sample of data.

Given by the question.

a. The wages of academician and non-academician workers in UPSI: This random variable can be approximated to a continuous distribution as wages can take on any numerical value within a range. However, it's worth noting that in practice, there may be discrete intervals or categories of wages, in which case a discrete distribution may be more appropriate.

b. The time taken to submit online quiz answer's document: This random variable can also be approximated to a continuous distribution as it can take on any numerical value within a range.

c. The prices of SAMSUNG mobile phones displayed at a phone shop: This random variable can be approximated to a continuous distribution as prices can take on any numerical value within a range.

d. The number of pumps at Shell petrol stations in Perak: This random variable can be approximated to a discrete distribution since the number of pumps can only take on integer values.

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Answer: The answer is D

Step-by-step explanation:

Pythagorean theorem: a²+b²=c²

x²+x²=14²

2x²=196

Evaluate...

x=7√2

The probability that, out of 18 independent samples received from one river, just two were contaminated is 0.8438.

In the following 18 samples to be evaluated,

Let X = the number of samples that now the pollutant is present in.

Thus, with p = 0.10 and n = 18, X is a binomial random variable.

Using the binomial theorem:

[tex](^{n} _{r} ) p^{x} q^{n-x}[/tex]

p = 0.10

q = 1 - 0.10 = 0.9

n = 18

The likelihood that only two samples out of 18 obtained in different ways from the same river were polluted

P(x = 2) = [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{18-2}[/tex]

=  [tex](^{18} _{2} ) (0.1)^{2} (0.9)^{16}[/tex]

= 153 x 0.01 x 0.1853

= 0.8438

Thus, the probability that, out of 18 separate samples received from one river, just two were contaminated is 0.8438.

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Answer:

One possible function that models the ocean tide is:

h(t) = A sin(ωt + φ) + B

where:

h(t) represents the height of the tide (in meters) at time t (in hours)

A is the amplitude of the tide (in meters)

ω is the angular frequency of the tide (in radians per hour)

φ is the phase shift of the tide (in radians)

B is the mean sea level (in meters)

This function is a sinusoidal function, which is a common type of function used to model periodic phenomena. The sine function has a natural connection to circles and periodic motion, making it a good choice for modeling the regular rise and fall of ocean tides.

The amplitude A represents the maximum height of the tide above the mean sea level, while B represents the mean sea level. The angular frequency ω determines the rate at which the tide oscillates, with one full cycle (i.e., a high tide and a low tide) occurring every 12 hours. The phase shift φ determines the starting point of the tide cycle, with a value of zero indicating that the tide is at its highest point at time t=0.

To determine specific values for A, ω, φ, and B, we would need to gather data on the tide height at various times and locations. However, typical values for these parameters might be:

1. A = 2 meters (representing a relatively large tidal range)

2. ω = π/6 radians per hour (corresponding to a 12-hour period)

3. φ = 0 radians (assuming that high tide occurs at t=0)

4. B = 0 meters (assuming a mean sea level of zero)

Using these values, we can write the equation for the tide as:

h(t) = 2 sin(π/6 t)

We can evaluate this equation for various values of t to get the height of the tide at different times. For example, at t=0 (the start of the cycle), we have:

h(0) = 2 sin(0) = 0

indicating that the tide is at its lowest point. At t=6 (halfway through the cycle), we have:

h(6) = 2 sin(π/2) = 2

indicating that the tide is at its highest point. We can also graph the function to visualize the rise and fall of the tide over time:

Tide Graph

Overall, this function provides a simple and effective way to model the ocean tide using trigonometric functions.

(please mark my answer as brainliest)

Answer:

8(3/4y - 2) + 6(-1/2x + 4) + 1 can be simplified as:

6(-1/2x) = -3x

8(3/4y) = 6y

8(-2) = -16

6(4) = 24

1 remains as 1.

So the expression becomes:

6y - 3x - 16 + 24 + 1

which simplifies to:

6y - 3x + 9

Therefore, the expressions that are equivalent to 8(3/4y - 2) + 6(-1/2x + 4) + 1 are:

6y - 3x + 9

The direction angle of the electric field at the point (x = 1.10 m, y = 0.400 m, z = 0) is approximately 74.5 degrees clockwise from the positive x-axis in the xy plane.

To calculate the electric field at the point (x = 1.10 m, y = 0.400 m, z = 0), we need to take the negative gradient of the electric potential V:

E = -∇V

where ∇ is the del operator, which is given by:

∇ = i(∂/∂x) + j(∂/∂y) + k(∂/∂z)

and i, j, k are the unit vectors in the x, y, and z directions, respectively.

To calculate the magnitude of the electric field at the point, we first need to find the partial derivatives of V with respect to x and y:

∂V/∂x = 2Axy - By^2

∂V/∂y = Ax^2 - 2Bxy

Substituting the values of A, B, x, and y, we get:

∂V/∂x = 2(5.00 V/m^3)(1.10 m)(0.400 m) - (8.00 V/m^3)(0.400 m)^2 = 0.44 V/m

∂V/∂y = (5.00 V/m^3)(1.10 m)^2 - 2(8.00 V/m^3)(1.10 m)(0.400 m) = -1.64 V/m

Next, we can calculate the magnitude of the electric field:

E = -∇V = -i(∂V/∂x) - j(∂V/∂y) - k(∂V/∂z)

= -i(0.44 V/m) + j(1.64 V/m) + 0k

= (0.44 i - 1.64 j) V/m

The magnitude of the electric field is given by:

|E| = sqrt((0.44 V/m)^2 + (-1.64 V/m)^2) = 1.70 V/m

Therefore, the magnitude of the electric field at the point (x = 1.10 m, y = 0.400 m, z = 0) is 1.70 V/m.

To calculate the direction angle of the electric field, we need to find the angle that the electric field vector makes with the positive x-axis in the xy plane.

The angle can be found using the arctan function:

θ = arctan(Ey/Ex)

Substituting the values of Ex and Ey, we get:

θ = arctan(-1.64 V/m / 0.44 V/m) = -1.30 radians

The negative sign indicates that the direction angle is measured counter clockwise from the negative x-axis, which is equivalent to measuring clockwise from the positive x-axis.

Converting to degrees, we get:

θ = -1.30 radians * (180 degrees / pi radians) = -74.5 degrees

Therefore, the direction angle is approximately 74.5 degrees clockwise in the xy plane.

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Answer:

We have two equations:

√x +2y^2 = 15 ----(1)

√4x - 4y^2=6 ----(2)

Let's solve for x:

From (1), we have:

√x = 15 - 2y^2

Squaring both sides, we get:

x = (15 - 2y^2)^2

Expanding, we get:

x = 225 - 60y^2 + 4y^4

From (2), we have:

√4x = 6 + 4y^2

Squaring both sides, we get:

4x = (6 + 4y^2)^2

Expanding, we get:

4x = 36 + 48y^2 + 16y^4

Substituting the expression for x from equation (1), we get:

4(225 - 60y^2 + 4y^4) = 36 + 48y^2 + 16y^4

Simplifying, we get:

900 - 240y^2 + 16y^4 = 9 + 12y^2 + 4y^4

Rearranging, we get:

12y^2 - 12y^4 = 891

Dividing both sides by 12y^2, we get:

1 - y^2 = 74.25/(y^2)

Multiplying both sides by y^2, we get:

y^2 - y^4 = 74.25

Let z = y^2. Substituting, we get:

z - z^2 = 74.25

Rearranging, we get:

z^2 - z + 74.25 = 0

Using the quadratic formula, we get:

z = (1 ± √(1 - 4(1)(74.25))) / 2

z = (1 ± √(-295)) / 2

Since the square root of a negative number is not real, there are no real solutions for z, which means there are no real solutions for y and x.

Therefore, the answer is "no solution".

Answer:

To find the Z-score that has 48.4% of the distribution's area to its left, we can use a standard normal distribution table or a calculator.

Using a standard normal distribution table, we can look for the closest probability value to 0.484. In the table, we find that the closest probability value is 0.4838, which corresponds to a Z-score of approximately 1.96.

Alternatively, we can use a calculator with a built-in normal distribution function. Using the inverse normal distribution function, also known as the quantile function, we can find the Z-score that corresponds to a given probability. For a probability of 0.484, we get:

To find the Z-score that has 48.4% of the distribution's area to its left, we can use a standard normal distribution table or a calculator.

Using a standard normal distribution table, we can look for the closest probability value to 0.484. In the table, we find that the closest probability value is 0.4838, which corresponds to a Z-score of approximately 1.96.

Alternatively, we can use a calculator with a built-in normal distribution function. Using the inverse normal distribution function, also known as the quantile function, we can find the Z-score that corresponds to a given probability. For a probability of 0.484, we get:

To find the Z-score that has 48.4% of the distribution's area to its left, we can use a standard normal distribution table or a calculator.

Using a standard normal distribution table, we can look for the closest probability value to 0.484. In the table, we find that the closest probability value is 0.4838, which corresponds to a Z-score of approximately 1.96.

Alternatively, we can use a calculator with a built-in normal distribution function. Using the inverse normal distribution function, also known as the quantile function, we can find the Z-score that corresponds to a given probability. For a probability of 0.484, we get:

invNorm(0.484)= 1.96

Therefore, the Z-score that has 48.4% of the distribution's area to its left is approximately 1.96.

The given statement "Star clusters with lots of bright, blue stars of spectral type O and B are generally younger than clusters that don't have any such stars." is True. The reason for this is that O and B stars are short-lived and burn through their fuel quickly.

The reason for this is that O and B stars burn through their fuel quickly, causing them to exhaust their nuclear fuel and end their lives in a relatively short period, typically within a few tens of millions of years.

On the other hand, stars of lower mass and cooler temperatures, like G and K type stars like our sun, have longer lifetimes and take billions of years to exhaust their nuclear fuel.

Therefore, clusters without any bright, blue stars are likely to have evolved for longer periods, allowing these short-lived stars to have already expired.

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Answer:

Let's assume that Mr. Nkalle invested an amount of x in Scheme A and (20900 - x) in Scheme B.

The simple interest earned on Scheme A in 2 years would be:

SI(A) = (x * 9 * 2)/100 = 0.18x

The simple interest earned on Scheme B in 2 years would be:

SI(B) = [(20900 - x) * 8 * 2]/100 = (3344 - 0.16x)

The total simple interest earned in 2 years is given as N$3508:

SI(A) + SI(B) = 0.18x + (3344 - 0.16x) = 3508

0.02x = 164

x = 8200

Therefore, Mr. Nkalle invested N$8200 in Scheme A and N$12700 (20900 - 8200) in Scheme B. So the amount invested in Scheme B was N$12700.

According to the question. A. No punctuation is missing.

Punctuation is the use of symbols to indicate the structure and organization of written language. It is used to help make the meaning of sentences clearer and to make them easier to read and understand. Punctuation marks can also be used to indicate pauses in speech, to create emphasis, and to indicate the speaker’s attitude. There are many different types of punctuation marks, each with its own purpose. The most commonly used punctuation marks are the period, comma, question mark, exclamation mark, quotation marks, and the apostrophe.

Quotation marks are used to enclose quoted material, while the apostrophe is used to indicate possession or to replace missing letters in a word or phrase. By using punctuation correctly, writers can ensure that their messages are correctly understood by their readers.

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Sin 30 =3/x

1/2=3/x

x=6

The function represented by the table is linear, as it has a constant rate of change and is represented by a straight line.

Function in mathematics is a relation between two sets, where one set is the input and the other set is the output. Functions are an important tool in mathematics and can be used to describe and model real-world phenomena. Functions take inputs, manipulate them and produce outputs. They can be used to represent relationships between two or more variables, or to represent a complex process. Functions allow us to break down complex problems into smaller, more manageable pieces and to study how changes in one variable affect other variables.

The function represented by the table is linear. It can be determined by the fact that the y-values change by the same amount every time the x-values increase by one unit. In this case, the y-values decrease by 2 each time the x-values increase by one unit. This is an example of a linear function.

Linear functions have the shape of a straight line and are characterized by having a constant rate of change. The constant rate of change is represented by the slope of the line, which in this case is -2. This means that for every one unit increase in the x-values, the y-values decrease by two.

A quadratic function is the opposite of a linear function, as it has a rate of change that is not constant. Quadratic functions are characterized by their parabolic shape and their rate of change increases as x-values increase. Exponential functions are characterized by their curved shape and increase exponentially as x-values increase.

In conclusion, the function represented by the table is linear, as it has a constant rate of change and is represented by a straight line.

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Uri paid a total of $85.47 for the landscaping service including tax and tip.

Taxes are compulsory payments made by a government organisation, whether local, regional, or federal, to people or businesses. Tax revenues are used to fund a variety of government initiatives, such as Social Security and Medicare as well as public infrastructure and services like roads and schools. Taxes are borne by whoever bears the cost of the tax in economics, whether this is the entity being taxed, such as a business, or the final users of the items produced by the firm. Taxes should be taken into consideration from an accounting standpoint, including payroll taxes, federal and state income taxes, and sales taxes.

Given that company charged $74 for the service plus 5% tax.

The tax is 5%, that is:

Tax = 5% of $74 = 0.05 x $74 = $3.70

Cost after tax = $74 + $3.70 = $77.70

Now, tip is 10%:

Tip = 10% of $77.70 = 0.10 x $77.70 = $7.77

Total cost = $77.70 + $7.77 = $85.47

Hence, Uri paid a total of $85.47 for the landscaping service including tax and tip.

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Answer:

[tex]\dfrac{4096\pi}{5}\approx 2573.593\; \sf (3\;d.p.)[/tex]

Step-by-step explanation:

The shell method is a calculus technique used to find the volume of a solid revolution by decomposing the solid into cylindrical shells. The volume of each cylindrical shell is the product of the surface area of the cylinder and the thickness of the cylindrical wall. The total volume of the solid is found by integrating the volumes of all the shells over a certain interval.

The volume of the solid formed by revolving a region, R, around a vertical axis, bounded by x = a and x = b, is given by:

[tex]\displaystyle 2\pi \int^b_ar(x)h(x)\;\text{d}x[/tex]

where:

[tex]\hrulefill[/tex]

We want to find the volume of the solid formed by rotating the region bounded by y = 0, y = √x, x = 0 and x = 16 about the y-axis.

As the axis of rotation is the y-axis, r(x) = x.

Therefore, in this case:

[tex]r(x)=x[/tex]

[tex]h(x)=\sqrt{x}[/tex]

[tex]a=0[/tex]

[tex]b=16[/tex]

Set up the integral:

[tex]\displaystyle 2\pi \int^{16}_0x\sqrt{x}\;\text{d}x[/tex]

Rewrite the square root of x as x to the power of 1/2:

[tex]\displaystyle 2\pi \int^{16}_0x \cdot x^{\frac{1}{2}}\;\text{d}x[/tex]

[tex]\textsf{Apply the exponent rule:} \quad a^b \cdot a^c=a^{b+c}[/tex]

[tex]\displaystyle 2\pi \int^{16}_0x^{\frac{3}{2}}\;\text{d}x[/tex]

Integrate using the power rule (increase the power by 1, then divide by the new power):

[tex]\begin{aligned}\displaystyle 2\pi \int^{16}_0x^{\frac{3}{2}}\;\text{d}x&=2\pi \left[\dfrac{2}{5}x^{\frac{5}{2}}\right]^{16}_0\\\\&=2\pi \left[\dfrac{2}{5}(16)^{\frac{5}{2}}-\dfrac{2}{5}(0)^{\frac{5}{2}}\right]\\\\&=2 \pi \cdot \dfrac{2}{5}(16)^{\frac{5}{2}}\\\\&=\dfrac{4\pi}{5}\cdot 1024\\\\&=\dfrac{4096\pi}{5}\\\\&\approx 2573.593\; \sf (3\;d.p.)\end{aligned}[/tex]

Therefore, the volume of the solid is exactly 4096π/5 or approximately 2573.593 (3 d.p.).

[tex]\hrulefill[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Power Rule of Integration}\\\\$\displaystyle \int x^n\:\text{d}x=\dfrac{x^{n+1}}{n+1}(+\;\text{C})$\\\end{minipage}}[/tex]

The difference of the medians as a multiple of the interquartile range is 0.5,So the correct answer is option (A) 0.5.

The median is a measure of central tendency that represents the middle value in a data set when the values are arranged in numerical order.

For example, consider the data set {3, 5, 2, 6, 1, 4}. When the values are ordered from smallest to largest, we get {1, 2, 3, 4, 5, 6}. The median in this case is the middle value, which is 3.

We can first find the medians and interquartile ranges of the two dot plots.

For Mr. Wilson's class:

Median = 12

Q1 = 10

Q3 = 14

IQR = Q3 - Q1 = 14 - 10 = 4

For Mr. Watson's class:

Median = 10

Q1 = 8

Q3 = 12

IQR = Q3 - Q1 = 12 - 8 = 4

The difference of the medians is |12 - 10| = 2. Therefore, the difference of the medians as a multiple of the interquartile range is:

$$\frac{2}{4} = \boxed{0.5}$$

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The average rate of change on the intervals [0, 3], [3, 5], [5, 7], and [7, 9] are 2, -1.5, 1, and -1.5, respectively.

It expresses how much the function changed per unit on average during that time period. It is computed by taking the slope of the straight line connecting the interval's endpoints on the function's graph.

To calculate the average rate of change for the intervals shown in the graph, we must first determine the slope of the line connecting the endpoints of each interval.

0-3 interval:

Because the interval's endpoints are (0, 1) and (3, 7), the slope of the line connecting them is:

slope = (y change) / (x change) = (7 - 1) / (3 - 0) = 2

pauses [3, 5]:

Because the interval's endpoints are (3, 7) and (5, 4), the slope of the line connecting them is:

slope = (y change) / (x change) = (4 - 7) / (5 - 3) = -1.5

[5–7] Interval:

Because the interval's endpoints are (5, 4) and (7, 6), the slope of the line connecting them is:

slope = (y change) / (x change) = (6 - 4) / (7 - 5) = 1

Interval 7 and 9:

Because the interval's endpoints are (7, 6) and (9, 3), the slope of the line connecting them is:

slope = (y change) / (x change) = (3 - 6) / (9 - 7) = -1.5

As a result, the average rate of change on the intervals [0, 3], [3, 5], [5, 7], and [7, 9] is 2, -1.5, 1, and -1.5.

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i) Find the mean, median, and mode of the frequency table as follows:

ii) The average that justifies the teacher's statement congratulating the class that 'over three quarters were above average' is the average mark of 10, which is 5.

The mean refers to the average or the quotient of the total values divided by the number of items.

The median is the middle value in the data, which occurs with marks 8 for the 13th and 14th students.

The mode is the value that occurs most frequently, which is 3 which occurs 6 times.

Frequency Table:

Mark   Frequency  Cumulative Frequency

3              6                            18 (0 + 3 x 6)

4              3                            30 (18 + 4 x 3)

5              1                            35 (30 + 5 x 1)

6              2                           47 (35 + 6 x 2)

7              0                           47 (47 + 7 x 0)

8              5                           87 (47 + 8 x 5)

9              5                         132 (87 + 9 x 5)

10            4                         172 (132 + 10 x 4)

Mean = 6.6 (172/26)

Median = 8

Mode = 3

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Answer:

The value of the derivative at (-2/3, 2√3/3) is zero.

Step-by-step explanation:

Given function:

[tex]f(x)=-3x\sqrt{x+1}[/tex]

To differentiate the given function, use the product rule and the chain rule of differentiation.

[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Product Rule of Differentiation}\\\\If $y=uv$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{7 cm}\underline{Differentiating $[f(x)]^n$}\\\\If $y=[f(x)]^n$, then $\dfrac{\text{d}y}{\text{d}x}=n[f(x)]^{n-1} f'(x)$\\\end{minipage}}[/tex]

[tex]\begin{aligned}\textsf{Let}\;u &= -3x& \implies \dfrac{\text{d}u}{\text{d}{x}} &= -3\\\\\textsf{Let}\;v &= \sqrt{x+1}& \implies \dfrac{\text{d}v}{\text{d}{x}} &=\dfrac{1}{2} \cdot (x+1)^{-\frac{1}{2}}\cdot 1=\dfrac{1}{2\sqrt{x+1}}\end{aligned}[/tex]

Apply the product rule:

[tex]\implies f'(x) =u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}[/tex]

[tex]\implies f'(x)=-3x \cdot \dfrac{1}{2\sqrt{x+1}}+\sqrt{x+1}\cdot -3[/tex]

[tex]\implies f'(x)=- \dfrac{3x}{2\sqrt{x+1}}-3\sqrt{x+1}[/tex]

Simplify:

[tex]\implies f'(x)=- \dfrac{3x}{2\sqrt{x+1}}-\dfrac{3\sqrt{x+1} \cdot 2\sqrt{x+1}}{2\sqrt{x+1}}[/tex]

[tex]\implies f'(x)=- \dfrac{3x}{2\sqrt{x+1}}-\dfrac{6(x+1)}{2\sqrt{x+1}}[/tex]

[tex]\implies f'(x)=- \dfrac{3x+6(x+1)}{2\sqrt{x+1}}[/tex]

[tex]\implies f'(x)=- \dfrac{9x+6}{2\sqrt{x+1}}[/tex]

An extremum is a point where a function has a maximum or minimum value.

From inspection of the given graph, the maximum point of the function is (-2/3, 2√3/3).

To determine the value of the derivative at the maximum point, substitute x = -2/3 into the differentiated function.

[tex]\begin{aligned}\implies f'\left(-\dfrac{2}{3}\right)&=- \dfrac{9\left(-\dfrac{2}{3}\right)+6}{2\sqrt{\left(-\dfrac{2}{3}\right)+1}}\\\\&=-\dfrac{0}{2\sqrt{\dfrac{1}{3}}}\\\\&=0 \end{aligned}[/tex]

Therefore, the value of the derivative at (-2/3, 2√3/3) is zero.

Answer:

According to the scale, 1 inch on the drawing represents 1.5 feet in real life. So, to find the actual length of the door, we need to multiply the length on the drawing by the scale factor:

4 inches x 1.5 feet/inch = 6 feet

Similarly, to find the actual width of the door, we need to multiply the width on the drawing by the scale factor:

7 inches x 1.5 feet/inch = 10.5 feet

Therefore, the actual measurements of the door are 6 feet by 10.5 feet.

Answer: D. Square

Step-by-step explanation:

The shape created by the cross section of the cut through a square pyramid is a square.

To see why, imagine the pyramid sitting on a table with its square base flat against the surface. The cut goes through the vertex, which is the point at the top of the pyramid. Since the cut is perpendicular to the base, it divides the pyramid into two smaller pyramids with congruent, but not identical, bases. Each of these smaller pyramids has a triangular base that is an isosceles right triangle. The two triangles share a common hypotenuse, which is the line of the cut.

The cross section of the cut is the shape formed where the two triangles meet along the hypotenuse. Since both triangles are congruent and the hypotenuse is the same for both, the cross section is a square. The sides of the square are equal to the base of the original pyramid, which is one of the legs of the isosceles right triangles formed by the cut. Therefore, the answer is D, a square.

The point of intersection of the two lines is (-3.4, -1.2).

Step 1: The slope of a line passing through two points (x1,y1) and (x2,y2) can be found using the formula:

m = (y2-y1)/(x2-x1)

Using this formula, we can find the slope of the first line:

m1 = (−2−3)/(-5 -5) = −5/(-10) = 1/2

And the slope of the second line:

m2 = (−4−4)/(2 -(-6)) = -8/4 = -2

Step 2: Find the y-intercept of each line

The y-intercept of a line in slope-intercept form (y = mx + b) is the value of y when x=0. We can use one of the two given points on each line to find the y-intercept:

For the first line passing through points (5,3) and (−5,−2):

y = mx + b

3 = (1/2)(5) + b

b = 3 - 5/2

b = 1/2

So the first line can be written as y = 1/2x + 1/2

For the second line passing through points (−6,4) and (2,−4):

y = mx + b

4 = (-2)(−6) + b

b = 4 - 12

b = -8

So the second line can be written as y = -2x - 8

Step 3: Each line in slope-intercept form (y = mx + b):

First line: y = 1/2x + 1/2

Second line: y = -2x - 8

Step 4: To find the point of intersection of the two lines, we need to solve the system of equations. We can solve for x by setting the two right-hand sides equal to each other:

1/2x + 1/2 = -2x - 8

(x + 1)/2 = -2x - 8

x + 1 = -4x - 16

5x = -16 - 1

5x = -17

x = -17/5

x = -3.4

Now that we know x, we can find y by substituting x=10 into one of the two equations:

y = -2x - 8

y = -2(-3.4) - 8

y = - 1.2

Thus, the point of intersection of the two lines is (-3.4, -1.2).

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Answer:

y = 4 sin(π x) + 5

Step-by-step explanation:

A sine function with a midline of y=5, an amplitude of 4, and a period of 2 can be written in the following form:

y = A sin(2π/ x) +

where A is the amplitude, is the period, is the vertical shift (midline), and x is the independent variable (usually time).

Substituting the given values, we get:

y = 4 sin(2π/2 x) + 5

Simplifying this expression, we get:

y = 4 sin(π x) + 5

Therefore, the sine function with the desired characteristics is:

y = 4 sin(π x) + 5

[tex]\begin{cases} x=\frac{2+i\sqrt{2}}{3}\implies 3x=2+i\sqrt{2}\implies 3x-2-i\sqrt{2}=0\\\\ x=\frac{2-i\sqrt{2}}{3}\implies 3x=2-i\sqrt{2}\implies 3x-2+i\sqrt{2}=0 \end{cases} \\\\\\ \stackrel{ \textit{original polynomial} }{a(3x-2-i\sqrt{2})(3x-2+i\sqrt{2})=\stackrel{ 0 }{y}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{ \textit{difference of squares} }{[(3x-2)-(i\sqrt{2})][(3x-2)+(i\sqrt{2})]}\implies (3x-2)^2-(i\sqrt{2})^2 \\\\\\ (9x^2-12x+4)-(2i^2)\implies 9x^2-12x+4-(2(-1)) \\\\\\ 9x^2-12x+4+2\implies 9x^2-12x+6 \\\\[-0.35em] ~\dotfill\\\\ a(9x^2-12x+6)=y\hspace{5em}\stackrel{\textit{now let's make}}{a=-\frac{1}{9}} \\\\\\ -\cfrac{1}{9}(9x^2-12x+6)=y\implies \boxed{-x^2+\cfrac{4}{3}x-\cfrac{2}{3}=y}[/tex]

The value of the triple integral[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex] by using spherical coordinates [tex]2\pi(e^{-1}-e^{-9})[/tex].

Given that the triple integral is-

[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex]

E is the region bounded by the spheres which are,

[tex]x^2+y^2+z^2=1\\\\x^2+y^2+z^2=9[/tex]

In spherical coordinates we have,

x = r cosθ sin ∅

y = r sinθ sin∅

z = r cos∅

dV = r²sin∅ dr dθ d∅

E contains two spheres of radius 1 and 3 () respectively, the bounds will be like this,

1 ≤ r ≤ 3

0 ≤ θ ≤ 2π

0 ≤ ∅ ≤ π

Then

[tex]\int \int\int _{E } \frac{e^{-(x^2+y^2+z^2)}}{\sqrt{(x^2+y^2+z^2}}\sqrt{dV}[/tex]

[tex]\int\int\int _{E} \frac{e^{-r^2}}{r}r^2Sin\phi drd\phi d\theta\\\\2\pi \int_{0}^{\pi} \int_1^3 re^{-r^2} dr d\phi\\\\2\pi \int_1^3 re^{-r^2} dr\\\\2\pi(e^{-1}-e^{-9})[/tex]

The complete question is-

Use spherical coordinates to evaluate the triple integral ∭ee−(x2 y2 z2)x2 y2 z2−−−−−−−−−−√dv, where e is the region bounded by the spheres x2 y2 z2=1 and x2 y2 z2=9.

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The pairs of events of random integers are pairs of events are,

even and odd , negative and positive integers, zero and non-zero integers.

Two events are mutually exclusive if they cannot occur at the same time.

Selecting a random integer from -200 to 200,

Any two events that involve selecting a specific integer are mutually exclusive.

For example,

The events selecting the integer -100

And selecting the integer 50 are mutually exclusive

As they cannot both occur at the same time.

Any pair of events that involve selecting a specific integer are mutually exclusive.

Here are a few examples,

Selecting an even integer and selecting an odd integer.

Selecting a negative integer and selecting a positive integer

Selecting the integer 0 and selecting an integer that is not 0.

But,

Events such as selecting an even integer and selecting an integer between -100 and 100 are not mutually exclusive.

As there are even integers between -100 and 100.

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Step 1: Arrange the arrays so that A and B are in the same order: A = [ 1 2 4 0 5 6 ], B = [ 7 3 2 5 1 9]

Step 2: To find C = A+B, add each element of A and B together.

C = [1+7, 2+3, 4+2, 0+5, 5+1, 6+9]

C = [8, 5, 6, 5, 6, 15]

Step 3: To find D = A-B, subtract each element of B from A.

D = [1-7, 2-3, 4-2, 0-5, 5-1, 6-9]

D = [-6, -1, 2, -5, 4, -3]