15/3 = 5 4_7 + 5 2_3
b=?

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]

Answer:

  (b)  (3,-1)

Step-by-step explanation:

One can determine the correct answer by checking to see if it satisfies the equations.

The first equation can be rewritten to ...

  x = 6 +3y . . . . . . . add 3y to both sides

This makes it easier to check answer choices:

  a) -1 = 6 +3(3) . . . . no

  b) 3 = 6 +3(-1) . . . . yes . . . . . (3, -1) is the solution

  c) 1 = 6 +3(-3) . . . . no

  d) -3 = 6 +3(1) . . . . no

__

Additional comment

We can further convince ourselves this is the correct choice by seeing if it satisfies the second equation:

  2x +2y = 4 . . . for (x, y) = (3, -1)

  2(3) +2(-1) = 4 . . . . yes

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]

The time he spent running is 13.80 seconds.

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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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

The division of the figure based on the information is 14.09.

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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in triangle ABC, AC=13, BC=84, and AB=85.  the measure of angle is C.5.1 degrees,

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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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]

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]

From 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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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].

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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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

(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.

D = 24 + 15 - t²/2

when the time, t  = 0

D = 24 m

When the object reaches O, its final velocity, v = 0

v = dD/dt

v = 15 - t

0 = 15 - t

t = 15 mins

v = 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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The equivalent expression of [tex]\sqrt{\frac{128x^5y^6}{2x^7y^5}}[/tex] is [tex]\frac{8\sqrt y}{x}[/tex]

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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The percent greater of 8 than 6 to the nearest percent is 25%

Percent 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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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

The fractional bar is a horizontal bar that divides the numerator and denominator of every fraction into these two halves.

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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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.

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.

Then 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.

According 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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The volume of a cylinder with a diameter of 16 feet and height of 10 feet is 2010.62 ft³

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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The key feature that the function of f(x) and g(x) has in common is the domain.

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 < ∞.

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 < ∞.

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Please see the blue curve of the image attached below to know the graph of the function g(x) = (1/3) · 2ˣ.

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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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

Final Answer: The inverse of f (x)=7x-4 is f^-1 (x)= (x+4)/7

The dimension that would give the maximum area is 20.8569

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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The true statements about terms in geometry that are undefinable are;

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 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 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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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.

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]

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:

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:

We are given the ratio of segments:

Substitute and solve for x:

Answer:

1777777777888876332222233

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.

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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