x - 6 = - 4
Answer: x = 2
x + 3 = -7
Answer: x = -10
5x = -2
Answer: x = -0.4
0.5x = 5
Answer: x = 10
The solutions to the equations as arranged are 2, -10, -0.4 and 10 respectively
Solving equationsGiven the following equations
x − 6 = -4
Add 6 to both sides
x - 6 + 6 = -4 + 6
x = -4 + 6
x = 2
For the equation x + 3 = -7
Subtract 3 from both sides
x + 3 -3 = -7 - 3
x = -10
For the equation 5x = -2
x = -2/5
x = -0.4
For the equation
0.5x = 5
x = 5/0.5
x =10
Hence the solutions to the equations as arranged are 2, -10, -0.4 and 10 respectively
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could anyone help me with this?
Answer:
Step-by-step explanation:
A=6x²
[tex]\frac{dA}{dt} =6\times 2x \times \frac{dx}{dt} \\\frac{dA}{dt}=12x \frac{dx}{dt}[/tex]
f(x)=2^x. What is g(x)?
The function g(x) is g(x)= (3x)^2
How to solve for g(x)?The complete question is in the image
From the graph in the image, we have:
f(x) = x^2
The function f(x) is stretched by a factor of 3 to form g(x).
This means that:
g(x) = f(3x)
So, we have:
g(x)= (3x)^2
Hence, the function g(x) is g(x)= (3x)^2
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inverse function of f(x)=x-7/x+4
Final Answer: The inverse of f (x)=7x-4 is f^-1 (x)= (x+4)/7
as part of a competition, diego must spin around in a circle 6 times and then run to a tree. the time he spends on each spin is represented by S AND THE TIME HE SPEND RUNNING is R. He gets to the tree 21 seconds after he starts spinning. If it takes diego 1.2 seconds to spin around each time, how many seconds did he spend running
The time he spent running is 13.80 seconds.
How much time did he spend running?
The equation that can be used to represent the time he gets to the tree is:
Time he gets to the tree = (time of each spin x total spins) + time he spent running
21 = (6 x 1.2) + r
21 = 7.20 + r
r = 21 - 7.20
r = 13.80 seconds
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Complete the equations to solve 1{,}860 \div61,860÷61, comma, 860, divided by, 6.
\phantom{=}\greenD{1{,}860}\div{\blueD6}=1,860÷6empty space, start color #1fab54, 1, comma, 860, end color #1fab54, divided by, start color #11accd, 6, end color #11accd
=(\greenD{1{,}800}\div\,=(1,800÷equals, left parenthesis, start color #1fab54, 1, comma, 800, end color #1fab54, divided by
) \, + \,(\greenD{60}\div\,)+(60÷right parenthesis, plus, left parenthesis, start color #1fab54, 60, end color #1fab54, divided by
))right parenthesis
= 300 +=300+equals, 300, plus
==equals
The division of the figure based on the information is 14.09.
How to illustrate the information?It should be noted that the question is simply to divide 860 by 61.
The division based on the information will be illustrated thus:
= 860 ÷ 61
= 14.09
In conclusion, the correct option is 14.09.
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Please look at the attachment and answer my question :(
Answer:
AB / BC = 2 / 3
A equals 9 and C = 13
AC = 13 - 9 = 4
AB + BC = 13
A) AB + BC = 4
B) AB / BC = 2/3 Therefore BC = AB / (2/3)
B) BC = 1.5 AB
A) BC = 4 - A/B
Multiplying equation A) by -1
A) -BC = -4 + AB then we add equation B)
B) BC = 1.5 AB then adding both equations
0 = 2.5 AB -4
2.5 AB = 4
AB = 1.6
Since A = 9 then the number at B is 9 + 1.6
equals 10.6
Step-by-step explanation:
Answer:
The number at B is 10.6
Step-by-step explanation:
Let the number at B be x.
We know that distance between two points on the number line is the absolute value of the difference of numbers at those points.
Then distances representing the lengths of segments AB and BC are:
AB = x - 9BC = 13 - xWe are given the ratio of segments:
AB/BC = 2/3Substitute and solve for x:
(x - 9)/(13 - x) = 2/33(x - 9) = 2(13 - x)3x - 27 = 26 - 2x3x + 2x = 26 + 275x = 53x = 53/5x = 10.6Hi I would like to know how I can solve this problem.
The [tex]n[/tex]-th term is
[tex]U_n = \dfrac14 n^2 (n+1)^2[/tex]
so the 39th term is
[tex]U_{39} = \dfrac14 39^2 40^2 = \boxed{608,400}[/tex]
Observe that
[tex]2^3 + 4^3 + 6^3 = 2^3 + 2^3\times2^3 + 2^3\times3^3 = 8\left(1^3+2^3+3^3)[/tex]
which suggests that
[tex]V_n = 8U_n = \boxed{2n^2(n+1)^2}[/tex]
i don’t understand!!!
In other words, 12 goes in the top box and 13 goes in the bottom. The fraction slash sign is not part of either box.
===========================================================
Explanation:
Cosine is the ratio of adjacent over hypotenuse.
cos(angle) = adjacent/hypotenuse
For the reference angle X, the adjacent leg is 36 units long. It's the leg closest or touching angle X.
The hypotenuse is always the longest side. It is always opposite the 90 degree angle. The hypotenuse in this case is 39 units.
Therefore,
cos(X) = 36/39 = (12*3)/(13*3) = 12/13
Alan is building a garden shaped like a rectangle with a semicircle attached to one short side. If he has 70 feet of fencing to go around it, what dimensions will give him the maximum area in the garden? Round the answers to the nearest tenth.
The dimension that would give the maximum area is 20.8569
How to solve for the maximum area
Let the shorter side be = x
Perimeter of the semi-circle is πx
Twice the Length of the longer side
[tex][70-(\pi )x -x][/tex]
Length = [tex][70-(1+\pi )x]/2[/tex]
Total area =
area of rectangle + area of the semi-circle.
Total area =
[tex]x[[70-(1+\pi )x]/2] + [(\pi )(x/2)^2]/2[/tex]
When we square it we would have
[tex]70x +[(\pi /4)-(1+\pi)]x^2[/tex]
This gives
[tex]70x - [3.3562]x^2[/tex]
From here we divide by 2
[tex]35x - 1.6781x^2[/tex]
The maximum side would be at
[tex]x = 35/2*1.6781[/tex]
This gives us 20.8569
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Write these decimals in order from smallest to largest: 0.507, 0.75, 0.5, 0.078
look at the pictures
The key feature that the function of f(x) and g(x) has in common is the domain.
What is the domain and range of a function?The domain of a function is the set of input or argument values for which the function is valid and well defined. The range is the set of the dependent variable for which a function is defined.
From the given information, we are to find the domain, range, x-intercept, and, y-intercept of the given equation:
[tex]\mathbf{f(x) = -4^x+5}[/tex]
[tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]
For [tex]\mathbf{f(x) = -4^x+5}[/tex];
The domain of a function [tex]\mathbf{-4^x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.
The range f(x) < 5. The x-intercepts, when y is zero = [tex]\mathbf{(\dfrac{In(5)}{2In(2)},0)}[/tex]The y-intercepts; when x is zero = (0,4)For [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex]
The domain of a function [tex]\mathbf{g(x) = x^3 +x^2 -4x+5}[/tex] has no undefined points or constraints. Thus, the domain is -∞ < x < ∞.
The range -∞ < f(x) < ∞. The x-intercepts, when y is zero = (-2.939, 0)The y-intercepts; when x is zero = (0,5)Learn more about the domain and range of a function here;
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a 300-acre farm produced 20,000 bushels of corn last year. what is the minimum rate of production, in bushels per acre, that is needed this year, so the farm's two-year production total will be at least 49,100 bushels of corn?
The minimum production rate of bushels per acre of corn needed this year is 97 for two year's production to reach the goal of 49,100.
How to calculate the minimum production rate?To calculate the minimum production rate for this year, we must subtract the production rate of the previous year, with what is expected to be obtained this year.
49,100 - 20,000 = 29,100Then we must divide the minimum that must be obtained by the total area, by the 300 acres that we have, to know how much corresponds to each acre.
29,100 ÷ 300 = 97According to the above, each acre must produce 97 bushels of corn to reach the goal of 49,100 for both years.
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x is directly proportional to y. When x = 4, y = 7. Work out the value
of y when x = 12
Answer:
y = 21
Step-by-step explanation:
When x is directly proportional to y:
When x increases, y also increases.
In this case, it is so you write,
1. Write down the formula: y=kx
k being the constant.
Let's use "When x = 4, y = 7" to work out the constant.
2. Substitute the 'when' values: 7=k×4
3. Rearrange to find the constant: 7÷4=k
4. Find k: k=1.75
Now let's see the new problem, 'what is y when x = 12'.
5. Substitute the values now but keep the constant: y = 1.75 × 12
6. Rearrange if needed.
7. Find the missing value: y = 21
Hope this helped and if you require further assistance from me please comment below! :)
Ps: If it was inversely proportional then the formula would be y = k / x
with k still being the constant.
Given the functions a(x) = 3x - 12 and b(x) = x-9, solve a[b(x)].
Oa[b(x)] = 3x²-21
O a[b(x)] = 3x² - 39
Oa[b(x)] = 3x - 21
Oa[b(x)] = 3x - 39
Answer:
Step-by-step explanation:
hello :
a(x) = 3x - 12 and b(x) = x-9, so
a[b(x)]=a(x-9) =3(x-9)-12
a[b(x)]=3x-9-12
a[b(x)]=3x+21
Which expression is equivalent to
xay
8
700
8√√√x
y
8√√y
Mark this and return
128x56
√ 2x75
? Assume x > 0 and y> 0.
Save and Exit
The equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]
How to determine the equivalent expression?The expression is given as:
[tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex]
Divide 128 by 2
[tex]\sqrt{\frac{64x^5y^6}{x^7y^5}}[/tex]
Apply the law of indices to the variables
[tex]\sqrt{\frac{64y^{6-5}}{x^{7-5}}}[/tex]
Evaluate the differences
[tex]\sqrt{\frac{64y}{x^2}}[/tex]
Take the square root of 64
[tex]8\sqrt{\frac{y}{x^2}}[/tex]
Take the square root of x^2
[tex]\frac{8\sqrt y}{x}[/tex]
Hence, the equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]
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3 5/6 + 2 4/9 in its simplest form
Answer:
>> [tex]6\frac{5}{18}[/tex]
Step-by-step explanation:
1) Add the whole numbers first.
[tex]5+\frac{5}{6} +\frac{4}{9}[/tex]
2) Find the Least Common Denominator (LCD) of [tex]\frac{5}{6} ,\frac{4}{9}[/tex] . In other words, find the Least Common Multiple (LCM) of [tex]6,9[/tex].
Method 1: By Listing Multiples
1. List the multiples of each number.
Multiples of 6 : 6, 12, 18, ...
Multiples of 9 : 9, 18, ...
2. Find the smallest number that is shared by all rows above. This is the LCM.
LCM = 18
3. Make the denominators the same as the LCD.
[tex]5+\frac{5\times 3}{6\times 3}+\frac{4\times 2}{9\times 2}[/tex]
4. Simplify. Denominators are now the same.
[tex]5+\frac{15}{18}+\frac{8}{18}[/tex]
5. Join the denominators.
[tex]5+\frac{15+8}{18}[/tex]
6. Simplify.
[tex]5+\frac{23}{18}[/tex]
7. Convert [tex]\frac{23}{18}[/tex] to mixed fraction.
[tex]5+1\frac{5}{18}[/tex]
8. Simplify.
[tex]6\frac{5}{18}[/tex]
Decimal Form: 6.277778
Cheers.
The addition of 3 5/6 + 2 4/9 is 113/18
What is fraction?The fractional bar is a horizontal bar that divides the numerator and denominator of every fraction into these two halves.
The number of parts into which the whole has been divided is shown by the denominator. It is positioned in the fraction's lower portion, below the fractional bar.How many sections of the fraction are displayed or chosen is shown in the numerator. It is positioned above the fractional bar in the upper portion of the fraction.Given:
3 5/6 + 2 4/9
Now, writing it into normal fraction
3 5/6 + 2 4/9
= 23/ 6 + 22/9
= 23/6 x 3/3 + 22/9 x 2/2
= 69/ 18 + 44/18
= 113/18
Hence, the addition is 113/18
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To the nearest percent how much greater is 8 than 6
The percent greater of 8 than 6 to the nearest percent is 25%
PercentagePercent greater = difference / higher chance value × 100
= 2/8 × 100
= 0.25 × 100
Percent greater = 25%
Therefore, the percent greater of 8 than 6 to the nearest percent is 25%
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Solve [tex]2 cosx=4cosx sin^2x[/tex]
There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].
How to solve a trigonometric equation
In this problem we must simplify the trigonometric equation by both algebraic and trigonometric means and clear the variable x:
2 · cos x = 4 · cos x · sin² x
2 · sin² x = 1
sin² x = 1/2
sin x = ± √2 /2
There are several solutions:
x₁ = π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₂ = 3π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₃ = 5π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex]
There are four solutions for the trigonometric equation 2 · cos x = 4 · cos x · sin² x are x₁ = π/4 ± 2π · i, x₂ = 3π/4 ± 2π · i, x₃ = 5π/4 ± 2π · i and x₄ = 7π/4 ± 2π · i, [tex]i \in \mathbb{N}_{O}[/tex].
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Simplify the following
Answer is 6
Firstly changing mixed fraction.
[tex] \frac{3}{2} - \frac{5}{4} + \frac{23}{4} [/tex]
By taking the LCM
[tex] \frac{6 - 5 + 23}{4} [/tex]
-5 + 23 (-) (+) = (-)
[tex] \frac{6 + 18}{4} [/tex]
[tex] \frac{24}{4} [/tex]
[tex]6[/tex]
Hope it helps you, any confusions you may ask!
Answer:
6 (work below)
Step-by-step explanation:
1 1/2 = 3/2
1 1/4 = 5/4
5 3/4 = 23/4
The least common multiple of the denominators is 4, so 3/2 will become 6/4.
6/4 - 5/4 = 1/4 + 23/4 = 24/4 = 6
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The lengths of the sides of the right triangle above are a, 3, and c. What is a in terms of c?
The expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]
Pythagorean theoremFrom the question, we are to determine the expression for a in terms of c
In the given right triangle, we can write that
[tex]c^{2} = a^{2} +3^{2}[/tex] (Pythagorean theorem)
Thus,
[tex]c^{2} = a^{2} +9[/tex]
[tex]a^{2}= c^{2} -9[/tex]
[tex]a^{2}= \sqrt{c^{2} -9}[/tex]
Hence, the expression for a in terms of c is [tex]a^{2}= \sqrt{c^{2} -9}[/tex]. The correct option is the third option [tex]a^{2}= \sqrt{c^{2} -9}[/tex]
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Evaluate the sum (for math nerds)
[tex]i {}^{0!} + i {}^{1!} + i {}^{2!} + i {}^{3!} + ... + i {}^{100!} [/tex]
Note that :
[tex]i = \sqrt[]{ - 1} [/tex]
Answer: i+96
Step-by-step explanation:
Note that [tex]i^{4k}[/tex], where k is an integer, is equal to 1.
This means that [tex]i^{4!}=i^{5!}=i^{6}=\cdots=i^{99!}+i^{100!}=1[/tex]
So, we can rewrite the sum as [tex]i^{1}+i^{1}+i^{2}+i^3+97(1)=i+i-1-i+97=i+96[/tex]
[tex]n![/tex] is divisible by 4 for all [tex]n\ge4[/tex]. This means, for instance,
[tex]i^{4!} = \left(i^4\right)^{3!} = 1^{3!} = 1[/tex]
[tex]i^{5!} = \left(i^4\right)^{5\times3!} = 1^{5\times3!} = 1[/tex]
etc, so that [tex]i^{n!} = 1[/tex] for all [tex]n\ge4[/tex].
Meanwhile,
[tex]i^{0!} = i^1 = i[/tex]
[tex]i^{1!} = i^1 = i[/tex]
[tex]i^{2!} = i^2 = -1[/tex]
[tex]i^{3!} = i^6 = (-1)^3 = -1[/tex]
Then the sum we want is
[tex]i^{0!} + i^{1!} + i^{2!} + i^{3!} + 97\times1 = i + i - 1 - 1 + 97 = \boxed{95+2i}[/tex]
What is the volume of the cylinder?
The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³
What is an equation?An equation is an expression that shows the relationship between two or more numbers and variables.
The volume of a cylinder with radius (r) and height (h) is:
Volume = πr²h
Given that h = 10 ft, r = 16/2 = 8 ft
The volume = π * 8² * 10 = 2010.62 ft³
The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³
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Suppose f(x)=2^x. What is the graph of g(x)=1/3f(x)?
Please see the blue curve of the image attached below to know the graph of the function g(x) = (1/3) · 2ˣ.
How to graph a transformed function
Herein we have an original function f(x). The transformed function g(x) is the result of compressing f(x) by 1/3. Then, we find that g(x) = (1/3) · 2ˣ. Lastly, we graph both function on a Cartesian plane with the help of a graphing tool.
The result is attached below. Please notice that the original function f(x) is represented by the red curve, while the transformed function g(x) is represented by the blue curve.
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in triangle ABC, AC=13, BC=84, and AB=85. find the measure of angle C.
in triangle ABC, AC=13, BC=84, and AB=85. the measure of angle is C.5.1 degrees,
How can the angle be found?Using the The Law of Cosines in any triangle which can be expressed as;
[tex]c^2 = a^2 + b^2 - 2ab * cos(C)[/tex]
Then if we expressed the given sides in the cosine formula we have
[tex]13^2 = 85^2 + 84^2 - 2 * 85 * 84 * cos(C)[/tex]
Then we have [tex]169 = 7225 + 7056 - 14280 * cos(C)[/tex]
[tex]14280 * cosC = (7225 + 7056 - 169)[/tex]
[tex]cosC = \frac{14212}{ 14280}[/tex]
cos 0.9947
[tex]C = cos^-1(0.9947)[/tex]
C = 5.1 degrees
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Suppose f(x) and g(x) are differentiable functions satisfying f(1) = f′(1) = 2. Let
h(x) = g(f(x)) − g(2x). Determine whether h′(1) > 1
Answer:
1777777777888876332222233
A candy shop sells a box of chocolates for $30. It has $29 worth of chocolates plus $1 for the box. The box includes two kinds of candy: caramels and truffles. Lita knows how much the different types of candies cost per pound and how many pounds are in a box. She said,
If x is the number of pounds of caramels included in the box and y is the number of pounds of truffles in the box, then I can write the following equations based on what I know about one of these boxes:
x + y = 3
8x + 12y + 1 = 30
Assuming Lita used the information given and her other knowledge of the candies, use her equations to answer the following:
How many pounds of candy are in the box?
What is the price per pound of the caramels
What does the term 12y in the second equation represent?
What does 8x + 12y + 1 in the second equation represent?
Answer:
There are 3 lbs of candy in the box.
The caramels are $8 per pound.
12y represents the total cost of truffles in the box.
8x + 12y + 1 represents the cost of caramels in the box + the cost of truffles in the box + the cost of the box.
Step-by-step explanation:
x = the number of pounds of caramels
y = the number of pounds of truffles
x + y = total lbs. and we know x + y = 3
We know x is the number of pounds of caramels.
8x = (cost per pound of caramels) × (number of pounds of caramels)
So 8 = cost of caramels
We know y = number of pounds of truffles. So 12 is the cost of truffles by pound, and 12y is the total cost of truffles in the box.
Based on all the above,
The cost of caramels in the box + the cost of truffles in the box + the cost of the box is = total cost of the box. 8x + 12y + 1 = 30
will give brainly .Which statements are true regarding undefinable terms in geometry? Check all that apply.
A point has no length or width.
A point indicates a location in a coordinate plane.
A plane has one dimension, length.
A line has a definite beginning and end.
A plane consists of an infinite set of lines.
A line consists of an infinite set of points.
The true statements about terms in geometry that are undefinable are;
A point has no length or widthA point indicates a location on the coordinate planeA line consists of an infinite set of points.A plane consists of an infinite set of points.How can the true statements be found?Three undefinable terms in geometry are;
1) Point
2) Line
3) Plane
The above terms do not have a formal definition but they can be described based on their properties.
The other geometric terms are defined based on the above undefinable terms.
A point can be described as a location in space that is dimensionless and can be specified on the coordinate plane as an ordered pairs (x, y).
True statements about a point are therefore;
A point has no length or widthA point indicates a location on the coordinate planeA line is infinitely long, that has no beginning or end. It has one dimension, with no thickness or height.
Given that a point is dimensionless, a line can be considered a set of points.
A true statement is therefore;
A line consists of an infinite set of points.A plane is a two dimensional geometric figure that have infinite length and width.
An infinite set of lines that forms a two dimensional figure can be used to describe a plane.
The true statement with regards to a plane is therefore;
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A particle travels so that its distance D (in metres) from its origin O is modelled by the equation D = 24 + 15t - [tex]\frac{t^{2} }{2}[/tex], where t is the time in minutes after the particle has started to move.
a. calculate the particle's distance from O when it first started to move.
b. determine the time when the particle first reaches O. Give your answer to 2 decimal places.
c. determine the particle's speed when it has been moving for 3 minutes. Give your answer in m [tex]S^{-1}[/tex]
(a) The particle's distance from O when it first started to move is 24 m.
(b) The time when the particle first reaches O is 15 mins.
(c) The particle's speed when it has been moving for 3 minutes is 0.2 m/s.
Particle's distance from O when it first started to move
D = 24 + 15 - t²/2
when the time, t = 0
D = 24 m
When the object first reaches OWhen the object reaches O, its final velocity, v = 0
v = dD/dt
v = 15 - t
0 = 15 - t
t = 15 mins
Speed of the particle after 3 minutesv = 15 - t
v = 15 - 3
v = 12 m/min
v = 12 m/min x 1min/60s = 0.2 m/s
Thus, the particle's distance from O when it first started to move is 24 m.
The time when the particle first reaches O is 15 mins.
The particle's speed when it has been moving for 3 minutes is 0.2 m/s.
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What point is 2/3 of the distance from point A(3, 1) to point B(3, 19)?
What point is 2/3 of the distance from point A(3, 1) to point B(3, 19)?
(3, 13)
(9/3, 13)
(2, 12)
(3, 19)
Answer: (3, 13)
Step-by-step explanation:
If we let the point be P, then AP:BP=2:1.
[tex]P=\left(\frac{(2)(3)+(1)(3)}{2+1}, \frac{(2)(19)+(1)(1)}{2+1} \right)=(3, 13)[/tex]
A random variable X has a gamma density function with parameters α= 8 and β = 2.
Without making any assumptions, derive the moment generating function of X and use to
determine the mean and variance of X.
I know you said "without making any assumptions," but this one is pretty important. Assuming you mean [tex]\alpha,\beta[/tex] are shape/rate parameters (as opposed to shape/scale), the PDF of [tex]X[/tex] is
[tex]f_X(x) = \dfrac{\beta^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{-\beta x} = \dfrac{2^8}{\Gamma(8)} x^7 e^{-2x}[/tex]
if [tex]x>0[/tex], and 0 otherwise.
The MGF of [tex]X[/tex] is given by
[tex]\displaystyle M_X(t) = \Bbb E\left[e^{tX}\right] = \int_{-\infty}^\infty e^{tx} f_X(x) \, dx = \frac{2^8}{\Gamma(8)} \int_0^\infty x^7 e^{(t-2) x} \, dx[/tex]
Note that the integral converges only when [tex]t<2[/tex].
Define
[tex]I_n = \displaystyle \int_0^\infty x^n e^{(t-2)x} \, dx[/tex]
Integrate by parts, with
[tex]u = x^n \implies du = nx^{n-1} \, dx[/tex]
[tex]dv = e^{(t-2)x} \, dx \implies v = \dfrac1{t-2} e^{(t-2)x}[/tex]
so that
[tex]\displaystyle I_n = uv\bigg|_{x=0}^{x\to\infty} - \int_0^\infty v\,du = -\frac n{t-2} \int_0^\infty x^{n-1} e^{(t-2)x} \, dx = -\frac n{t-2} I_{n-1}[/tex]
Note that
[tex]I_0 = \displaystyle \int_0^\infty e^{(t-2)}x \, dx = \frac1{t-2} e^{(t-2)x} \bigg|_{x=0}^{x\to\infty} = -\frac1{t-2}[/tex]
By substitution, we have
[tex]I_n = -\dfrac n{t-2} I_{n-1} = (-1)^2 \dfrac{n(n-1)}{(t-2)^2} I_{n-2} = (-1)^3 \dfrac{n(n-1)(n-2)}{(t-2)^3} I_{n-3}[/tex]
and so on, down to
[tex]I_n = (-1)^n \dfrac{n!}{(t-2)^n} I_0 = (-1)^{n+1} \dfrac{n!}{(t-2)^{n+1}}[/tex]
The integral of interest then evaluates to
[tex]\displaystyle I_7 = \int_0^\infty x^7 e^{(t-2) x} \, dx = (-1)^8 \frac{7!}{(t-2)^8} = \dfrac{\Gamma(8)}{(t-2)^8}[/tex]
so the MGF is
[tex]\displaystyle M_X(t) = \frac{2^8}{\Gamma(8)} I_7 = \dfrac{2^8}{(t-2)^8} = \left(\dfrac2{t-2}\right)^8 = \boxed{\dfrac1{\left(1-\frac t2\right)^8}}[/tex]
The first moment/expectation is given by the first derivative of [tex]M_X(t)[/tex] at [tex]t=0[/tex].
[tex]\Bbb E[X] = M_x'(0) = \dfrac{8\times\frac12}{\left(1-\frac t2\right)^9}\bigg|_{t=0} = \boxed{4}[/tex]
Variance is defined by
[tex]\Bbb V[X] = \Bbb E\left[(X - \Bbb E[X])^2\right] = \Bbb E[X^2] - \Bbb E[X]^2[/tex]
The second moment is given by the second derivative of the MGF at [tex]t=0[/tex].
[tex]\Bbb E[X^2] = M_x''(0) = \dfrac{8\times9\times\frac1{2^2}}{\left(1-\frac t2\right)^{10}} = 18[/tex]
Then the variance is
[tex]\Bbb V[X] = 18 - 4^2 = \boxed{2}[/tex]
Note that the power series expansion of the MGF is rather easy to find. Its Maclaurin series is
[tex]M_X(t) = \displaystyle \sum_{k=0}^\infty \dfrac{M_X^{(k)}(0)}{k!} t^k[/tex]
where [tex]M_X^{(k)}(0)[/tex] is the [tex]k[/tex]-derivative of the MGF evaluated at [tex]t=0[/tex]. This is also the [tex]k[/tex]-th moment of [tex]X[/tex].
Recall that for [tex]|t|<1[/tex],
[tex]\displaystyle \frac1{1-t} = \sum_{k=0}^\infty t^k[/tex]
By differentiating both sides 7 times, we get
[tex]\displaystyle \frac{7!}{(1-t)^8} = \sum_{k=0}^\infty (k+1)(k+2)\cdots(k+7) t^k \implies \displaystyle \frac1{\left(1-\frac t2\right)^8} = \sum_{k=0}^\infty \frac{(k+7)!}{k!\,7!\,2^k} t^k[/tex]
Then the [tex]k[/tex]-th moment of [tex]X[/tex] is
[tex]M_X^{(k)}(0) = \dfrac{(k+7)!}{7!\,2^k}[/tex]
and we obtain the same results as before,
[tex]\Bbb E[X] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=1} = 4[/tex]
[tex]\Bbb E[X^2] = \dfrac{(k+7)!}{7!\,2^k}\bigg|_{k=2} = 18[/tex]
and the same variance follows.