Is the graph shown below that of a function?

Is The Graph Shown Below That Of A Function?

Answers

Answer 1

Answer:

yes

Step-by-step explanation:


Related Questions

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

Answers

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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Find the value of x in the picture below. (round to nearest tenth if needed) THANK YOU FOR HELPING ME:)

Answers

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

what's the distance between theses two. (-5, 1) (2, 4)

Answers

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.

Helppp I really need this question answered

Answers

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 ^^

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.

Answers

[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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The radius of a circle is 8 inches. What is the area?

r=8 in

Give the exact answer in simplest form.

Answers

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....

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​

Answers

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.

(4c+3)+(5b+8) simplify this answer

Answers

Answer:

4c + 5b + 11

Step-by-step explanation:

4c + 3 + 5b + 8

4c + 5b + 11

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

Answers

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

What is the value of x in the equation 13−2(+4)=8+1 13 x − 2 ( x + 4 ) = 8 x + 1 ?

Answers

Answer: x= 12/103

alternate form x =0.116505

Step-by-step explanation:

Try Photo math! It explains step by step!

Hope this helps!!!!

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​

Answers

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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If I fail this homework, I may get beat.

Answers

Answer:

Don't click the link its a virus

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

Answers

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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ANSWERRRRRR :puppy eyes emoji:

Answers

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.

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.

Answers

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 images

From 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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What is the value of y in the equation 4 + y=-3?
7
1
-1
-7

Answers

Answer: -7

Because if y is -7 then the equation would be 4 - 7 = -3 which is true :)

Answer this question to get marked as barinliest!!!!

Answers

Area= units to the second
Volume= units to the third
Circumference= units to the second
Circumferences just units

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

Answers

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.

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

Answers

For scientific tools to be measured at such a height...you must first begin multiplication and do the simple division.

Please help :) thank you to whoever does help!?

Answers

Answer:[tex]\boxed{\boxed{\sf{a=5}}}[/tex]Solution Steps:

____________________________

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].

____________________________

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)

Answers

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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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.

Answers

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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There are 96 children in a room.
40 of them are girls.
Find the fraction of the children that are boys.
Write your answer in its simplest form.
I will need it by this midnight

Answers

Answer:

7/12

Step-by-step explanation:

Basically to find the boys fraction I subtracted 40 from 96 and got 56. Your answer for the number of boys would be 56/96. I divided 4 from each and got 14/24. lastly I divide by 2 and got 7/12.

The blue segments below is a diameter of oo. What is the length of the radius of the circle?

Answers

Answer:

C. 5.1 units

Step-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]

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​

Answers

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

Answers

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, 20

What 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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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.

Answers

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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the other one i need help with asap!!

Answers

Answer:

(3x-2)(2x-3)

Step-by-step explanation:

Good luck!

Which comparison is not correct?

-2 > -7
1 < -9
-3 > -8
6 > 5

Answers

Answer:

1 < -9

Step-by-step explanation:

A positive number can't be less than a negative

1 < -9
.......................

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?

Answers

Answer:

Step-by-step explanation:

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