Use the Law of Cosines to find the values of x°and y°. Round to the nearest tenth.

To show that `(n + 3)7 ∈ Θ(n7)`, we need to prove that `(n + 3)7 = Θ(n7)`.This can be done by showing that `(n + 3)7 = O(n7)` and `(n + 3)7 = Ω(n7)` .Now, let's prove the two parts separately:

Proof for `(n + 3)7 = O(n7)`.

We want to prove that there exists a positive constant c and a non-negative constant k such that `(n + 3)7 ≤ cn7` for all `n ≥ k`.Using the Binomial theorem, we can expand `(n + 3)7` as:```
(n + 3)7
= n7 + 7n6(3) + 21n5(3)2 + 35n4(3)3 + 35n3(3)4 + 21n2(3)5 + 7n(3)6 + 37
≤ n7 + 21n6(3) + 21n5(3)2 + 35n4(3)3 + 35n3(3)4 + 21n2(3)5 + 7n(3)6 + n7
≤ 2n7 + 21n6(3) + 21n5(3)2 + 35n4(3)3 + 35n3(3)4 + 21n2(3)5 + 7n(3)6
≤ 2n7 + 84n6 + 441n5 + 2205n4 + 10395n3 + 45045n2 + 153609n + 729
```Thus, we can take `c = 153610` and `k = 1` to satisfy the definition of big-Oh notation. Hence, `(n + 3)7 = O(n7)`.Proof for `(n + 3)7 = Ω(n7)`We want to prove that there exists a positive constant c and a non-negative constant k such that `(n + 3)7 ≥ cn7` for all `n ≥ k`.Using the Binomial theorem, we can expand `(n + 3)7` as:```
(n + 3)7
= n7 + 7n6(3) + 21n5(3)2 + 35n4(3)3 + 35n3(3)4 + 21n2(3)5 + 7n(3)6 + 37
≥ n7
```Thus, we can take `c = 1` and `k = 1` to satisfy the definition of big-Omega notation. Hence, `(n + 3)7 = Ω(n7)`.

As we have proved that `(n + 3)7 = O(n7)` and `(n + 3)7 = Ω(n7)`, therefore `(n + 3)7 = Θ(n7)`.Thus, we have shown that `(n + 3)7 ∈ Θ(n7)`.From the proof, we can see that we used the Binomial theorem to expand `(n + 3)7` and used algebraic manipulation to bound it from above and below with suitable constants. This technique can be used to prove the time complexity of various algorithms, where we have to find the tightest possible upper and lower bounds on the number of operations performed by the algorithm.

Hence, we have shown that `(n + 3)7 ∈ Θ(n7)` for non-negative integer n.

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On solving the provided question we cans say that quadratic equation - x^2-5x-4 = 0 and x = 1 , 3

A quadratic equation is x ax2+bx+c=0, which is a quadratic polynomial in a single variable. a 0. The Fundamental Theorem of Algebra ensures that it has at least one solution since this polynomial is of second order. Solutions may be simple or complicated. An equation that is quadratic is a quadratic equation. This indicates that it has at least one word that has to be squared. The formula "ax2 + bx + c = 0" is one of the often used solutions for quadratic equations. where are numerical coefficients or constants a, b, and c. where the variable "X" is unidentified.

here, we have

quadratic equation -

x^2-5x-4 = 0

x(x-1) -3(x-3) =

(x-1)(x-3) = 0

x = 1 , 3

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The definition of math is the overall group of sciences that study numbers, shapes and their relationships. An example of math is arithmetic, the study of addition, subtraction, multiplication and division. ... (uncountable, North America) Arithmetic calculations; (see do the math).

1) \(m\angle KER=m\angle MEN\) (given)

2) \(\angle 3 \cong \angle 3\) (reflexive property)

3) \(m\angle 3=m\angle 3\) (congruent angles have equal measure)

4) \(m\angle 1=m\angle 3\) (subtraction property of equality)

Step-by-step explanation:

the answer would be 6 meters

btw the origin is known as (0,0)

hope this helps bro<33

Answer:

-1 and -6.

Step-by-step explanation:

-1 and -6

-1*-6 = 6

-1 + (-6) = -1-6= -7.

Answer:

wait.

Step-by-step explanation:

what do u really mean by zombies

In math, protractors are essential tools for measuring and determining vertical and adjacent angles.

In the study of geometry, angles play a fundamental role, and accurately measuring them is crucial for solving various mathematical problems. When it comes to vertical and adjacent angles, a key tool used by both students and professionals is the protractor. A protractor is a DIY (do-it-yourself) tool that allows for precise angle measurement and identification.

With a protractor, one can easily determine the size of vertical angles, which are formed by intersecting lines or rays that share the same vertex but point in opposite directions. These angles have equal measures. Similarly, adjacent angles are formed when two angles share a common side and a common vertex but do not overlap. By using a protractor, one can measure the individual angles and determine their relationship to each other.

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Answer:

that seems very hard i dont think there is a inverse maybe there is but it dont seem like it

Yes, it is possible for a plot of a partial expansion of f[x] to share ink with the plot of f[x] all the way from -[infinity] to + [infinity].

Explanation:

We can approximate f(x) as a Fourier series, as follows:

\($$f(x) = \sum_{n=0}^{\infty}a_n\cos\left(\frac{n\pi x}{L}\right)+\sum_{n=1}^{\infty}b_n\sin\left(\frac{n\pi x}{L}\right)$$\)

If f(x) is an odd function, the cosine terms are gone, and if f(x) is an even function, the sine terms are gone.

We can create an approximation for f(x) using only the first n terms of the Fourier series, as follows:

\($$f_n(x) = a_0 + \sum_{n=1}^{n}\left[a_n\cos\left(\frac{n\pi x}{L}\right)+b_n\sin\left(\frac{n\pi x}{L}\right)\right]$$\)

For any continuous function f(x), the Fourier series converges uniformly to f(x) on any finite interval, as given by the Weierstrass approximation theorem.

However, if f(x) is discontinuous, the Fourier series approximation does not converge uniformly.

Instead, it converges in the mean sense or the L2 sense. The L2 norm is defined as follows:

\($$\|f\|^2 = \int_{-L}^{L} |f(x)|^2 dx$$\)

Hence, it is possible for a plot of a partial expansion of f(x) to share ink with the plot of f(x) all the way from -[infinity] to + [infinity].

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Answer:

Rs. 3923.08

Step-by-step explanation:

First principal invested = Rs. 10000

Interest rate = 5%

The interest is compounded yearly.

Time = 1 year

1 year compound interest is equal to simple interest.

Formula for simple interest:

\(SI = \dfrac{PRT}{100}\)

Interest on first sum = \(\frac{10000\times 5\times 1}{100} = Rs\ 500\)

Another sum is on fixed deposit 8% compounded half yearly.

Let the sum = Rs \(x\)

Formula for compound interest is given as:

\(CI = P(1+\frac{R}{100})^T - P\)

It is compounded half - yearly, therefore T = 2

\(CI = x (1+\frac{8}{100})^2 - x\\\Rightarrow CI = x(1.08)^2-x = 0.1664x\)

As per question statement:

\(0.1664x - 500 = 152.80\\\Rightarrow x = \dfrac{652.80}{0.1664} = Rs\ 3823.08\)

Answer:

Here you go...

Step-by-step explanation:

You have 4, and you have 5.

You have 4+2 and you have 5+1

So, you take a 1 from the 2 in the first problem, which gives u 4+1 that equals 5.

Therefore, if u put the first equation in a simpler form then you get 5+1 with the other equation. So it would look like 5+1=5+1.

FURTHERMORE, 4+2= 5+1 (because you simplify 2 into a 1).

So. 5+1 equals 5+1. (which is common sense... 5+1= 6.)

Answer:

#1 is the AA postulate

Step-by-step explanation:

It is the AA postulate because both triangles have congruent angles, making the triangles similar.

1. By using Angle-Angle (AA) Similarity Postulate triangle PRQ and STU are similar.

2. By using Angle-Angle (AA) Similarity Postulate triangle ABC and DEC are similar.

If two triangles have an equal number of corresponding sides and an equal number of corresponding angles, then they are comparable. Similar figures are described as items with the same shape but varying sizes, such as two or more figures.

Given:

1. From Figure

In triangle PRQ and STU

<RPQ = <UST

and, <PQR  = <STU

Thus, By using Angle-Angle (AA) Similarity Postulate

triangle PRQ and STU are similar.

2.  From Figure

In triangle ABC and DEC

<ACB  = <ECD (Vertical opposite angle)

and, <BAC  = <CDE (Alternate Interior Angle)

Thus, By using Angle-Angle (AA) Similarity Postulate

triangle ABC and DEC are similar.

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The Power series representation of f(x) = \(x^8tan^-^1x^3\) is \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^1^1}{2n+1}\)

The Maclaurin series expansion method can be used to get the power series of such functions if the function has a composite part, such as the multiplication or quotient form.

An infinite series in mathematics known as a "power series" can be compared to a polynomial with an infinite number of terms.

The Maclaurin series will be used in this instance to expand the function:

\(tan^-^1x\) = \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^2^n^+^1}{2n+1}\)

and for    \(tan^-^1x^3 = $$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{(x^3)^2^n^+^1}{2n+1}\)

=>  \(tan^-^1x^3 = $$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^3}{2n+1}\)  -------(i)

and f(x) = \(x^8tan^-^1x^3\)

now we need to multiply \(x^8\) in the above equation to get the power series expansion of f(x)

\(x^8tan^-^1x^3\) = \(x^8\) \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^3}{2n+1}\)

=> \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^3^+^8}{2n+1}\)

=>  \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^1^1}{2n+1}\)

so the expansion of f(x) = \(x^8tan^-^1x^3\) is \($$ \Sigma_{n=0}^\infty $$ (-1)^n\frac{x^6^n^+^1^1}{2n+1}\)

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Answer:

A ≈ 28.5

Step-by-step explanation:

a, b, c

P = a + b + c

Semiperimeter s = \(\frac{a+b+c}{2}\)

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)

~~~~~~~~~~~~~~~

\(P_{ABC}\) = 4.3 + 2.89 + 6.81 = 14

s = 14 ÷ 2 = 7

\(A_{ABC}\) = \(\sqrt{7(7-4.3)(7-2.89)(7-6.81)}\) = √14.75901 ≈ 3.84

\(P_{BCD}\) = 8.59 + 7.58 + 6.81 = 22.98

s = 22.98 ÷ 2 = 11.49

\(A_{BCD}\) = \(\sqrt{11.49(11.49-8.59)(11.49-7.58)(11.49-6.81)}\) = √609.7343148 ≈ 24.6928

\(A_{ABCD}\) = 3.84 + 24.6928 ≈ 28.5

Answer:

Yup your good

Step-by-step explanation:

Answer:

Yes the answer is correct.

Answer:

C

Step-by-step explanation:

The add to 180 because that's a straight line    

The expression (x + 1)(x - 3) represents the Area of base of the prism.

a crystal is a polyhedron containing a n-sided polygon base, a respectable halfway point which is a deciphered duplicate of the first, and n different countenances, fundamentally all parallelograms, joining relating sides of the two bases. Translations of the bases exist in every cross-section that runs parallel to the bases.

According to question:

The volume of a rectangular prism is given by the formula V = Bh, where B is the area of the base and h is the height of the prism. In this case, the base is a square with edge length (x + 1), so its area is (x + 1)^2. The volume of the prism is given as (x + 1)^2(x - 3).

We can find the height of the prism by dividing the volume by the area of the base:

B = V/h = (x + 1)^2(x - 3)/(x + 1) = (x + 1)(x - 3)

Therefore, the expression (x + 1)(x - 3) represents the Area of base of the prism.

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Answer:

13

Step-by-step explanation:

.

Answer:

Alex is 14 years old.

Step-by-step explanation:

You can take William's age and divide it by 3, because William is 3 times the age of Alex:

31 divided by 3 = 10.33.

So, we will just say 10 for now.

Now, we add 4 to the 10 because WIlliam is 4 years older than three times his age.

So, in conclusion, Alex is 14 years old.

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The correlation coefficient is -0.29.

Correlation refers to a process for establishing whether or not relationships exist between two given variables.

In statistics, correlation coefficients are used to quantify the strength of the association between two variables. Like Pearson's correlation, which is frequently employed in linear regression, there are other varieties of correlation coefficients. It is widely used in statistics and is highly well-liked.

X      Y        X²          Y²           XY

0       8        0           64            8

1        11        1            121           11

3       10       9           100          30

4        13       16         169          52

7        10       49         100         70

10       6        100       36           60

∑X = 25

∑Y = 58

∑X² = 175

∑Y²= 590

∑XY= 231

r = N×∑XY−(∑X∑Y) / √[N∑x2−(∑x)2][N∑y2−(∑y)2]

Putting all the values,

r = 6*231 - (25*58) / √(6*175-625)(6*590-3364)

r = 1368 - 1450 / √(425)(176)

r = -82/ √ 74800

r = -82/ 273.49

r= -0.29

Therefore, the correlation coefficient is -0.29.

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Answer:

The computation shows that the cost of each bond is A. $3142.50

How to solve the cost?

From the complete information given, the price of the bond is given as $2856.80 each.

Therefore, with the commission, the price will be:

= $2856.80 + (10% × $2856.80)

= $2856.80 + $285.68

= $3142.50

In conclusion, the cost of each bond is $3142.50.

Step-by-step explanation:

or  b

Answer:

Not proportional

Step-by-step explanation:

1/5 = 0.2

2/8 = 0.25

6/18 = 0.33

etc...

The vertical asymptotes for the function f(x) = (x^2)(x-6)(x+3)^-1 are x=6 and x=-3.

The vertical asymptotes for the function f(x) = (x^2)(x-6)(x+3)^-1 are x=6 and x=-3. This is because the denominator can be factored into two linear expressions: x-6 and x+3. When either of these expressions equals 0, the fraction becomes undefined and a vertical asymptote is created. To find the asymptotes, set each linear expression in the denominator equal to 0 and solve for x. For x-6, x=6; for x+3, x=-3. Therefore, the vertical asymptotes of the function are x=6 and x=-3. When x is close to either of these values, the fraction will become very large, making the graph approach the vertical asymptotes without ever touching them.

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The floor area of the engineer's model of the sugar factory is 10.825 square feet.

To determine the floor area of the engineer's model of a sugar factory in square feet, we need to convert the given measurements from inches to feet. Since there are 12 inches in a foot, we can divide both dimensions by 12 to convert them.

The length of the model in feet is 30 inches / 12 = 2.5 feet, and the width is 52 inches / 12 = 4.33 feet.

To find the floor area, we multiply the length by the width:

Area = Length × Width

= 2.5 feet × 4.33 feet

= 10.825 square feet

It's important to note that the given measurements are not a standard aspect ratio or scale for a sugar factory. The given dimensions may be scaled down for the model's convenience, so the calculated floor area is only applicable to the scale of the model.

In actuality, a sugar factory would have much larger dimensions.

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Both c. f is continuous on the interval [-2,3], where f(-2)=0 and f(3)=20 and d. f is continuous on the interval [-2,3], where f(-2)=15 and f(3)=30 options are correct. given below is the explanation of the result.

using Intermediate value theorem:(statement: suppose that f∈c[a,b] and f(a)≠f(b) then given a number λ lies between f(a) and f(b) there exist a point c ∈(a,b) such that f(c)=λ)

know according to the Intermediate value theorem both option c and d are correct here because either f(a)<f(b) for a number λ lies between f(a) and f(b) there exist a point c ∈(a,b) such that f(c)=λ) here if we take  interval [-2,3], where f(-2)=0 and f(3)=20 the theorem is applicable. if we have f(a)>f(b) for a number λ lies between f(a) and f(b) there exist a point c ∈(a,b) such that f(c)=λ so if we take the interval [-2,3], where f(-2)=15 and f(3)=30,the above stated theorem is applicable.

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Answer:

ready-steady paint

Step-by-step explanation:

if he needs 12 tins, and purchased from paint -O  mine, he would spend (12/3) X 7.50 = 4 x 7.50 = £30

from ready steady, he can buy 4 for £11. he needs 12.

so he will spend (12/4) X 11 = 3 X 11 = £33. but he can get 15% off. 15% off is the same as multiplying by 0.85.

33 X 0.85 = £28.05.

so he his better purchasing from ready steady paint

To find the rate of change of the side of the square, we can use the formula for the area of a square: A = s^2, where A is the area and s is the side length.

Given that the area is increasing at a rate of 32 cm^2 per second, we can differentiate both sides of the equation with respect to time (t) to find the rate of change of the area: dA/dt = 2s * ds/dt.

Now, we can substitute the given rate of change of the area (32 cm^2/s) and the given side length (2 cm) into the equation to find the rate of change of the side length: 32 = 2(2) * ds/dt.

Simplifying the equation, we have: 32 = 4 * ds/dt.

Dividing both sides by 4, we get: ds/dt = 8 cm/s.

Therefore, the rate of change of the side length of the square when it is 2 cm is 8 cm/s.

- We used the formula for the area of a square, A = s^2, to relate the area and side length.
- By differentiating both sides of the equation with respect to time, we found an expression for the rate of change of the area in terms of the rate of change of the side length.
- We substituted the given values into the equation and solved for the rate of change of the side length.
- Finally, we concluded that the rate of change of the side length is 8 cm/s.

When the side length of the square is 2 centimeters, the rate of change of the side length is 8 centimeters per second.

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