Describe the end behavior of the following function:
F(x)=x^5-x^3+x²
A. The graph of the function starts high and ends high.
B. The graph of the function starts low and ends low.
C. The graph of the function starts high and ends low.
D. The graph of the function starts low and ends high.

Answer:

B. 126 [tex]\pi[/tex]

Step-by-step explanation:

The volume for a cylinder is: Area of the base x Height

The base is a circle with a diameter of 6cm.

The formula to find the area of the base is: radius x radius x pi

In this case, the radius is 3 (6/2) so the area will be 9pi.

Now that we know that area of the base, all we need to do is multiply the height, which is 14cm.

9 pi x 14 = 126 pi

So B. 126 pi is the correct answer!

Answer:

126π

option B is the correct answer

Using the Geometric mean theorem and the Pythagorean theorem:

x = 3√3, and y = 6.

The Geometric mean theorem or the right triangle altitude theorem states that the geometric mean of the the two segments equals he length of the altitude of a right triangle.

The geometric mean theorem is expressed by the equation, h = √(ab), where:

Given the following:

h = x

a = 9

b = 3

x = √(9 × 3)

x = 3√3

Find y using the Pythagorean theorem

Based on the Pythagorean theorem, we would have the following:

y = √(x² + 3²)

y = √((3√3)² + 3²)

y = √(27 + 9)

y = √(36)

y = 6

Thus, in the simplest radical form, the value of x = 3√3, and y = 6.

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Answer:

Wayne used 2/5 more flour than sugar

Step-by-step explanation:

If you add 2/5 to 2/5 you get 4/5

Based on the type I error, if a hypothesis test leads to the probability of a type I error decreasing, then the a. probability of incorrectly accepting the null hypothesis decreases.

A type I error refers to when we are are engaged in a hypothesis test and end up rejecting the Null hypothesis even though it is true.

If we are in hypothesis tests and the probability of a type 1 error decreases, then it means that the probability of us correctly rejecting the null hypothesis when it is false increases.

It also means that the probability of us correctly accepting the null hypothesis when it is true increases as well.

This can further be said as the probability of us accepting the null hypothsis when it is false, decreases. There are therefore less chances of us incorrectly accepting the null hypothesis.

Options for this question include:

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Answer:

35%

Step-by-step explanation:

All the prime numbers from 1 to 20 are:

2 3 5 7 11 13 17 19

7 / 20 = 35 / 100 = 35%

Answer:

First we write y and its derivatives as power series:

y=∑n=0∞anxn⟹y′=∑n=1∞nanxn−1⟹y′′=∑n=2∞n(n−1)anxn−2

Next, plug into differential equation:

(x+2)y′′+xy′−y=0

(x+2)∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

x∑n=2∞n(n−1)anxn−2+2∑n=2∞n(n−1)anxn−2+x∑n=1∞nanxn−1−∑n=0∞anxn=0

Move constants inside of summations:

∑n=2∞x⋅n(n−1)anxn−2+∑n=2∞2⋅n(n−1)anxn−2+∑n=1∞x⋅nanxn−1−∑n=0∞anxn=0

∑n=2∞n(n−1)anxn−1+∑n=2∞2n(n−1)anxn−2+∑n=1∞nanxn−∑n=0∞anxn=0

Change limits so that the exponents for  x  are the same in each summation:

∑n=1∞(n+1)nan+1xn+∑n=0∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−∑n=0∞anxn=0

Pull out any terms from sums, so that each sum starts at same lower limit  (n=1)

∑n=1∞(n+1)nan+1xn+4a2+∑n=1∞2(n+2)(n+1)an+2xn+∑n=1∞nanxn−a0−∑n=1∞anxn=0

Combine all sums into a single sum:

4a2−a0+∑n=1∞(2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an)xn=0

Now we must set each coefficient, including constant term  =0 :

4a2−a0=0⟹4a2=a0

2(n+2)(n+1)an+2+(n+1)nan+1+(n−1)an=0

We would usually let  a0  and  a1  be arbitrary constants. Then all other constants can be expressed in terms of these two constants, giving us two linearly independent solutions. However, since  a0=4a2 , I’ll choose  a1  and  a2  as the two arbitrary constants. We can still express all other constants in terms of  a1  and/or  a2 .

an+2=−(n+1)nan+1+(n−1)an2(n+2)(n+1)

a3=−(2⋅1)a2+0a12(3⋅2)=−16a2=−13!a2

a4=−(3⋅2)a3+1a22(4⋅3)=0=04!a2

a5=−(4⋅3)a4+2a32(5⋅4)=15!a2

a6=−(5⋅4)a5+3a42(6⋅5)=−26!a2

We see a pattern emerging here:

an=(−1)(n+1)n−4n!a2

This can be proven by mathematical induction. In fact, this is true for all  n≥0 , except for  n=1 , since  a1  is an arbitrary constant independent of  a0  (and therefore independent of  a2 ).

Plugging back into original power series for  y , we get:

y=a0+a1x+a2x2+a3x3+a4x4+a5x5+⋯

y=4a2+a1x+a2x2−13!a2x3+04!a2x4+15!a2x5−⋯

y=a1x+a2(4+x2−13!x3+04!x4+15!x5−⋯)

Notice that the expression following constant  a2  is  =4+  a power series (starting at  n=2 ). However, if we had the appropriate  x -term, we would have a power series starting at  n=0 . Since the other independent solution is simply  y1=x,  then we can let  a1=c1−3c2,   a2=c2 , and we get:

y=(c1−3c2)x+c2(4+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(4−3x+x2−13!x3+04!x4+15!x5−⋯)

y=c1x+c2(−0−40!+0−31!x−2−42!x2+3−43!x3−4−44!x4+5−45!x5−⋯)

y=c1x+c2∑n=0∞(−1)n+1n−4n!xn

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The answer choices which represents the slope of the line described by the points, (1,-1) and (5,-7) is; Choice A; -3/2.

The line in discuss as described in the task content is passing through the points; (1,-1) and (5,-7).

Hence, it follows conventionally from the slope formula that the slope of the line is;

slope, m = (-7-(-1))/(5-1)

m = -6/4

m = -3/2.

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The range of the means is 0.5

Mean of S1

= 5 + 6+ 6+ 2

= 19/4

= 4.75

Mean of S2

= 4 + 8 + 4 + 3

= 19/4

= 4.75

Mean of s3

= 4 + 2 + 6+ 5

= 17/4

= 4.25

Range of sample means = 4.75 - 4.25

= 0.5

c. the true statements here is that

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Answer:

g(2) = 16; g(-2) = 4; g(a) = 3a + 10; 3(a+2) = 3a + 16

Step-by-step explanation:

We want to substitute the value in the parentheses on the left-hand side for the value of x on the right-hand side

Thus, g(2) = 3(2) + 10 = 6 + 10 = 16

g(-2) = 3(-2) + 10 = -6 + 10 = 4

g(a) = 3(a) + 10 = 3a + 10

g(a + 2) = 3(a + 2) + 10 = 3a + 6 + 10 = 3a + 16

The expression [tex]$-500 \times 18+15000=-9000+15000=6000[/tex] which best fit exists in the Linear model.

Given: Monthly Rate = $20

Number of customers = 5000

If there exists a decrease of $1 in the monthly rate, the number of customers increases by 500.

Let us decrease the monthly rate by $1.

Monthly Rate = $20 - $1 = $19

Number of customers = 5000 + 500 = 5500

Let us decrease the monthly rate by $1 more.

Monthly Rate = $19 - $1 = $18

Number of customers = 5500 + 500 = 6000

Linear change in the number of customers whenever there exists a decrease in the monthly rate.

We have 2 pairs of values here,

x = 20, y = 5000

x = 19, y = 5500

The equation in slope-intercept form: y = mx + c

The slope of a function: [tex]$&m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\[/tex]

[tex]$&m=\frac{5500-5000}{19-20} \\[/tex]

[tex]$&\Rightarrow-500[/tex]

So, the equation is y = -500x + c

Putting x = 20, y = 5000:

[tex]$&5000=-500 \times 20+c \\[/tex]

[tex]$&\Rightarrow c=5000+10000=15000 \\[/tex]

[tex]$&\Rightarrow \mathbf{y}=-500 \mathbf{x}+15000[/tex]

Whether (18,6000) satisfies it.

Putting x = 18

[tex]$-500 \times 18+15000=-9000+15000=6000[/tex]

Therefore, the expression [tex]$-500 \times 18+15000=-9000+15000=6000[/tex] which best fits exist Linear model.

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Answer:

39C7 or 15380937

Step-by-step explanation:

There are 13 clubs in a 52 card deck. We can remove these from the possibilities. 52-13=39. The rest is simple, choose 7 from 39 using the choose function is 39!/(32!*7!)=15380937. Or just say 39C7 as your answer.

Answer:

Step-by-step explanation:

overall the answer is wrong i say it is the first one  

Answer:

The first one is theost plausible because root 16 is 4, 30 + 4 is 34, the actual answer is 25.38

Answer:

b

Step-by-step explanation:

the opposite sides of a parallelogram are parallel and congruent

thus option b is the correct one

The factor of [tex]$a^{2} b+2 a^{2}+3 b+6$[/tex] exists [tex]$\left(a^{2}+3\right)(b+2)$$[/tex].

Let the given factor be [tex]$a^{2} b+2 a^{2}+3 b+6$[/tex]

To factor this we use the grouping method

We group the first two terms and last two terms then we factor out the greatest common factor (GCF) from each group.

[tex]$\left(a^{2} b+2 a^{2}\right)+(3 b+6)$$[/tex]

Take out GCF from each group

[tex]$a^{2}(b+2)+3(b+2)$$[/tex]

Now factor out b+2, we get

[tex]$\left(a^{2}+3\right)(b+2)$$[/tex]

The factor of [tex]$a^{2} b+2 a^{2}+3 b+6$[/tex] exists [tex]$\left(a^{2}+3\right)(b+2)$$[/tex].

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Using a linear function, it is found that they will have 405 MB of free space when they first have the same amount of space left.

A linear function is modeled by:

y = mx + b

In which:

The free space for each of Nereo and Azeneth can be modeled by a linear function, in which the slope is negative with the amount that each second of video takes and the y-intercept is the initial free space. Her the functions for the amount of free space after x seconds are given by:

They will have the same amount of free space when:

N(x) = A(x)

1000 - 7x = 1255 - 10x

3x = 255

x = 255/3

x = 85.

The space will be of:

N(85) = 1000 - 7 x 85 = 405 MB.

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Answer:

y= 4

Step-by-step explanation:

(y-4) (y+7)=0

(4-4) (4+7)=0

(0) (11)=0

11*0=0

Measures of central tendency are called such because they all find approximately the same "center" of a data set (option B).

A measure of central tendency attempts to describe a data set by determining the central value of the data set. Measures of central tendency are mean, median and mode

Mean is the average of a set of numbers. It is determined by adding the numbers together and dividing it by the total number.

Median can be described as the number that occurs in the middle of a set of numbers that are arranged either in ascending or descending order.

Mode refers to a value that appears most frequently in a data set.

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1) The Inequalities for Tammy and Keith are respectively; For Tammy: 10x + 5y > 200 and for Keith: 12x + 3y > 180

2) The possible combination of the number of lawns they could have each mowed and the number of yards they could have raked are;

(11, 17), (14, 5), (16, 0), (18, 2)

Tammy earns $10 for each lawn mowed.

Tammy earns $5 for each yard raked.

We are told that She wants to earn more than $200 from her part-time jobs. Thus, the inequality for Tammys would be;

10x + 5y > 200

Keith earns $12 for each lawn mowed

Keith earns $3 for each yard raked.

We are told that Keith wants to earn more than $180 for each part time job. Thus, the inequality is;

12x + 3y > 180

B) To know the possible combination of the number of lawns they could have each mowed and the number of yards they could have raked, i have drawn a graph of both inequalities and the possible points from there are;

(11, 17), (14, 5), (16, 0), (18, 2)

The complete question is;

Tammy and Keith each work two part-time jobs in the summer mowing lawns andraking yards . Tammy earns $10 for each lawn she mows and $5 for each yard sherakes . She wants to earn more than $200 from her part-time jobs . Keith earns $12 for each lawn he mows and $3 for each yard he rakes. He wants to earn more than $180 for each part time job.

A) Write a system of Inequalities to model the number of lawns they each mow (x) and the number of yards they each rake.

B.) What is a possible combination of the number of lawns they could have each mowed and the number of yards they could have raked? Explain how you chose the combination and why you know that combination is correct.

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Step-by-step explanation:

I answered this already yesterday.

the composite figure is actually the combination of 2 figures :

1. a 7cm × 6cm × 2cm box (purple)

2. a 8cm × 7cm × 6cm triangular shaped half-box (pink) with 10cm length of the rectangular "roof".

2 sides are completely blocking each other, so they are not part of the combined surface area.

let's start with the purple box. its contribution to the surface area is :

top and bottom 7×2 rectangles

front and back 6×2 rectangles

no left (blocked by the half-box)

right 6×7 rectangle

so, we get

2 × 7×2 = 2×14 = 28 cm²

2 × 6×2 = 2×12 = 24 cm²

6×7 = 42 cm²

in total that is : 94 cm²

the half-box contributes to the surface area :

top 10×7 rectangle

bottom 8×7 rectangle

front and back 8×6/2 triangles

no left (due to the triangular shape)

no right (blocked by the box)

so, we get

10×7 = 70 cm²

8×7 = 56 cm²

2 × 8×6/2 = 2×24 = 48 cm²

in total that is : 174 cm²

and so, the total surface area of the composite figure is

174 + 94 = 268 cm²

The solution of the assumed equation is

θ = 135 + 360k

and

θ = -45 + 360k (or 315 + 360k)

Assuming the equation is

csc^2(θ) = 2cot(θ) + 4

and not

Assuming the equation to be:

csc^2(θ) = cot^2(θ) + 1

Solving these equations usually begins with algebra and/or trigonometry. ID for transforming equations to have one or more equations of the form:

trigfunction(expression) = number

It is not always easy to understand how to perform the desired transformation. If it's not obvious, first use the trigonometric identity to reduce the number of different arguments or functions in the equation. In this expression, all arguments are θ. Therefore, there is no need to reduce the number of arguments. But he has two different functions, csc and cot.

csc^2(θ) = cot^2(θ) + 1

Substituting the right side of this equation into the left side of the equation, we get: :

cot^2 (θ) + 1 = 2 cot(θ) + 4

Now that we have only one function cot and one argument θ, we are ready to find the form we need. Subtracting the entire right-hand side from both sides gives:

cot^2(θ) - 2cot(θ) - 3 = 0

The left-hand side factorizes:

(cot(θ)-3)(cot (θ) ) + 1 ) = 0

Using the properties of the zero product,

cot(θ) = 3 or cot(θ) = -1

These two equations are now in the desired form.

The next step is to write the general solution for each equation. The general solution represents all solutions of the equation.

cot(θ) = 3

Note that 3 is not a specific angle value for cot. That's why you need a calculator. Your computer probably doesn't have the crib button, so you'll need to switch it to tan

Since tan is the reciprocal of cot, if cot = 3...

tan(θ) = 1/3

Inverse tan tan^-1(1/3) can be used to find the reference angle. You should get a reference angle of 18.43494882 degrees. Using this reference angle and cot (and tan) being positive in the 1st and 3rd quadrants, we get the general solutions

θ = 18.43494882 + 360k

and

θ = 180 + 18.43494882 + 360k

.

θ = 198.43494882 + 360k

where

cot(θ) = -1

-1 should be recognized as a special angle value for cot. So you don't need a calculator. This reference angle is 45 degrees. Using this reference angle, cot is negative in the 2nd and 4th quadrants, so

θ = 180 - 45 + 360k

and

θ = -45 + 360k (or 360 - 45 + 360k) the general solution for

must get. to:

θ = 135 + 360k

and

θ = -45 + 360k (or 315 + 360k)

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A number line ranges from minus 5 to 5 with increment of 1 unit. Left, Point F is plotted on the number line at minus 3. Right, point G is plotted on the number line at 1.50, then the distance between F and G is 4.50.

Calculating the Distance between F and G:

It is given that, on a number line,

Point F is located at -3

Point G is at 1.50

Since, 1.5 > (-3) and each number is at a 1 unit difference form the adjacent number on the number line,

The difference between F and G = 1.50 - ( -3 )

This implies that the distance between F and G = 1.50 + 3

= 4.50 units

The numbers lying between the points F and G on the number line are: -2, -1, 0, and 1

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Answer: 4 1/2

Step-by-step explanation:

Got it right

The value added to the equation [tex]$x^2-6x=1[/tex] exists [tex]$x^2-6x+9=10[/tex].

A perfect square exists as a number that can be described as the product of an integer by itself or as the second exponent of an integer.

The perfect square trinomial exists

[tex](a[/tex] ± [tex]b)^2[/tex] = [tex]a ^2[/tex] ± 2ab + [tex]b ^2[/tex]

[tex]$x^2-6x=1[/tex]

[tex]x^2-2*3x=1[/tex]

then [tex]$2ab = 2 * 3*x = 2 * x *3[/tex]

The value of a = x and b = 3

[tex]$b^2=3^2=9[/tex]

[tex]$x^2-6x+9=1+9[/tex]

[tex]$x^2-6x+9=10[/tex]

The value added to the equation [tex]$x^2-6x=1[/tex] exists [tex]$x^2-6x+9=10[/tex].

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For starters,

[tex]\cos(6t+\pi) = \cos(6t) \cos(\pi) - \sin(6t) \sin(\pi) = -\cos(6t)[/tex]

Now by the fundamental theorem of calculus, integrating both sides gives

[tex]\displaystyle \frac{ds}{dt} = s'(0) + \int_0^t 36 \cos(6u) \, du = 100 + 6 \sin(6t)[/tex]

Integrating again, we get

[tex]\displaystyle s(t) = s(0) + \int_0^t (100 + 6\sin(6u)) \, du = \boxed{100t - \cos(6t) + 1}[/tex]

Alternatively, you can work with antiderivatives, then find the particular constants of integration later using the initial values.

[tex]\displaystyle \int \frac{d^2s}{dt^2} \, dt = \int 36\cos(6t) \, dt \implies \frac{ds}{dt} = 6\sin(6t) + C_1[/tex]

[tex]\displaystyle \int \frac{ds}{dt} \, dt = \int (6\sin(6t) + C_1) \, dt \implies s(t) = -\cos(6t) + C_1t + C_2[/tex]

Now,

[tex]s(0) = 0 \implies 0 = -1 + C_2 \implies C_2 = 1[/tex]

and

[tex]s'(0) = 100 \implies 100 = 0 + C_1 \implies C_1 = 100[/tex]

Then the particular solution to the IVP is

[tex]s(t) = -\cos(6t) + 100t + 1[/tex]

just as before.

Answer:

B. Property right are assigned to the party who is being

Answer:

=> -2.22

Step-by-step explanation:

=> 2.35 + ( -4.57)

=> 2.35 - 4.57

=> - 2.22

The five number summary and interquartile range for the data-set is given by:

The five number summary is composed by the minimum and maximum value, and the first quartile, median and third quartile. As for each of these data, they are explained below.

In this problem, we have that:

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Answer:

They are the same

Step-by-step explanation:

LA cyl =

[tex]2\pi \: rh[/tex]

SA sphere =

[tex]4\pi {r}^{2} [/tex]

If the sphere is inscribed in the cylinder, they have the same radius. The height of the cylinder is the diameter (Or 2× radius) of the sphere.

If you substitute 2r for h, then both formulas are

[tex] 4\pi{r}^{2} [/tex]