Can someone please help me with my math homework (m^2+2m-8)/(m+4) thanks and have a good day/night/evening.

Answer: C. 2x+3

Step-by-step explanation:

[tex]\displaystyle\\V=\frac{\pi r^2h}{3} \\\\V=\frac{\pi r^2(6)}{3}\\\\V=2\pi r^2\\\\\\8\pi x^2+24\pi x+18\pi \\\\V=2\pi (4x^2+12\pi +9)\\\\V=2\pi ((2x)^2+2(2x)(3)+3^2)\\\\V=2\pi (2x+3)^2\\\\THUS,\\[/tex]

[tex]2\pi r^2=2\pi (2x+3)^2[/tex]

Divide both parts of the equation by 2π:

[tex]r^2=(2x+3)^2[/tex]

Extract the square root of both parts of the equation:

[tex]r=2x+3[/tex]

They each will pass 12 scales.

What do you mean by algebraic expression?

The concept of algebraic expressions is the use of letters or alphabets to represent numbers without providing their precise values. We learned how to express an unknown value using letters like x, y, and z in the fundamentals of algebra. Here, we refer to these letters as variables. Variables and constants can both be used in an algebraic expression. A coefficient is any value that is added before a variable and then multiplied by it.

It is given that Becky has passed 4 scales, whereas her friend Charlotte has passed 8 of them.

Going forward, Becky has committed to passing 4 scales per week, and Charlotte has committed to passing 2 per week.

Let number of weeks be x

Becky has passed 4x + 4

Charlotte has passed 2x + 8

According to question:

4x + 4 = 2x + 8

4x - 2x = 8 - 4

2x = 4

x = 4/2 = 2

Becky has passed 4x + 4 = 4(2) + 4 = 8 + 4 = 12

Charlotte has passed 2x + 8 = 2(2) + 8 = 4 + 8 = 12

Therefore, They each will pass 12 scales.

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The Maclaurin series of  ln (1 + x³)  is

[tex]= x^3-\dfrac{1}{2}x^6+\frac{1}{3}x^9+\ldots \:[/tex]

The function is given in the question as

h(x) = ln (1 + x³)

Taylor (Maclaurin) series ofln (1 + x³) up to n = 3

A Maclaurin series is given by f(x),

[tex]f\left(x\right)=f\left(a\right)+\frac{f^'\left(a\right)}{1!}\left(x-a\right)+\frac{f^{''}\left(a\right)}{2!}\left(x-a\right)^2+\frac{f^{'''}\left(a\right)}{3!}\left(x-a\right)^3+\ldots[/tex]

[tex]f\left(x\right)=f\left(0\right)+\frac{f^'\left(0\right)}{1!}\left(x\right)+\frac{f^{''}\left(0\right)}{2!}\left(x\right)^2+\frac{f^{'''}\left(0\right)}{3!}\left(x\right)^3+\ldots[/tex]

We need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

f⁽⁰⁾(x) = f(x) = ln (1 + x³)

Evaluate the function at the point: f(0)=0

1st derivative: f(1)(x)=(ln(x3+1))′=3x²/(x³+1)  

1st derivative at the given point: (f(0))′=0

2nd derivative: f(2)(x)=(f(1)(x))′=(3x²/(x³+1) )′=3x(2−x³)/(x⁶+2x³+1) 

Evaluate the 2nd derivative at the given point: (f(0))′′=0

3rd derivative: f(3)(x)=(f(2)(x))′= (3x(2−x³)/(x⁶+2x³+1) )′= 6(x⁶−7x³+1)x9+3x⁶+3x³+1

Evaluate the 3rd derivative at the given point: (f(0))′′′=6

Now, use the calculated values to get a polynomial:

f(x)≈0/0!x⁰+0/1!x¹+0/2!x²+6/3!x³+0/4/!x⁴+0/5!x⁵

Thus, the Maclaurin series of  ln (1 + x³)  is

[tex]= x^3-\dfrac{1}{2}x^6+\frac{1}{3}x^9+\ldots \:[/tex]

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Answer:

A = 29°

Step-by-step explanation:

set

(3x-4) + (9x+17) + (6x-31) = 180

Solve for x

18x - 18= 180

18x = 198

x = 11

So A will be

A = 3x - 4

A = 3(11) - 4

A = 29°

Answer:

Step-by-step explanation:

The length and width of a rectangle are 36 and 30 yard respectively.

Its the sum of length of the sides used to made the given figure.

We are given that width of a rectangle is 6 yards less than the length. The perimeter is 132 yards.

Therefore, the perimeter = (2 x length)+(2 x width)

The length = 6 + w

Hence, we have;

132 = (2(6 + w))+(2w)

132 = 12 + 2w + 2w

132 = 12 + 4w

120 = 4w

w = 30

Length = 36 yard.

Thus, the length and width of a rectangle are 36 and 30 yard respectively.

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Answer:

(1, -2)

Step-by-step explanation:

The line is tangent to the parabola, so it intersects in only 1 point.

Answer: (1, -2)

Answer:

The probability of a head of a die is 1/6 and that of a side of a coin

Answer: 54

Step-by-step explanation: 3 x 30seconds = 90 seconds (1 1/2)

3 x 18 = 54

The berries is the bowl is 6 berries fewer than the original amount

The amount of berries in the bowel is solved using addition and subtraction property of equality

36 berries are in a bowl James eats 21 of the berries

= 35 berries - 21 berries

= 15 berries

he put 14 more berries in the bowl

= 15 berries + 14 berries

= 29 berries

How many fewer is solved by

= 35 berries - 29 berries

= 6 berries

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Answer: The answer is 2

Step-by-step explanation:

3j + 4 = 10  J = 2

3x2=6

6+4= 10

[tex]\begin{array}{llll} ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} \end{array} ~\hfill \begin{array}{llll} \textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \end{array} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{6x^{-1}y^{-1}}{\sqrt[4]{xy^4}}\implies \cfrac{6x^{-1}y^{-1}}{\left( xy^4 \right)^{\frac{1}{4}} }\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^{4\cdot \frac{1}{4}}}\implies \cfrac{6x^{-1}y^{-1}}{x^{\frac{1}{4}} y^1}\implies \cfrac{6x^{-1}y^{-1}x^{-\frac{1}{4}} y^{-1}}{1}[/tex]

[tex]6x^{-1}x^{-\frac{1}{4}}y^{-1} y^{-1}\implies 6x^{-1-\frac{1}{4}} y^{-1-1}\implies {\Large \begin{array}{llll} \underset{\textit{\small a}}{6} \stackrel{\textit{\small ~~ b}}{x^{-\frac{5}{4}}} \stackrel{\textit{\small ~~ c}}{y^{-2}} \end{array}}[/tex]

The linear function is C(x) = 500 + 1.25x and the revenue function is R(x) = 3.50x.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

From the data:

The linear function:

C(x) = 500 + 1.25x

For revenue:

R(x) = 3.50x

R(50) = 3.50(50) = $175

Using the C(x) and revenue function we can find the number of slushies and profit.

Thus, the linear function is C(x) = 500 + 1.25x and the revenue function is R(x) = 3.50x.

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The answer is 10.67. It can be calculated by integrating and then applying the limits.

What is the integration of [tex]x^n[/tex]?

The integration of power of a variable is given by the following formula:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

If limits are there then limits can be applied after integration.

Consider the given function.

[tex]\int^3_1(x^2-4x+9)dx[/tex]

Now, apply integration on each term as follows:

[tex]\int^3_1x^2dx-4\int^3_1xdx+9\int^3_1dx[/tex]

Now, use the following formula to integrate:

[tex]\int x^ndx=\frac{x^{n+1}}{n+1}[/tex]

Now, the integration will look like as follows:

[tex][x^3/3]^3_1-4[x^2/2]^3_1+9[x/1]^3_1[/tex]

Now, apply limits to find the answer as follows:

[tex][3^3/3-1^3/3]-4[3^2/2-1^2/2]+9[3-1]\\[/tex]

[tex][9-\frac{1}{3}]-4[9/2-1/2]+18\\10.67[/tex]

Hence, the answer is 10.67. It can be calculated by integrating and then applying the limits.

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The perimeter of the image of the triangle is approximately equal to 87 units.

In this problem we find the coordinates of the vertices of a triangle: A(x, y) = (- 4, 3), B(x, y) = (4, 3), C(x, y) = (- 4, - 6), of which we must determine the perimeter of its image, which is the result of a scale factor of 3. The perimeter is the sum of the lengths of the triangle. The length of each side of the image of the triangle can be found by Pythagorean theorem and definition of dilation:

l = √[(Δx)² + (Δy)²]

Where:

And the perimeter of the image of the triangle is described below:

p = k · (AB + BC + AC)

Where k is the scale factor.

First, determine the lengths of sides AB, BC and AC:

Side AB

AB = √[[4 - (- 4)]² + (3 - 3)²]

AB = 8

Side BC

BC = √[(- 4 - 4)² + (- 6 - 3)²]

BC = √[(- 8)² + (- 9)²]

BC = √145

Side AC

AC = √[(- 4 - 4)² + (- 6 - 3)²]

AC = 9

Second, determine the perimeter of the image:

p = 3 · (8 + √145 + 9)

p = 3 · (17 + √145)

p = 51 + 3√145

p ≈ 87.125

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The athletic director can therefore draw the conclusion that the average academic standing of football players is comparable to the average standing of other student athletes.

The following formula is used to determine the test statistic:

SRS, s = 0.54, n = 40 > 30,

x = 3.18

H0: μ = 3.10

H1: μ > 3.10

α = 0.05

The t-distribution table and our alpha level of.05 are then used to determine our crucial values.

Our critical values are 2.093 for a two specimens with 19 levels of freedom and α = 0.05 level of significance.

t = x -  μ / s/√n

t = 3.18 - 3.10/ 54/√20

t = 0.66

As a result, It is clear that sampling error is the only possible explanation for the discrepancy between both the sample mean as well as the hypothesized value.

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Answer:

7

Step-by-step explanation:

seven is the value e you would get

The probability that the total showing is less than 10 is 5/6.

A mathematical expression is made up of terms (constants and variables) separated by mathematical operators. A mathematical equation is used to equate two expressions. Equation modelling is the process of writing a mathematical verbal expression in the form of a mathematical expression for correct analysis, observations and results of the given problem. The formula to calculate the probability of occurrence of an event 'A' can be written as -  

P(A) = n(A)/n(S)

where -

n(A) = Number of outcomes favorable to event A.

n(S) = Total number of outcomes

Given are two dices rolled.

The total number of favorable outcomes can be calculated as -

[a] = (1,1) to (1,6) = 6

[b] = (2,1) to (2,6) = 6

[c] = (3,1) to (3,6) = 6

[d] = (4,1) to (4,5) = 5

[e] = (5,1) to (5,4) = 4

[f] = (6,1) to (6,3) = 3

n(A) = [a] + [b] + [c] + [d] + [e] + [f] =  6+6+6+5+4+3 = 30

n(S) = 36

So, the probability of occurrence of an event 'A' -

P(A) = n(A)/n(S)

P(A) = 30/36

P(A) = 5/6

Therefore, the probability that the total showing is less than 10 is 5/6.

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The statement that best describes a sequence of transformations in which the blue figure is the image of the red figure is given as follows:

A. Rotate 90° clockwise about the origin and then dilate with respect to the origin using a scale factor of 1/2.

The horizontal side lengths on the blue figure are given as follows:

8 - 4 = 6 - 2 = 4.

The vertical side lengths on the transformed red figure is of:

4 - 2 = 3 - 1 = 2.

Hence the scale factor of the dilation is given as follows:

2/4 = 1/2.

Which means that options C and D are not correct.

The rules for each rotation are given as follows:

On the second quadrant, of the blue figure, we have that:

On the first quadrant, of the red figure, we have that:

Hence the rotation was clockwise, and option A is correct.

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Answer:

Step-by-step explanation:

24

Answer: 24!

Step-by-step explanation:

hope this help

The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5.

Determine the known quantities and designate the unknown quantity as a variable while trying to set up or construct a linear equation to fit a real-world application.

In other words, an equation is a set of variables that are constrained through a situation or case.

Let's say that number is x,

Quotients of 20 and x will be given as 20/x

20/x - 4 = 0

20/x = 4

x = 20/4 = 5

Hence"The equation associated with the quotient of twenty and a number, decreased by 4, is equal to zero is 20/x - 4 = 0 and that number will be 5".

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Answer:

Slope- y=1/4x-6

Y intercept= (0,-6)

Hope this helps!

Answer:

15d10mo

Step-by-step explanation:

Simplify:

(9+6)(mod10)

=15d10mo

It is a uniform rectilinear movement which is one in which an object moves in a straight line, in one only direction, with a constant speed.

When we spoke of constant speed we mean that the movement retains the same speed, that is; that the object does not move faster, or slower and always at the same speed.

We obtain the data, as specified by the exercise.

Data:

 V = 28 m/s

 T = 10 s

 D = ?

We have that the uniform motion formula is:

     [tex]\large\displaystyle\text{$\begin{gathered}\sf V=\dfrac{d}{t}, \to where \end{gathered}$}[/tex]

We solve the formula, to calculate the distance and we obtain.And we also substitute our data.

[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=v*t} \end{gathered}$} }[/tex]

[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=28 \ \frac{m}{\not{s}}*10 \not{s} } \end{gathered}$}}[/tex]

[tex]\boxed{\boxed{\large\displaystyle\text{$\begin{gathered}\sf \bf{d=280 \ m } \end{gathered}$}}}[/tex]

Answer: $18/hour

12 x 1.5 = 18