Can someone help me please Assap?

The definite integral is [tex]& \frac{243}{5}(\ln 3)^2-\frac{2}{5}\left[\ln 3\left(\frac{243}{5}\right)-\frac{243}{25}\right]-\frac{2}{125}[/tex].

The integral is [tex]$\int_1^3 x^4(\ln x)^2 d x$[/tex]

First evaluate indefinite integral, [tex]$\int x^4(\ln x)^2 d x$[/tex]

The definite integral of a real-valued function f(x) with respect to a real variable x on an interval [a, b] is expressed as [tex]\int_a^b f(x) d x=F(b)-F(a)[/tex]

Here,

∫ = Integration symbol

a = Lower limit

b = Upper limit

f(x) = Integrand

dx = Integrating agent

Thus, ∫ab f(x) dx is read as the definite integral of f(x) with respect to dx from a to b.

Indefinite integrals are implemented when the boundaries of the integrand are not specified.

Use integration by parts:

[tex]$\int u d v=u v-\int v d u$[/tex]

Let [tex]$u=(\ln x)^2$[/tex] and [tex]$a v=x^4 d x$[/tex]

This implies [tex]$d u=\frac{2(\ln x)}{x} d x$[/tex] and [tex]$v=\int x^4 d x=\frac{x^5}{5}$[/tex]

Equation (1) becomes,

[tex]$$\begin{aligned}& \int x^4(\ln x)^2 d x=(\ln x)^2 \frac{x^5}{5}-\int \frac{x^5}{5} \times \frac{2(\ln x)}{x} d x \\& \int x^4(\ln x)^2 d x=(\ln x)^2 \frac{x^5}{5}-\frac{2}{5} \int x^4 \ln x d x\end{aligned}$$[/tex]

Again, use integration by parts to evaluate [tex]$\int x^4 \ln x d x$[/tex]

Let [tex]$u=\ln x, d v=x^4 d x$[/tex]

This implies [tex]$d u=\frac{1}{x}$[/tex] and  [tex]$v=\int x^4 d x=\frac{x^5}{5}$[/tex]

Let [tex]$u=\ln x, d v=x^4 d x$[/tex]

This implies [tex]d u=\frac{1}{x}$[/tex] and [tex]$v=\int x^4 d x=\frac{x^5}{5}$[/tex]

[tex]$$\begin{aligned}\int x^4 \ln x d x & =\ln x\left(\frac{x^5}{5}\right)-\int\left(\frac{x^5}{5}\right) \frac{1}{x} d x \quad \text { since } \int u d v=u v-\int u d u \\& =\ln x\left(\frac{x^5}{5}\right)-\frac{1}{5} \int x^4 d x \\& =\ln x\left(\frac{x^5}{5}\right)-\frac{x^5}{25}\end{aligned}$$[/tex]

From equation (2),

[tex]$$\int x^4(\ln x)^2 d x=(\ln x)^2 \frac{x^5}{5}-\frac{2}{5}\left[\ln x\left(\frac{x^5}{5}\right)-\frac{x^5}{25}\right]$$[/tex]

The required definite integral is,

[tex]$$\begin{aligned}\int_1^3 x^4(\ln x)^2 d x & =\left[(\ln x)^2 \frac{x^5}{5}-\frac{2}{5}\left[\ln x\left(\frac{x^5}{5}\right)-\frac{x^5}{25}\right]\right]^3 \\& =\left[(\ln 3)^2 \frac{3^5}{5}-\frac{2}{5}\left[\ln 3\left(\frac{3^5}{5}\right)-\frac{3^5}{25}\right]\right]-\left[(\ln 1)^2 \frac{1^5}{5}-\frac{2}{5}\left[\ln 1\left(\frac{1^5}{5}\right)-\frac{1^5}{25}\right]\right] \\& =\frac{243}{5}(\ln 3)^2-\frac{2}{5}\left[\ln 3\left(\frac{243}{5}\right)-\frac{243}{25}\right]-\frac{2}{125}\end{aligned}$$[/tex]

Therefore, the required definite integral is [tex]& =\frac{243}{5}(\ln 3)^2-\frac{2}{5}\left[\ln 3\left(\frac{243}{5}\right)-\frac{243}{25}\right]-\frac{2}{125}[/tex].

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The  △ABC and △DEF are congruent by SAS if  BC = EF. Hence option D is the correct option.

What are congruent triangles?

If all three corresponding sides are equal and all three corresponding angles are identical in measure, two triangles are said to be congruent. These triangles can be moved, rotated, flipped, and turned to look exactly the same. They coincide if they are relocated. ' is the symbol of congruence.

Given that in △ABC and △DE:

AB = DE

∠A ≅ ∠D

SAS rule: Two triangles are congruent if two sides and an included angle of one triangle are equivalent to two sides and an included angle of the second triangle.

The opposite side of ∠A is BC and the opposite side ∠D is EF.

Thus the additional information, that is needed is BC = EF.

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The type of function that represents function g(x) is given as follows:

Logarithmic.

To define the function in this problem, we must observe a pattern with the inputs and the outputs.

The input and output pairs are given as follows:

From the pattern, we can see that each output is obtained as the power of 3 to which the input is elevated.

This means that the function is a logarithmic function, more specifically the logarithm of base 3 of x.

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The measure of congruent angles X and Y of the hexagon are 135.

let the measure of the congruent angles of  X and Y be a

Then by the angle sum property of a hexagon sum of all the angle will be

U +V +W +X +Y +Z = 720

then 100 + 110+ 149 + 91 + a + a =720

2a = 720 - 450

2a = 270

Then a = 135

Hence the congruent angles X and Y are 135.

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The equation connecting h and v is h = 56/v and the value of h when v = 5 is 11.2

From the question, we have the following parameters that can be used in our computation:

Variation = inverse variation

h = 8, v = 7

An inverse variation is represented as

k = vh

Where k is the constant of proportion

Substitute the known values in the above equation, so, we have the following representation

k = 7 * 8

Evaluate

k = 56

So, we have the following representation

vh = 56

Make h the subject

h = 56/v

Substitute the known values in the above equation, so, we have the following representation

h = 56/5

Evaluate

h = 11.2

Hence, the solution is 11.2

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Answer: 0.125 Miles

Step-by-step explanation:

X+2.25 +2.125 = 4.5
X+4.375 =4.5
4.5-4.375= X = 0.125

200

Portions of quantities can be represented as percentages and decimals, which can help solve different types of questions.

Multiplying Decimals

To find any percent of a quantity, you can multiply the quantity by the percentage as a decimal. We can convert percentages to decimals easily. By moving the decimal point 2 places to the left, we can change 40% into a decimal. So, 40% is equal to 0.4. Now, multiply this by 500.

So, Leo has put together 200 pieces.

Multiples of 10

Since 40% is a multiple of 10, we can solve using another method. It is easy to find 10% of any quantity, just move the decimal point 1 place to the left. So, 10% of 500 is 50.

Then, since 40 is 4 times 10, we can say that 40% of 500 is 4 times 10% of 500.

This proves that Leo has done 200 pieces.

Answer

0^0 =1

or

0^0 can be undefined.

The transformation is an example of a translation because the figure slides in one direction, but does not flip, turn, or change size.

Now, According to the question:

Type of Transformation:

There are four types of transformation in geometry, namely translation, reflection, rotation, and dilation. In translation, it slides the figure in any direction while in reflection, it flips the figure over a point or a line. Also, in rotation the figure turns, while in dilation the figure is either enlarged or reduced.

The type of transformation can be defined as moving a figure to a new location, without any change in the figure's size or shape being called translation. In translation, the figure merely slides or moves in the same direction and distance.

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Answer:

For me, it always helps to break down equations like this. Divide it up into two separate parts.

4^2*4^2=x

4^2=16

Replace both 4^2 with 16

16*16=256

The statement 115 < -225 because 115 is located to the right of -225 on the number line is not true.

The horizontal straight lines in mathematics known as number lines are where integers are arranged in equal intervals. A number line can be used to represent every number in a sequence. This line continues forever at both ends.

On a number line as we go from left to right the numbers increase. For the given numbers, 115 > -135> -225 as 115 is located on the right of the number line.

Hence, statement one 115 < -225 because 115 is located to the right of -225 on the number line is not true.

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h = kv is the formula connecting the height reached and the initial velocity

A variation is a relation between a set of values of one variable and a set of values of other variables.

From the information, the stone is thrown upwards, the height, h, it reaches is proportional to the square of the initial velocity, v.

Since the height, h is proportional to the square of the initial velocity, v.

We can write the formula as follows:

h α v

h = kv   (where k is constant of proportionality)

Therefore, the formula connecting the height reached and the initial velocity is h = kv  

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The equation of the line which passes through points (2 3) and (-4 1) is 3y - x = 7.

What is the equation of the line?

The formula for a straight line is y=mx+c where c is the height at which the line intersects the y-axis, often known as the y-intercept, and m is the gradient.

Here, we have

Given: points (2 3) and (-4 1)

We have to find the equation of a line.

we need to determine the slope of the line. The slope can be found by using the formula

m = (y₂ - y₁) /(x₂ - x₁)

Now, we put the given points and get the value of m.

m = (1 - 3) /(-4 - 2)

m = 1/3

Now, we can use the point-slope formula to find an equation for the line. The point-slope formula states:

(y - y₁) = m(x - x₁)

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

3y - 9 = x - 2

3y - x = 7

Hence,  the equation of the line which passes through points (2 3) and (-4 1) is 3y - x = 7.

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No, it is not possible to draw a triangle with side lengths of 2.5 cm, 4.2 cm, and 8 cm.

The triangle inequality theorem states that for any triangle, the sum of the lengths of the two shorter sides must be greater than the length of the longest side.

In other words, for a triangle with sides of lengths a, b, and c, where c is the longest side, the following must be true: a + b > c, b + c > a, and a + c > b.

In the case of the triangle with side lengths of 2.5 cm, 4.2 cm, and 8 cm, the sum of the lengths of the two shorter sides (2.5 cm + 4.2 cm) is 6.7 cm, which is less than the length of the longest side (8 cm).

Therefore, it is not possible to form a triangle with these side lengths, as the triangle inequality theorem is not satisfied.

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Difference in the weekly earnings of Dana and Monique is $127.5.

Multiplication of two numbers is defined as the addition of one of the number repeatedly until the times of the other number.

a × b means that a is added to itself b times or b is added to itself a times.

Dana :

Number of days Dana grooms dogs = 4 days a week

Number of dogs Dana grooms in a day = 15 dogs

Number of dogs Dana grooms in a week = 15 × 4 = 60 dogs

Earning of Dana per dog = $12.75

Earnings of Dana in a week = 60 × $12.75 = $765

Monique :

Number of days Monique grooms dogs = 5 days a week

Number of dogs Monique grooms in a day = 10 dogs

Number of dogs Monique grooms in a week = 10 × 5 = 50 dogs

Earning of Monique per dog = $12.75

Earnings of Monique in a week = 50 × $12.75 = $637.5

Difference in their earnings = $765 - $637.5 = $127.5

Hence there is a difference of $127.5 in their earnings.

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Let's call Mindy's height "M". According to the problem, we know that:

Camille's height = 66 inches

Camille's height = 2 * Mindy's height - 20 inches

We can substitute the first equation into the second equation to solve for Mindy's height:

66 inches = 2 * M - 20 inches

We can add 20 inches to both sides of the equation to get:

66 inches + 20 inches = 2 * M

This gives us:

86 inches = 2 * M

To find Mindy's height, we can divide both sides of the equation by 2:

M = 86 inches / 2

So Mindy is 43 inches tall.

Answer:

Mindy is  46 inches

Step-by-step explanation:

Let m = Mindy's height

66 = 2m - 20  Add 20 to both sides

66 + 20 = 2m - 20 + 20

86 = 2m  Divide both sides by 2

[tex]\frac{86}{2}[/tex] = [tex]\frac{2m}{2}[/tex]

46 = m

To determine the limit of f(x) as x approaches -7, we need to analyze the behavior of the function as x gets closer and closer to -7.

Since f(x) is defined differently for x < 7 and x >=7, we need to look at both cases separately.

For x < 7, f(x) = -3x + 0

For x >= 7, f(x) = 5x + 1

If we substitute x = -7 into the first equation, we get f(-7) = -3(-7) + 0 = 21

If we substitute x = -7 into the second equation, we get f(-7) = 5(-7) + 1 = -36

We can see that the function is defined differently for x < 7 and x >=7, and it does not have the same value when x = -7.

Therefore, the limit of f(x) as x approaches -7 does not exist (DNE).

A graph of f(x) with two different lines for x<7 and x>=7, one with slope -3 and y-intercept 0, the other with slope 5 and y-intercept 1. The two lines do not intersect at x = -7, this illustrates the DNE limit.

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equation for a tractrix, hence it proves that the trace of the tractrix is given as (t, t tan t).

The trace of the tractrix is given as:

Trace = (x, y) = (t, t tan t)

To show that the trace is that of a tractrix, we will use the equation for a tractrix, which is given as:

x = t – t tanh t

where t is the angle that the y-axis makes with the vector.

Substituting the value of x in the trace equation, we get:

t – t tanh t = t

Simplifying the equation, we get:

t tanh t = 0

This is the equation for a tractrix, hence it proves that the trace of the tractrix is given as (t, t tan t).

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The result of the given multiplication of the mixed fraction is found as 78/3.

A number that combines an integer as well as a legal fraction is referred to as a mixed fraction.

The given expression:

= 4 1/3 x 6

Solving the mixed fraction.

= (4*3 + 1)/3 x 6

= (12 + 1)/3 x 6

= 13/3 x 6

= 78/3

Thus, the result of the given multiplication of the mixed fraction is found as 78/3.

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The optimal values of x and y to maximize the profit are 10 and 10, respectively, and the maximum profit is $200.

Let X and Y represent the number of units of product A and B that we want to produce. We want to maximize the profit, so our objective function is:

Profit = 10x + 15y

The constraints are:

X + Y ≤ 20

2X + Y ≤ 30

X ≥ 0, Y ≥ 0

To find the optimal values of x and y, we must first solve the system of equations. By subtracting the first equation from the second equation, we get:

X = 10

Substituting this value into the first equation, we get:

Y = 10

Now that we have found the values of x and y, we can plug them into the objective function to find the maximum profit:

Profit = 10(10) + 15(10) = 200

Therefore, the optimal values of x and y to maximize the profit are 10 and 10, respectively, and the maximum profit is $200.

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[tex]26 sin 58°[/tex] is the expression that represents the length of segment PR.

The word "length" is used to describe an object's size or the separation between two points. The length of an object or the separation between two points is measured by its length. It is employed to determine an object's size or the separation between two points.

Since triangle JKL and triangle PQR are similar, we can use the similarity ratios to find the length of PR.

We know that the ratio of PR to JK is the same as the ratio of PQ to JL. We know that [tex]PQ = 26 and JL = 26 + 13 = 39[/tex], so the ratio is [tex]26/39[/tex].

So, [tex]PR = (26/39) * 13 = 26sin( JL/2) = 26sin58°[/tex]

So the answer is [tex]26 sin 58°[/tex]

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Answer:

Correct options are B) and C)

Step-by-step explanation:

The given equation is 2x+y=10.

Clearly, x=3 and y=4 satisfy this equation.

Therefore , x=3 and y=4 is one of the solution of given equation.

The equation to calculate slope from two points (x₁, y₁) and (x₂, y₂) on a line is m = (y₂ - y₁) / (x₂ - x₁).

Here,

m = the line's slope

x1 is equal to the initial point's x-coordinate.

y1 is the first point's y-coordinate.

x2 is equal to the second point's x-coordinate.

The second point's x-coordinate is equal to y2.

It only takes applying the slope formula rise/run to find the slope between two places. With various forms of information about the line available, various formulas can be used to determine the slope. Particularly when two points on the line are given, the formula for finding the slope from two points is utilized.

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Number of cups of flour required to make the same recipe for 3 cups of almond as per the given relation is equal to 7.5 cups of flour.

As given in the question,

For the given recipe cups of flour and cups of almond given is :

2 cups of almond used for 5 cups of flour

The relation is written in the form of expression :

2 cups of almond = 5 cups of flour

⇒ 1 cups of almond = ( 5 / 2 ) cups of flour

⇒ 3 cups of almond =  [ ( 5 / 2 ) × 3 ] cups of flour

⇒ 3 cups of almond =  ( 15 / 2 ) cups of flour

⇒ 3 cups of almond = 7.5 cups of flour

Therefore, the cups of flour required for 3 cups of almond is equal to 7.5 cups of flour.

The above question is incomplete , the complete question is :

A recipe call for 2 cup of almond for 5 cup of flour. Using the same recipe, how many cup of flour will you need for 3 cup of almond?

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0 is not a prime number. A prime number is defined as a positive integer greater than 1 that has no positive integer divisors other than 1 and itself. If we look at 0 it does not meet this requirement it means it cannot be a prime number.

0 does not meet this definition because it is not greater than 1 and it does have positive integer divisors other than 1 and itself (for example, it is divisible by 2, 4, and so on). Additionally, the concept of prime numbers is defined only for positive integers, 0 is not positive.

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the values of x will be:

Simplifying procedures is one way to achieve uniformity in job efforts, expenses, and time. It reduces diversity and variation that is pointless, harmful, or unneeded. Parenthesis, exponents, multiplication, division, addition, and subtraction are all referred to as PEMDAS. The order of the letters in PEMDAS informs you what to calculate first, second, third, and so on, until the computation is finished, given two or more operations in a single statement.

Given, Two functions, " x ÷ 5/6 = 1 and 4/5" and " x ÷ 2 and 1/10 = 1 and 2/3"

For  x ÷ 5/6 = 1 and 4/5

Simplifying the value,

x ÷ 5/6 = 1 4/5

x ÷ 5/6 = 9/5

x ÷ 5/6 = 9/5

Cross multiply

x = 9/5 * 5/6

x = 1.5

For  x ÷ 2 and 1/10 = 1 and 2/3

Simplifying the value,

x ÷ 2 1/10 = 1 2/3

x ÷21/10 = 5/3

x = 3.5

Therefore, the values of x will be:

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[tex]x=\cfrac{cos(2y)}{e^y}\hspace{5em}x\left( \frac{\pi }{8} \right)=\cfrac{1}{e^{\frac{\pi }{8}}\sqrt{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{dx}{dy}\implies \stackrel{quotient~rule}{\cfrac{-2sin(2y)e^y-cos(2y)e^y}{(e^y)^2}}\implies \cfrac{-2sin(2y)-cos(2y)}{e^y} \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dx}{dy} \right|_{y=\frac{\pi }{8}}\implies \cfrac{-2sin(\frac{\pi }{4})-cos(\frac{\pi }{4})}{e^y}\implies \cfrac{-3}{e^{\frac{\pi }{8}}\sqrt{2}}\impliedby m[/tex]

now, since we have the value of "x" at π/8 and we know that at that point y = π/8, and we also have the slope at π/8, let's plug all those fellows in the point-slope formula

[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\cfrac{\pi }{8}~~ = ~~\cfrac{-3}{e^{\frac{\pi }{8}}\sqrt{2}}\left(x-\cfrac{1}{e^{\frac{\pi }{8}}\sqrt{2}} \right)[/tex]

[tex]y-\cfrac{\pi }{8}~~ = ~~\cfrac{-3}{e^{\frac{\pi }{8}}\sqrt{2}}x+\cfrac{3}{2e^{\frac{\pi }{4}}} \implies \boxed{\cfrac{3}{e^{\frac{\pi }{8}}\sqrt{2}} {\Large \begin{array}{llll} x \end{array}} + {\Large \begin{array}{llll} y \end{array}}-\cfrac{\pi }{8}-\cfrac{3}{2e^{\frac{\pi }{4}}}~~ = ~~0}[/tex]