X, Y and Z form the vertices of a triangle. XY = 12.4m, XZ = 10.4m and YZ = 8.7m. Find the angle ∠ YXZ rounded to 1 DP.

Proof of identity tanθ = secθ/cscθ is shown below.

We have to give that,

Verify the identity,

tanθ = secθ/cscθ

Now, We can prove as,

Since,

sec θ = 1 / cos θ

csc θ = 1 / sin θ

tan θ = sin θ / cos θ

LHS,

tan θ = sin θ / cos θ

RHS,

secθ/cscθ = (1 / cos θ) / (1 / sin θ)

secθ/cscθ = (sin θ / cos θ)

secθ/cscθ = tan θ

Hence, We prove that,

tanθ = secθ/cscθ

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Answer:

the range is 3

Step-by-step explanation:

We define the variables:

• x = time spent texting (in hours),

• y_O = observed time spent exercising (in hours),

• y_P = predicted time spent exercising (in hours).

• r = residual (in hours) = y_O - y_P.

From the statement, we have the following equation for the line of best fit:

1) For x = 4.0 we have:

• y_O = 4.50 (from the table),

• y_P = 5.36 (by using the formula),

• r = 4.50 - 5.36 = 0.86.

2) For x = 5.0 we have:

• y_O = 6.50 (from the table),

• y_P = 4.96 (by using the formula),

• r = 6.50 - 4.96 = 1.54.

1) For x = 4.0 we have: y_O = 4.50, y_P = 5.36, r = -0.86

2) For x = 5.0 we have: y_O = 6.50, y_P = 4.96, r = 1.54

1. The amounts of flour, considering 4 cups of sugar for each recipe, are given as follows:

2. The constant for each recipe is given as follows:

A proportional relationship is modeled as follows:

y = kx.

In which k is the constant of proportionality.

The variables for this problem are given as follows:

From the table, the constant is given as follows:

k = (3/4)/(1/2)

k = 3/2.

Meaning that for each cup of sugar, 3/2 cups of flour are needed.

Hence the relation is:

y = 3x/2.

The amounts of flour, considering 4 cups of sugar for each recipe, are obtained as follows:

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Answer:

A. 40-30i is your answer.

Answer:

1. Yes it does. It's a function because it pass in the vertical line test, which tell us that if you can draw an vertical line through the graph and this line touches the graph in only one point, it is a function.

2. B. I and II only

3. A. y = 8/7 x +5

Answer: 63 ÷ (7-4) x 2=42

Step-by-step explanation:

Answer:

The answer is 42

Step-by-step explanation:

We can solve this by using Order of Operations

1. Solve whats inside the parentheses (7-4) =3

63÷3×2

2. Now solve from left to right

63÷3 =21

21 ×2= 42

And we get a final answer of 42.

Hope this helps!

Answer:

52

Step-by-step explanation:

I'm pretty sure the answer is 52.

In Jim and Dave each got 5 and David got 4 then :

65/5 = 13

65 - 13 = 52

or

5 4 5

10 8 10

15 12 15

20 16 20

25 20 25

30 24 30

35 28 35

40 32 40

45 36 45

50 40 50

55 44 55

60 48 60

65 52 65

The conversion of a fraction number with denominator 5 and numerator 2 , 2/5 in decimals is equals to the 0.4 value.

A decimal number can be defined as a number whose whole number part and fractional part are separated by a decimal point. Writing 2/5 as a decimal number by converting the denominator to powers of 10. We multiply the numerator and denominator by a number so that the denominator is a power of 10.

2/5 = (2 × 2) / (5 × 2) = 4/10

Now move the decimal point to the left as many places as there are zeros in the denominator, which is a power of 10.

The decimal moved one place to the left because the denominator was 10. Therefore, 4/10 = 0.4. Hence, required value is 0.4.

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Kindly check below.

1) Examining the expressions inside the tiles, we can tell that:

2) Based on this we can tell the following relationship of equivalence:

Answer:

w^2z^4/y^2

Step-by-step explanation:

Answer:

Step-by-step explanation:

Given:

A shelf is built on a wall. The angles of the triangle are given as (x + 3)°, (2x - 18)° and 90°

We need to determine the value of x.

Value of x:

By triangle sum property, the angles in a triangle always add up to 180°

Thus, we have;

Simplifying, we get;

Subtracting both sides by 75, we get;

Dividing both sides of the equation by 3, we get;

Thus, the value of x is 35.

Answer:

-2 , -2, parallel

Step-by-step explanation:

same slope, different y intercept, parallel

Answer:

since this is on a graph, and perfectly lined up, you can just count the squares on the length of XY

Answer:

10) -1.5

11) 1

Step-by-step explanation:

Hope this helps! Pls give brainliest!

Answer:

smaller number is 6 and larger number is 12.

Step-by-step explanation:

Let the one number is y.

The another number is y + 6.

2 (y + 6) + y = 30

2 y + 12 + y = 30

3 y = 30 - 12

3 y = 18

y = 6

So, the smaller number is 6 and the larger number if 6 + 6 = 12 .  

Answer:

1st, 3rd

and 4th - "4th" from different answer :)

Step-by-step explanation:

Have a good day :)

Answer:

1,3 and 4

Step-by-step explanation:

The most appropriate model to represent the data in the table is (d)

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

The table of values

In the above table of values, we can see that

To show as the number of miles change by day

A linear or line graph has to be plotted

Hence, the most appropriate model to represent the data is (d)

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The coordinates for point B are 7 = √(x - 5)² + (y - 4)².

What is the distance formula?

The distance formula is used to calculate the length of the line which joins the two points in an x−y plane. The distance between two points is the length of the line joining the two points. If the two points lie on the same horizontal or same vertical line.

Here, we have

Given: Segment AB Has point A located At (5, 4). If The Distance from A To B Is 7 units.

To find the distance between any two the following formula is used.

D = √(x₂ - x₁)² + (y₂ - y₁)²

7 = √(x - 5)² + (y - 4)²

Hence, the coordinates for point B are 7 = √(x - 5)² + (y - 4)².

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The Laplace transform of the given function is,

L{f(t)} = (8/s) - 4e^{-3s/2}/s - 6e^{-2s}/s

Given function is f(t) = {8 12 1 Jo, 0 ≤ t < 3/2, 3/2 ≤ t < 2, 2 ≤ t < ∞ respectively.

We have to find Laplace transform of the given function.

For first interval 0 ≤ t < 3/2,

f(t) = 8u(t) - 8u(t-3/2)

For second interval 3/2 ≤ t < 2,

f(t) = 12u(t-3/2) - 12u(t-2)

For third interval 2 ≤ t < ∞,

f(t) = Jo(u(t-2))

Hence, we can write the Laplace transform of the given function as,

L{f(t)} = L{8u(t) - 8u(t-3/2)} + L{12u(t-3/2) - 12u(t-2)} + L{Jo(u(t-2))}

Where, L is Laplace transform.

Let's calculate each Laplace transform stepwise,

1. L{8u(t) - 8u(t-3/2)}L{8u(t)} = 8/L{u(t)}L{u(t)}

= 1/sL{u(t-3/2)}

= e^{-3s/2}/s

Therefore,

L{8u(t) - 8u(t-3/2)} = 8[1/s - e^{-3s/2}/s]

2. L{12u(t-3/2) - 12u(t-2)}L{12u(t-3/2)}

= 12e^{-3s/2}/sL{12u(t-2)}

= 12e^{-2s}/s

Therefore,

L{12u(t-3/2) - 12u(t-2)} = 12[e^{-3s/2}/s - e^{-2s}/s]

3. L{Jo(u(t-2))}L{Jo(u(t-2))} = ∫_{0}^{∞}δ(t-2)e^{-st}dtL{Jo(u(t-2))}

= e^{-2s}

Hence, the Laplace transform of the given function is,

L{f(t)} = 8[1/s - e^{-3s/2}/s] + 12[e^{-3s/2}/s - e^{-2s}/s] + e^{-2s}

= (8/s) - 4e^{-3s/2}/s - 6e^{-2s}/s

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The value of x is 44° if the given figure below are two parallel lines and

the third line is intersecting them.

Two lines in the same plane that are equally spaced apart and never cross each other are referred to in geometry as parallel lines. Both vertical and horizontal can be used. A zebra crossing, rows of notebooks, and nearby railroad tracks are just a few instances of parallel lines that we encounter every day. Coplanar, straight lines that don't intersect anywhere are considered parallel lines in geometry. Curves do not touch or intersect when they are parallel to one another and keep a predetermined minimum distance between them.

From the diagram,

x+136°=180°

x=180°-136°

x=44°

Therefore, the value of x is 44° if the given figure below are two parallel

lines and the third line is intersecting them.

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To calculate the time it takes to triple your money at a 12.5 percent interest rate, we can use the formula for compound interest and we obtain the answer as 9.33(Approximately)

FV = PV * (1 + r)^n

Where FV is the future value, PV is the present value, r is the interest rate, and n is the number of compounding periods.

In this case, we want to find the value of n when the future value (FV) is three times the present value (PV). Let's assume the initial amount is $1.

3 * 1 = 1 * (1 + 0.125)^n

Simplifying the equation, we have:

3 = 1.125^n

To solve for n, we need to take the logarithm of both sides of the equation:

log(3) = n * log(1.125)

n = log(3) / log(1.125)

Using a calculator, we find that n is approximately 9.33 years.

Therefore, the correct answer is: 9.33 years.

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Answer:

its C

Step-by-step explanation:

The answer is (c) Manny paid the correct amount.

A value or ratio that may be stated as a fraction of 100 is referred to as a percentage in mathematics. If we need to compute a percentage of a number, we should divide it by its whole and then multiply it by 100. The proportion therefore refers to a component per hundred. Per 100 is what the term percent signifies. The letter "%" stands for it.

Let's first calculate the total cost of Manny's purchases before any discounts or taxes are applied:

Pair of jeans: $32.65

3 t-shirts: 3 × $14.89 = $44.67

Pair of sneakers: $39.99

Total before discounts or taxes: $32.65 + $44.67 + $39.99 = $117.31

Now we can apply the 10% discount to the total:

Discounted total: $117.31 × 0.9 = $105.58

Next, we can calculate the sales tax on the discounted total:

Sales tax: $105.58 × 0.08 = $8.45

Finally, we can add the sales tax to the discounted total to find Manny's total cost:

Total cost: $105.58 + $8.45 = $114.03

Since Manny paid exactly $114.03 for his purchase, he paid the correct amount. Therefore, the answer is (c) Manny paid the correct amount.

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The given equation is 5x - 2x² + 3x - 1/3 + 4 = -1 . The sum of the roots of the quadratic equation ax² + bx + c = 0. The sum of the roots of the equation is 4.

Step by step answer:

Step 1: Rearrange the equation5x - 2x² + 3x + 1/3 + 4 + 1 = 0 Multiplying the whole equation by 3, we get,15x - 6x² + 9x + 1 + 12 + 3 = 0

Step 2: Simplify the equation-6x² + 24x + 16 = 0 Dividing the whole equation by -2, we get,3x² - 12x - 8 = 0

Step 3: Find the roots of the quadratic equation

3x² - 12x - 8

= 0ax² + bx + c

= 0x

= [-b ± √(b² - 4ac)] / 2a

Here, a = 3,

b = -12,

c = -8x

= [12 ± √(12² - 4(3)(-8))] / 2(3)x

= [12 ± √216] / 6x

= [12 ± 6√6] / 6x

= 2 ± √6

Therefore, the roots of the quadratic equation are 2 + √6 and 2 - √6

Step 4: Find the sum of the roots  The sum of the roots of the quadratic equation ax² + bx + c = 0 is given by the formula, Sum of roots = -b/a   Here,

a = 3 and

b = -12

Sum of roots = -b/a= -(-12) / 3

= 4

Hence, the sum of the roots of the equation is 4.

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The equation of the line is y = 2x + 8 that is parallel to the given line y = 2x + 2 and passes through the point (−3, 2)

An equation is an expression that shows the relationship between two or more numbers and variables.

The equation of a straight line is:

y = mx + b

Where m is the rate of change and b is the y intercept.

Two lines are parallel if they have the same slope.

Let us assume that the line is parallel to y = 2x + 2 and passes through the point (−3, 2).

The slope of the line y = 2x + 2 is 2, Since it is parallel, the slope is also 2. hence:

y - y₁ = m(x - x₁)

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

y - 2 = 2(x + 3)

y = 2x + 8

The equation of the line is y = 2x + 8

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An equation of the line that is parallel to the given line and passes through the point (−3, 2) is: 4x + 3y = -6.

Mathematically, the slope of any straight line can be calculated by using this formula;

Slope, m = (Change in y-axis, Δy)/(Change in x-axis, Δx)

Slope, m = (y₂ - y₁)/(x₂ - x₁)

By critically observing the graph (see attachment), we can logically deduce that the line passes through the following points:

Points (x, y) = (3, -1)

Points (x, y) = (0, 3)

Substituting the given parameters into the formula, we have;

Slope, m = (3 + 1)/(0 - 3)

Slope, m = 4/-3

Slope, m = -4/3.

In Geometry, two (2) lines are parallel under the following conditions:

m₁ = m₂ ⇒ -4/3 = -4/3

Note: m represents the slope.

At point (-3, 2), an equation of the other line can be calculated by using the point-slope form:

y - y₁ = m(x - x₁)

Where:

Substituting the given points into the formula, we have;

y - y₁ = m(x - x₁)

y - 2 = -4/3(x - (-3))

y - 2 = -4/3(x + 3)

y - 2 = -4x/3 - 4

y - 2 = -4x/3 - 4 + 2

y = -4x/3 - 2

Multiplying all through by 3, we have:

3y = -4x - 6

Rearranging the equation, we have:

4x + 3y = -6

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Step-by-step explanation:

Answer:

\(y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}\)

Step-by-step explanation:

Given the second-order differential equation. Solve by using variation of parameters.

\(y''+2y'+y=e^{-t}\ln(t)\)

(1) - Solve the DE as if it were homogeneous to find the homogeneous solution

\(y''+2y'+y=e^{-t}\ln(t) \Longrightarrow y''+2y'+y=0\\\\\text{The characteristic equation} \rightarrow m^2+2m+1=0, \ \text{solve for m}\\\\m^2+2m+1=0\\\\\Longrightarrow (m+1)(m+1)=0\\\\\therefore \boxed{m=-1,-1}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Solutions to Higher-order DE's:}}\\\\\text{Real,distinct roots} \rightarrow y=c_1e^{m_1t}+c_2e^{m_2t}+...+c_ne^{m_nt}\\\\ \text{Duplicate roots} \rightarrow y=c_1e^{mt}+c_2te^{mt}+...+c_nt^ne^{mt}\\\\ \text{Complex roots} \rightarrow y=c_1e^{\alpha t}\cos(\beta t)+c_2e^{\alpha t}\sin(\beta t)+... \ ;m=\alpha \pm \beta i\end{array}\right}\)

Notice we have repeated/duplicate roots, form the homogeneous solution.

\(\boxed{\boxed{y_h=c_1e^{-t}+c_2te^{-t}}}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now using the method of variation of parameters, please follow along very carefully.

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(1 of 2):}}\\ \text{Given a DE in the form} \rightarrow ay''+by"+cy=g(t) \\ \text{1. Obtain the homogenous solution.} \\ \Rightarrow y_h=c_1y_1+c_2y_2+...+c_ny_n \\ \\ \text{2. Find the Wronskain Determinant.} \\ |W|=$\left|\begin{array}{cccc}y_1 & y_2 & \dots & y_n \\y_1' & y_2' & \dots & y_n' \\\vdots & \vdots & \ddots & \vdots \\ y_1^{(n-1)} & y_2^{(n-1)} & \dots & y_n^{(n-1)}\end{array}\right|$ \\ \\ \end{array}\right}\)

\(\boxed{\left\begin{array}{ccc}\text{\underline{Variation of Parameters Method(2 of 2):}}\\ \text{3. Find} \ W_1, \ W_2, \dots, \ W_n.\\ \\ \text{4. Find} \ u_1, \ u_2, \dots, \ u_n. \\ \Rightarrow u_n= \int\frac{W_n}{|W|} \\ \\ \text{5. Form the particular solution.} \\ \Rightarrow y_p=u_1y_1+u_2y_2+ \dots+ u_ny_n \\ \\ \text{6. Form the general solution.}\\ y_{gen.}=y_h+y_p\end{array}\right}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

(2) - Finding the Wronksian determinant

\(|W|= \left|\begin{array}{ccc}e^{-t}&te^{-t}\\-e^{-t}&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t}-te^{-t})-(te^{-t})(-e^{-t})\\\\\Longrightarrow (e^{-2t}-te^{-2t})-(-te^{-2t})\\\\\therefore \boxed{|W|=e^{-2t}}\)

(3) - Finding W_1 and W_2

\(W_1=\left|\begin{array}{ccc}0&y_2\\g(t)&y_2'\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}0&te^{-t}\\e^{-t} \ln(t)&e^{-t}-te^{-t}\end{array}\right|\\\\\Longrightarrow 0-(te^{-t})(e^{-t} \ln(t))\\\\\therefore \boxed{W_1=-t\ln(t)e^{-2t}}\)

\(W_2=\left|\begin{array}{ccc}y_1&0\\y_1'&g(t)\end{array}\right| \ \text{Recall:} \ g(t)=e^{-t} \ln(t)\\\\\Longrightarrow \left|\begin{array}{ccc}e^{-t}&0\\-e^(-t)&e^{-t} \ln(t)\end{array}\right|\\\\\Longrightarrow (e^{-t})(e^{-t} \ln(t))-0\\\\\therefore \boxed{W_2=\ln(t)e^{-2t}}\)

(4) - Finding u_1 and u_2

\(u_1=\int \frac{W_1}{|W|}; \text{Recall:} \ W_1=-t\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{-t\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow -\int t\ln(t)dt \ \text{(Apply integration by parts)}\\\\\\\boxed{\left\begin{array}{ccc}\text{\underline{Integration by Parts:}}\\\\uv-\int vdu\end{array}\right }\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=tdt \rightarrow v=\frac{1}{2}t^2 \\\\\)

\(\Longrightarrow -\Big[(\ln(t))(\frac{1}{2}t^2)-\int [(\frac{1}{2}t^2)(\frac{1}{t}dt)]\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\int (t)dt\Big]\\\\\Longrightarrow -\Big[\frac{1}{2}t^2\ln(t)-\frac{1}{2}\cdot\frac{1}{2}t^2 \Big]\\\\\therefore \boxed{u_1=\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t)}\)

\(u_2=\int \frac{W_2}{|W|}; \text{Recall:} \ W_2=\ln(t)e^{-2t} \ \text{and} \ |W|=e^{-2t} \\\\\Longrightarrow \int\frac{\ln(t)e^{-2t}}{e^{-2t}} dt\\\\\Longrightarrow \int \ln(t)dt \ \text{(Once again, apply integration by parts)}\\\\\text{Let} \ u=\ln(t) \rightarrow du=\frac{1}{t}dt \\\\\text{an let} \ dv=1dt \rightarrow v=t \\\\\Longrightarrow (\ln(t))(t)-\int[(t)(\frac{1}{t}dt )] \\\\\Longrightarrow t\ln(t)-\int 1dt\\\\\therefore \boxed{u_2=t \ln(t)-t}\)

(5) - Form the particular solution

\(y_p=u_1y_1+u_2y_2\\\\\Longrightarrow (\frac{1}{4}t^2-\frac{1}{2}t^2\ln(t))(e^{-t})+(t \ln(t)-t)(te^{-t})\\\\\Longrightarrow\frac{1}{4}t^2e^{-t}-\frac{1}{2}t^2\ln(t)e^{-t}+ t^2\ln(t)e^{-t}-t^2e^{-t}\\\\\therefore \boxed{ y_p=\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}\)

(6) - Form the solution

\(y_{gen.}=y_h+y_p\\\\\therefore\boxed{\boxed{y(t)=c_1e^{-t}+c_2te^{-t}+\frac{1}{2}t^2\ln(t)e^{-t}-\frac{3}{4} t^2e^{-t}}}\)

Thus, the given DE is solved.

The endpoint of the line segment is (43, 58).

To find the endpoint of the line segment, we need to first find the coordinates of the point that divides the line segment in a 1:5 ratio.

Let's call the endpoint we're looking for (x, y).

We know that the point (13, 14) divides the line segment into two parts with a ratio of 1:5. This means that the distance from (8, 19) to (13, 14) is one-sixth of the distance from (8, 19) to (x, y).

Using the distance formula, we can calculate the distance between the two points:

distance between (8, 19) and (13, 14) = \(\sqrt{(13-8)^{2}+(14-19)^{2} }\) = \(\sqrt{74}\)

We also know that the distance from (8, 19) to (x, y) is six times the distance from (8, 19) to (13, 14). So:

distance between (8, 19) and (x, y) = 6 * distance between (8, 19) and (13, 14)

= 6 * \(\sqrt{74}\)

Now we can use the midpoint formula to find the coordinates of the point that divides the line segment in a 1:5 ratio:

midpoint = ((1/6)*x + (5/6)*13, (1/6)*y + (5/6)*14)

= ((x+65)/6, (y+70)/6)

We know that the midpoint of the line segment is (13, 14), so:

(x+65)/6 = 13 and (y+70)/6 = 14

Solving for x and y, we get:

x = 43 and y = 58

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Answer: False

Step-by-step explanation:

Researchers typically report the adjusted R-squared value in addition to the regular R-squared value, not because they lack confidence in the actual R-squared, but because the adjusted R-squared provides additional information about the goodness of fit of a statistical model. The regular R-squared value measures the proportion of the variance in the dependent variable that is explained by the independent variables in the model. However, it can be biased and increase as more predictors are added to the model, even if the additional predictors do not contribute significantly to the prediction.

The adjusted R-squared, on the other hand, takes into account the number of predictors in the model and penalizes the addition of irrelevant predictors. It provides a more conservative measure of the goodness of fit by adjusting for the number of predictors and the sample size. Researchers often use the adjusted R-squared to evaluate and compare different models with varying numbers of predictors or to assess the overall explanatory power of a model while considering its complexity.

In summary, researchers report the adjusted R-squared value to address the limitations of the regular R-squared and to provide a more accurate assessment of the model's goodness of fit.