The given permutation (62848X) can be expressed as the product of disjoint cycles as (1 6 2 8 4) and as the product of transpositions as (1 6)(6 2)(2 8)(8 4).
To express the given permutation (62848X) as a product of disjoint cycles, we start by examining each element and its corresponding image under the permutation.
1 maps to 6.
6 maps to 2.
2 maps to 8.
8 maps to 4.
4 maps to 8 (since X represents a fixed point, meaning it remains unchanged).
Now, let's write these mappings as disjoint cycles:
(1 6 2 8 4)
The cycle notation indicates that 1 maps to 6, 6 maps to 2, 2 maps to 8, 8 maps to 4, and 4 maps back to 1.
Next, we'll express this permutation as a product of transpositions. A transposition swaps two elements.
We can achieve this by breaking down the cycle (1 6 2 8 4) into transpositions:
(1 6)(6 2)(2 8)(8 4)
Each pair of adjacent elements within the cycle forms a transposition. For example, (1 6) represents the transposition that swaps 1 and 6, (6 2) swaps 6 and 2, and so on.
Thus, the given permutation (62848X) can be expressed as the product of disjoint cycles as (1 6 2 8 4) and as the product of transpositions as (1 6)(6 2)(2 8)(8 4).
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What is the value of y in the equation 4 + y=-3?
7
1
-1
-7
Answer: -7
Because if y is -7 then the equation would be 4 - 7 = -3 which is true :)
Find the value of x in the picture below. (round to nearest tenth if needed) THANK YOU FOR HELPING ME:)
Answer:
12 I believe
Step-by-step explanation:
[tex]12\sqrt{2}[/tex] x sin(45) = 12
I am pretty positive I am correct, I am sorry if I am off a little
Open Ended Questions
Show all work to receive full credit
16. Use the model for vertical motion given by the equation h = -16t^2 + vt +s, where h is the
height in feet, t is the time in seconds, v is the initial upward velocity, and s is the initial height
in feet.
A. Rick tossed a baseball in the air from a height of 15 feet with an initial upward velocity of 8
feet per second. How long was the ball in the air before it hit the ground?
B. How high was the ball after 1 second?
Answer:
Step-by-step explanation:
what's the distance between theses two. (-5, 1) (2, 4)
Answer: squareroot of 58
You can solve this problem simply by using the distance formula . Using the distance formula we can solve this problem by just placing the numbers and then solving the equation.
x/2 + 4 < 18
What is the value of x?
And what does the point on the number line look like?
Someone help me
Worth 29 points
Answer:
x<28
Step-by-step explanation:
Isolate x
First, subtract 4 on both sides
x/2+<14
Then, multiply both sides by 2 to get x alone
x<28
On a number line, there would be an open circle (not filled in dot) on 28, and the entire left side of the number line would be filled in
Answer:
x<28
Step-by-step explanation:
x/2+4<18
multiply the 2 on both sides to get rid of it
x+8<36
isolate the x
x<28
on the number line, it's an open circle with the arrow pointing to the left.
If Thomas maintains a rate of 20 miles per hour riding his bicycle in a rally that covers 24 miles, how long will it take him to complete the race?
O 2 hours
O 1.2 hours
O 0.8 hours
O 2.4 hours
Answer:
1h=20m
60/4
0,25min+1h
1+0,25
=1,2
He will complete the race in 1.2 hours time.
Hence, option B is correct.
What is Measurement unit?A measurement unit is a standard quality used to express a physical quantity. Also it refers to the comparison between the unknown quantity with the known quantity.
Given that,
The distance completing in 1 hour = 20 miles
So 1 miles of distance is covered in = 1/20 hours
It is also given that,
The distance of rally = 24 miles.
Therefore,
The time taken to cover 24 miles = 24/20
= 1.2
Thus,
The required time = 1.2 hours
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A wire of 44cm long is cut into two parts. Each part is bent to form a square. Given that the total area of the two squares is 65cm^2, find the perimeter of each square.
Hi pls helppppp
Answer:
4225
Step-by-step explanation:
65x65=4225
John is cutting 3 wooden sticks to build part of a kite frame. The part he is building must be a right triangle.
Select all the possible lengths, in inches, of the sticks John could cut to make a right triangle.
A. 6, 8, 10
B. 2, 5, 10
C. 2, 3, 5
D. 12, 16, 20
E. 3, 4, 13
Answer:
I found 2 answers for this one
Step-by-step explanation:
B- 2,5,10
D-12,16,20
The possible length in inches of the stick that will make a right angle triangle are as follows
6, 8, 1012, 16, 20What is a right angle triangle?A right angle triangle has one of its angles as 90 degrees.
The right angle triangle must obeys the Pythagorean theorem.
c² = a² + b²
where
c = hypotenusea and b are the other legs.Therefore, the sides that makes a right angle are as follows:
6, 8 , 10 : 6² + 8² = 10²
12, 16, 20 : 12² + 16² = 20²
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The blue segments below is a diameter of oo. What is the length of the radius of the circle?
Answer:
C. 5.1 unitsStep-by-step explanation:
Given Information :
Diameter : 10.2 units
Radius (r) = ?
[tex]r = \frac{d}{2} [/tex]
Input the values
[tex]r = \frac{10.2}{2} \\ \\ r = 5.1[/tex]
What is the value of x in the equation 13−2(+4)=8+1 13 x − 2 ( x + 4 ) = 8 x + 1 ?
Answer: x= 12/103
alternate form x =0.116505
Step-by-step explanation:
Try Photo math! It explains step by step!
Hope this helps!!!!
the other one i need help with asap!!
Answer:
(3x-2)(2x-3)
Step-by-step explanation:
Good luck!
The radius of a circle is 8 inches. What is the area?
r=8 in
Give the exact answer in simplest form.
Answer:
Radius = 8 inches
Area = [tex]\pi {r}^{2} [/tex]
[tex]area = \pi( {8}^{2})[/tex]
[tex] = 64\pi \: square \: inches[/tex]
A = [tex]201.06 ^{2} \: inches[/tex]
Step-by-step explanation:
Hope it is helpful....
Use the table to determine a reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3? One-half One-fourth 3 DNE
The reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]
Given the limit of a function expressed as:
[tex]\lim_{x \to 3}\frac{2x^2-x+15}{x^3-5x-12}[/tex]
First, we need to substitute x = 3 into the function to have:
[tex]=\frac{2(3)^2-3+15}{3^3-5(3)-12}\\=\frac{18-3+15}{27-15-12}\\=\frac{0}{0} (indeterminate)[/tex]
Apply l'hospital rule on the function:
[tex]=\lim_{x \to 3}\frac{\frac{d}{dx} (2x^2-x+15)}{\frac{d}{dx} (x^3-5x-12)}\\=\lim_{x \to 3}\frac{4x-1}{3x^2-5}\\[/tex]
Subtitute x = 3 into the result
[tex]=\frac{4(3)-1}{3(3)^2-5}\\=\frac{12-1}{27-5}\\=\frac{11}{22}\\=\frac{1}{2}[/tex]
Hence the reasonable estimate for limStartFraction 2 x squared minus x + 15 Over x cubed minus 5 x minus 12 EndFraction as x approaches 3 is [tex]\dfrac{1}{2}[/tex]
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Answer:
1/2
Step-by-step explanation:
i am in your walls
Which comparison is not correct?
-2 > -7
1 < -9
-3 > -8
6 > 5
Answer:
1 < -9
Step-by-step explanation:
A positive number can't be less than a negative
Answer this question to get marked as barinliest!!!!
Introduction
Scientists have established a timeline of events after the Big Bang, based on astronomical observations and our understanding of the physical laws of the universe, such as gravity and the speed of light. In this lab activity, you will gather evidence to support the Big Bang theory.
Problem:
How can models demonstrate theories of our expanding universe?
Hypothesis:
Review the virtual lab demonstration in the lesson and stop the video when prompted to formulate a hypothesis. Hypothesize (or predict) what will happen to the distances between the labeled circles when you blow up the balloon ¼ full, ½ full, and ¾ full. Remember to include independent and dependent variables in your hypothesis.
The carbon dioxide represents how galaxies will spread out.
Materials:
Watch the virtual lab demonstration video within the lesson. No additional materials are needed.
Variables:
For this investigation:
List the independent variable(s):
List the dependent variable(s):
List the controlled variable(s):
Procedures:
1. Watch the virtual lab demonstration video within the lesson and record your observations in Table 1.
2. Using your expanding universe data from Table 1, construct a line graph using the volume of the below on the X axis and the distance between points on the Y axis. Be sure to include units and add titles to the graphs. Refer to the graph example and graphing tutorial in the lesson if needed.
3. Complete the Questions and Conclusion section of the lab report.
Data and Observations:
Table 1: Expanding Universe Observations
Galaxies Distance: Uninflated balloon (centimeters)
Distance: ¼ full (centimeters) Distance: ½ full (centimeters) Distance: ¾ full (centimeters)
A to B
A to C
A to D
B to C
B to D
C to D
Construct a line graph using the expanding the universe data from table 1. The volume will be plotted on the x-axis. The distance between the points will be plotted on the y-axis. Be sure to include units and add titles to the graph. Refer to the graph example and graphing tutorial in the lesson if needed.
Place your graph here.
Questions and Conclusion
1. How does the density and distribution of your “stars” change as the balloon expands?
2. How does your expanding balloon model represent an expanding universe?
3. What are some shortcomings of using this model as a replica of universe expansion?
4. How does the model you created help to show that the Steady State theory is inaccurate?
5. Suggest a way that a scientist could create an even more accurate model of universe expansion.
6. What will happen to the gravitational force between stars as the universe continues to expand?
In conclusion, how did your prediction of distances between points compare to your experimental results? All I truly need is the variables question 3 and 5 and I’m good :) thank you <3 this is for science but I didn’t know which one to pick so I picked a random one lol
Helppp I really need this question answered
Answer:
32.5 units²
Step-by-step explanation:
We can find the area by dividing the figure into two shapes. If we solve for the area of a square and a triangle, we can add them together.
First, we can solve for the area of the 5x5 square. To find the area, we can multiply the two dimensions.
5 · 5 = 25 units²
Next, we can find the area of the triangle. It has an equal height to the square, so the height of the triangle is 5 units. The width isn't specified. Instead, we are shown that the width of both the square and the triangle equals 8 units. If we subtract the width of the square from the total, we can find the width of the triangle, too.
8 - 5 = 3
So, the height and the width of the triangle are 5x3. To find the area we can multiply these together, and then divide the product by two.
5 · 3 = 15
[tex]\frac{15}{2}[/tex] = 7.5
The area of the triangle is 7.5 units².
Finally, we can add the area of the triangle and the square together.
25 + 7.5 = 32.5
The area of the figure, then, is 32.5 units².
The answer is the last option.
I hope this helps ^^
Let f(x)= e = 1+x. - a) Show that f has at least one real root (i.e. a number c such that f(c) = 0). b) Show that f cannot have more than one real root.
The function f(x) = e^(1+x) has at least one real root. The function f(x) = e^(1+x) cannot have more than one real root.
To show that f(x) has at least one real root, we need to find a value of x for which f(x) equals zero. Let's set f(x) = 0 and solve for x:
e^(1+x) = 0
Since e^(1+x) is always positive for any real value of x, there is no value of x that makes f(x) equal to zero. Hence, f(x) = e^(1+x) does not have any real roots. Therefore, we cannot show that f(x) has at least one real root.
b)
To show that f(x) cannot have more than one real root, we need to demonstrate that there cannot be two distinct real values, say c1 and c2, such that f(c1) = f(c2) = 0. Let's assume that f(x) = 0 at two distinct values, c1 and c2:
e^(1+c1) = e^(1+c2) = 0
However, this equation is not possible since e^(1+c1) and e^(1+c2) are always positive for any real values of c1 and c2. Therefore, f(x) = e^(1+x) cannot have more than one real root.
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Find the equations of the images of the following lines when reflected in the x-axis. a.y= 3x b.y= -x c. x = 0.
The equations of the images are after the transformations are
a. y = -3x
b. y = x
c. x = 0
How to determine the equations of the imagesFrom the question, we have the following parameters that can be used in our computation:
a. y = 3x
b. y = -x
c. x = 0.
The rule of the lines when reflected in the x-axis is
(x, y) = (x, -y)
This means that the functions are negated
So, we have the images to be
a. y = -3x
b. y = x
c. x = 0
Hence, the equations of the images are
a. y = -3x
b. y = x
c. x = 0
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Use strong induction to show that the square root of 18 is irrational. You must use strong induction to receive credit on this problem. Use strong induction to show that every integer amount of postage 30 cents or more can be formed using just 6-cent and 7-cent stamps.
[tex]\(\sqrt{k+1}\)[/tex] is irrational.
By the principle of strong induction, we can conclude that the square root of 18 is irrational.
What is irrationality?
In mathematics, irrationality refers to a property of certain numbers that cannot be expressed as a fraction of two integers or as a terminating or repeating decimal.
To prove that the square root of 18 is irrational using strong induction, we need to show that for every positive integer [tex]\(n\), if \(n > 1\) and \(\sqrt{n}\) is irrational, then \(\sqrt{n+1}\)[/tex] is also irrational.
[tex]\textbf{Base Case:}[/tex]
For [tex]\(n = 2\)[/tex], we have [tex]\(\sqrt{2}\).[/tex] It is a known fact that [tex]\(\sqrt{2}\)[/tex] is irrational. Thus, the base case holds true.
[tex]\textbf{Inductive Step:}[/tex]
Assume that for some positive integer k, if[tex]\(2 \leq k\) and \(\sqrt{k}\)[/tex] is irrational, then [tex]\(\sqrt{k+1}\)[/tex] is also irrational.
Now, consider the case for [tex]\(n = k+1\)[/tex]). We want to prove that [tex]\(\sqrt{k+1}\)[/tex] is irrational.
Since [tex]\(k \geq 2\)[/tex], we have [tex]\(k+1 > 2\)[/tex]. Therefore,[tex]\(\sqrt{k+1}\) is greater than \(\sqrt{2}\).[/tex]
Assume, for contradiction, that [tex]\(\sqrt{k+1}\)[/tex] is rational. Then, we can write \[tex](\sqrt{k+1}\)[/tex] as a fraction [tex]\(\frac{p}{q}\),[/tex] where p and q are positive integers with no common factors other than 1.
By squaring both sides, we get [tex]\(k+1 = \left(\frac{p}{q}\right)^2 = \frac{p^2}{q^2}\).[/tex]
Rearranging the equation, we have[tex]\(p^2 = (k+1)q^2\).[/tex]
Since [tex]\(k+1\)[/tex] is a positive integer and [tex]\(q^2\)[/tex] is also a positive integer, [tex]\(p^2\)[/tex]must be a multiple of [tex]\(k+1\)[/tex].
This implies that p must also be a multiple of [tex]\(k+1\).[/tex] Let \(p = m(k+1)\), where m is a positive integer.
Substituting this into the equation, we have [tex]\((m(k+1))^2 = (k+1)q^2\)[/tex].
Simplifying, we get [tex]\(m^2(k+1) = q^2\).[/tex]
This implies that [tex]\(q^2\)[/tex] is a multiple of [tex]\(k+1\)[/tex], which means [tex]\(q\)[/tex] is also a multiple of [tex]\(k+1\).[/tex]
However, this contradicts our assumption that p and q have no common factors other than 1.
Hence, our assumption that [tex]\(\sqrt{k+1}\)[/tex] is rational must be false.
Therefore, [tex]\(\sqrt{k+1}\)[/tex] is irrational.
By the principle of strong induction, we can conclude that the square root of 18 is irrational.
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PLEASE PLEASE PLEASE HELP 7 points
Answer:
a) 2x+(x+36)=90
Step-by-step explanation:
b) A1+A2=90°. (A=angle)
2x+(x+36)=90
2x+x+36=90
3x+36=90
3x=90-36
x=54/3
x=18
then A1=2x=2*18=36°
A2=x+36=18+36=54°
If I fail this homework, I may get beat.
Answer:
Don't click the link its a virus
Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30.
a. Find the probability that the number of successes is greater than 310.
P(X ˃ 310) = _____ (round to four decimal places as needed and show work)
b. Find the probability that the number of successes is fewer than 250.
P(X ˂ 250) = _____ (round to four decimal places as needed and show work)
P(X < 250) = P(X ≤ 249) = 0 (approximately) Hence, P(X ˃ 310) = 0 and P(X ˂ 250) = 0.
Given a random sample size of n = 900 from a binomial probability distribution with P = 0.30. The probability that the number of successes is greater than 310 and the probability that the number of successes is fewer than 250 are to be found.
Solution: a)We know that P(X > 310) can be found using normal approximation.
We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.
Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630. Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.
Using the normal approximation formula, z = (X - μ) / σwhere X = 310, μ = np and σ = √(npq), we getz = (310 - 270) / √(900*0.30*0.70)z = 4.25
Using the z-table, the probability of z being greater than 4.25 is almost zero.
Therefore, P(X > 310) = P(X ≥ 311) = 0 (approximately)
b)We know that P(X < 250) can be found using normal approximation. We have to check whether np and nq are greater than or equal to 10 or not, where p=0.30, q=0.70 and n=900.
Here, np = 900*0.30 = 270 and nq = 900*0.70 = 630.
Since np and nq are greater than or equal to 10, we can use normal approximation for this binomial distribution.
Using the normal approximation formula,z = (X - μ) / σwhere X = 250, μ = np and σ = √(npq), we getz = (250 - 270) / √(900*0.30*0.70)z = -4.25Using the z-table, the probability of z being less than -4.25 is almost zero.
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Given data: n = 900, P = 0.30.
a. The probability that the number of successes is greater than 310 is 0.0000.
b. The probability that the number of successes is fewer than 250 is 0.0174.
a. The formula for finding probability of binomial distribution is:
P(X > x) = 1 - P(X ≤ x)
P(X > 310) = 1 - P(X ≤ 310)
Mean μ = np
= 900 × 0.30
= 270
Variance σ² = npq
= 900 × 0.30 × 0.70
= 189
Standard deviation
σ = √σ²
= √189
z = (x - μ) / σ
z = (310 - 270) / √189
z = 4.32
Using normal approximation,
P(X > 310) = P(Z > 4.32)
= 0.00001673
Using calculator, P(X > 310) = 0.0000(rounded to four decimal places)
b. P(X < 250)
Mean μ = np
= 900 × 0.30
= 270
Variance σ² = npq
= 900 × 0.30 × 0.70
= 189
Standard deviation
σ = √σ²
= √189
z = (x - μ) / σ
z = (250 - 270) / √189
z = -2.12
Using normal approximation, P(X < 250) = P(Z < -2.12) = 0.0174.
Using calculator, P(X < 250) = 0.0174(rounded to four decimal places).
Therefore, the probability that the number of successes is greater than 310 is 0.0000 and the probability that the number of successes is fewer than 250 is 0.0174.
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Please help :) thank you to whoever does help!?
____________________________
1.) Fill in the formula: F = [tex]B^2-C^2=A^2[/tex]F = [tex]12^2-13^2=A^2[/tex]F = [tex]144-169=A^2[/tex]2.) Subtract:[tex]169-144=25[/tex]3.) Find the square root of 25:[tex]\sqrt{25} =5[/tex]So your answer would be C, [tex]\bold{a=5}[/tex].
____________________________
What are the solutions of the system?
Sy=2x+6
\y= r2 +5 +6
O (0, -3) and (6, 0)
O(-3,0) and (-2, 0)
O (-3, 0) and (0, 6)
O
(0, 6) and (-2, 0)
ہے
1 2
The solution to the system is {eq}(x, y) = (2, 1){/eq}. For the given system, we have the solutions (-3, 0) and (0, 6).
To solve a system of equations, we must find the value of each variable that satisfies both equations. One of the most common methods for solving systems of equations is called substitution.
In substitution, we solve for one variable in one equation and then plug that expression into the other equation.
For example, if we have the system {eq}2x + y = 5,
\quad 4x - y = 7 {/eq},
we can solve for y in the first equation: {eq}y = 5 - 2x {/eq} Then we substitute this expression for y into the second equation:
{eq}4x - (5 - 2x) = 7 {/eq}
Simplifying gives {eq}6x = 12 {/eq}, or {eq}x = 2 {/eq} Once we have a value for one variable, we can substitute it into either equation to find the value of the other variable.
Using {eq}y = 5 - 2x {/eq}, we have {eq}y = 5 - 2(2) = 1 {/eq}These solutions represent the values of x and y that satisfy both equations in the system. To check,
we can substitute each solution into both equations to ensure they are valid. If both equations are satisfied, then we have found the correct solutions.
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Let k be a constant and consider the function f(x,y,z) = kx? - kry + y2 -2yz - 22. (Thus, for example, if k = 4, then f(xy.z) = 4x2 - 4xy + 2y2-2yz -22) For what values (if any) of the constant k does / have a (nondegenerate) local maximum at (0.0.0)? For what values of k does / have a (nondegenerate) local minimum at (0.0.0)? Be sure to explain your reasoning.
The values of k for which the function f(x, y, z) = kx² - kry + y² - 2yz - 22 has a nondegenerate local maximum at (0, 0, 0) are when k > 0.
To find the critical points of the function, we need to calculate the partial derivatives with respect to each variable:
∂f/∂x = 2kx ∂f/∂y = -kr + 2y - 2z ∂f/∂z = -2y
2kx = 0 => x = 0 (Equation 1) -kr + 2y - 2z = 0 => r = y - z (Equation 2) -2y = 0 => y = 0 (Equation 3)
From Equation 3, we can see that y = 0. Substituting this into Equation 2, we get:
r = 0 - z r = -z (Equation 4)
The Hessian matrix is given by:
H = | ∂²f/∂x² ∂²f/∂x∂y ∂²f/∂x∂z | | ∂²f/∂y∂x ∂²f/∂y² ∂²f/∂y∂z | | ∂²f/∂z∂x ∂²f/∂z∂y ∂²f/∂z² |
Calculating the second-order partial derivatives:
∂²f/∂x² = 2k ∂²f/∂y² = 2 ∂²f/∂z² = 0 ∂²f/∂x∂y = 0 ∂²f/∂y∂z = -2 ∂²f/∂z∂x = 0
Thus, the Hessian matrix becomes:
H = | 2k 0 0 | | 0 2 -2 | | 0 -2 0 |
D = ∂²f/∂x² ∂²f/∂y² ∂²f/∂z² + 2∂²f/∂x∂y ∂²f/∂y∂z ∂²f/∂z∂x - (∂²f/∂x² ∂²f/∂y∂z ∂²f/∂z∂x + ∂²f/∂y² ∂²f/∂z∂x ∂²f/∂x∂y ∂²f/∂z²)
Substituting the partial derivatives we calculated earlier:
D = (2k)(2)(0) + 2(0)(-2)(0) - (2k)(-2)(0) - (2)(0)(0) D = 0
If the determinant D is zero, the second derivative test is inconclusive. In such cases, we need to consider the eigenvalues of the Hessian matrix.
To find the eigenvalues, we solve the characteristic equation:
det(H - λI) = 0
where λ is the eigenvalue and I is the identity matrix. Substituting the values from the Hessian matrix:
| 2k-λ 0 0 | | 0 2-λ -2 | | 0 -2 -λ |
The characteristic equation becomes:
(2k - λ)((2 - λ)(-λ) - (-2)(0)) - (0)((2 - λ)(-2) - (0)(0)) = 0 (2k - λ)(λ² - 2λ) = 0
From this equation, we can see that one eigenvalue is (2k - λ) = 0, which implies λ = 2k.
For our case, we have one eigenvalue (λ = 2k). Thus, the sign of λ depends on the value of k.
When k < 0, the point (0, 0, 0) is a nondegenerate local minimum. When k = 0, the second derivative test is inconclusive, and further analysis would be required to determine the nature of the critical point.
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A data point far from the mean of both the x's and y's is always:
a) an influential point and an outlier
b) a leverage point and an influential point
c) an outlier and a leverage point
d) None of the above
The correct answer is c) an outlier and a leverage point.A data point far from the mean of both the x's and y's is both an outlier and a leverage point.
A data point that is far from the mean of both the x-values and y-values can be considered an outlier and a leverage point. An outlier is a data point that significantly deviates from the overall pattern of the data. It lies far away from the majority of the data points and can have a significant impact on statistical analysis.
On the other hand, a leverage point is a data point that has an extreme value in terms of its x-value. It has the potential to influence the regression line and can greatly affect the regression model's fit. Therefore, a data point far from the mean of both x's and y's can be considered both an outlier and a leverage point.
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(4c+3)+(5b+8) simplify this answer
Answer:
4c + 5b + 11
Step-by-step explanation:
4c + 3 + 5b + 8
4c + 5b + 11
a cylinder has a volume of 500cm³ and a diameter of 18cm. which of the following is the closest to the height of the cylinder
Step-by-step explanation:
Volume of Cylinder =
[tex]500 {cm}^{3} = \pi {r}^{2} h[/tex]
given d = 18
r = 1/2 x d = 9cm,
[tex]\pi( {9}^{2} )h = 500 \\ 81\pi \: h = 500 \\ h = \frac{500}{81\pi} cm[/tex]
I will leave the answer in terms of Pi as I am not sure how you want to leave your answer as.
ANSWERRRRRR :puppy eyes emoji:
Answer:
A rhombus is a special type of parallelogram
Step-by-step explanation:
Similarities in a rhombus and parallelogram:
Both pairs of opposite sides are equal and parallel.Diagonals bisect each other and are unequal.Opposite angles are equal.None of the angles is 90 degrees.A rhombus is a special type of parallelogram in that
All its sides are equal.The diagonals bisect each other at right angles.Each diagonal bisects the angle at the vertices.