Use a maclaurin series in this table to obtain the maclaurin series for the given function. F(x) = x cos(5x)

Answer:

y=2x

Step-by-step explanation:

hope this helps

:) :3

f(n) = ϴ(g(n)) is not true in cases B(sqrt(n)log(n), C(\(3^n 5^n\)), and D(sin(n)+3 cos(n)+1).

A) \(log(n^200) log(n^2)\)

Here, f(n) = \(log(n^200)\) and g(n) = \(log(n^2)\). Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = \([log(n^200) / log(n^2)]\) = 100

This means that as n approaches infinity, the ratio f(n) / g(n) is constant, and so we can say that f(n) = ϴ(g(n)). Therefore, f(n) = ϴ(g(n)) is true in this case.

B) sqrt(n) log(n) Here, f(n) = sqrt(n) and g(n) = log(n). Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = [sqrt(n) / log(n)]

As log(n) grows much slower than sqrt(n) as n approaches infinity, this limit approaches infinity. Therefore, we cannot say that f(n) = ϴ(g(n)) is true in this case.

C) 3^n 5^n

Here, f(n) = \(3^n\) and g(n) = \(5^n\) . Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = \([3^n / 5^n]\)

As \(3^n\) grows much slower than \(5^n\) as n approaches infinity, this limit approaches zero. Therefore, we cannot say that f(n) = ϴ(g(n)) is true in this case.

D) sin(n) + 3 cos(n) + 1

Here, f(n) = sin(n) + 3 and g(n) = cos(n) + 1. Now, if we take the limit of f(n) / g(n) as n approaches infinity, then:

f(n) / g(n) = [sin(n) + 3] / [cos(n) + 1]

As this limit oscillates between positive and negative infinity as n approaches infinity, we cannot say that f(n) = ϴ(g(n)) is true in this case.

Therefore, f(n) = ϴ(g(n)) is not true in cases B, C, and D.

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Answer:

Step-by-step explanation:

CX can be written in the form pa + qb

where p and q are fractions in terms of n.

c) (1) What is the value of p?

(i) What is the value of q?

(1)

m

(1)

i need help with c

Answer: \(x < -\frac{24}{5}\)

Step-by-step explanation:

You have:

\(-\frac{5}{6} x-3 > 1\)

Add 3 to both sides to get:

\(-\frac{5}{6} x > 4\)

Multiply both sides by 6 to get:

\(-{5} x > 24\)

Divide both sides by -5. When you divide an inequality by a negative number, you must switch the signs. Therefore:

\(x < -\frac{24}{5}\)

Answer:

The equation is saying multiply or add negative 5/6 to x, then subtract by positive 3 which makes the number go upwards, but not making it positive, and lastly, the equation is saying the result will be greater than 1.

One of the values for x is 5, because...

\(-\frac{5}{6} * 5 = -4\frac{1}{6} - 3 = 7\frac{1}{6} > 1\)

With that information, the statement is now complete and true. If you make 5 or anything more than 5 the value for x, the statement will still be true.

Note that I just used 5 as one of the solutions. There are many more than that.

Answer:  y = (1/3)x - 5

This is the same as writing \(y = \frac{1}{3}x-5\)

======================================================

Explanation:

Assume we have a linear pattern matching the form y = mx+b.

The first row has x = 12 lead to y = -1

The second row has x = 3 lead to y = -4

This produces the two points (12,-1) and (3,-4)

Let's find the slope through those points.

m = (y2-y1)/(x2-x1)

m = (-4-(-1))/(3-12)

m = (-4+1)/(3-12)

m = -3/(-9)

m = 1/3

The slope is 1/3.

The y intercept is b = -5 since x = 0 leads to y = -5

We go from y = mx+b to y = (1/3)x - 5

--------------------------

To verify we have the correct equation, plug in each left side value as x values.

Let's try x = 12

y = (1/3)x - 5

y = (1/3)*12 - 5

y = 4 - 5

y = -1

This shows that the input x = 12 maps to the output y = -1

i.e.  12 ↦ −1 is the case here.

Repeat for x = 3

y = (1/3)x - 5

y = (1/3)*3 - 5

y = 1 - 5

y = -4

We can see that 3 ↦ −4 occurs.

I'll let you check the others, but you should find that the other mappings hold true as well. Therefore, the answer has been fully confirmed.

The function for the surface area of the closed box with a square base is \(S(x) = 4x^2 + 8xh\), where x represents the length of the side of the square base and h represents the height of the box. This function takes into account the areas of the square base and the four rectangular sides.

To determine the surface area function, we need to consider the different components of the box's surface. The box has a square base, so the area of each side of the base is \(x^2\). Since there are four sides to the base, the total area of the base is \(4x^2\). Additionally, there are four identical rectangular sides with dimensions x by h, resulting in a total area of 4xh.

Combining the areas of the base and the four sides, we have the surface area function \(S(x) = 4x^2 + 8xh\), which represents the total surface area of the closed box.

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1. In a triangle line parallel to one side will divides other two sides in the same ratio

2. 8/x = 12/18

   x = 12

Finding the slope between two points on a line and solving for the y-intercept in the slope-intercept equation y=mx+b allows us to formulate an equation for that line.

A linear equation has the slope-intercept form y = mx + b. Variables in the equation are x and y. When x is 0, the integers m and b provide the line's slope (m) and the value of y. (b).

Because (0,y) is the location where the line crosses the y-axis, the value of y when x is 0 is referred to as the y-intercept.

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Marcus can type 120 words per one and a half minutes while Rhys can only 100 words so Marcus' rate is 20 words per one and a half minutes greater than Rhys.

In mathematics, the arithmetic operation has four main operators such as addition, subtraction, multiplication, and division.

Given that,

Rhys can type 100 words in one and a half minutes.

Rate of Rhys = 100 words / 1.5 minutes.

Marcus can type 40 words in half a minute

Rate of Marcus = 40 words / 0.5 minutes

Multiply and divide by 1.5

Rate of Marcus = 40(1.5)/0.5words/1.5minutes

Rate of Marcus = 120 words / 1.5 minutes.

So Rate of Marcus > Rate of Rhys

Hence "Marcus' rate is 20 words per one and a half minutes greater than Rhys".

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9514 1404 393

Answer:

  $100

Step-by-step explanation:

The amount of simple interest earned is given by the formula ...

   I = Prt

where P is the amount investe, r is the annual rate, and t is the number of years.

Solving for P, we find ...

  P = I/(rt) = $5.00/(0.005·10) = $100

Susan deposited $100.

Answer:

2 \(\frac{5}{2}\)

Step-by-step explanation:

\(\frac{25}{1} \\\) × \(\frac{10}{100}\) =  \(\frac{10}{4}\) or

Simplified: \(\frac{5}{2}\)

2 whole number \(\frac{1}{2}\)

ANSWERS

a) 58

b) 62.2

EXPLANATION

a) The median is the middle value. It separates the data set in two. To find the median we have to put the data in order, from least to greatest:

The number of data is odd, this means that the median will be one of the values of the data set and not a mean between two of them. If there are 9 values, the median is the 5th value - so we have 4 values to the left and 4 values to the right.

The first 4 values are 44,49,53,54, so the median is 58.

b) The mean is the sum of all values of the data set divided by the amount of data:

Rounded to one decimal place, the mean is 62.2

Answer:

its showing how its congruent

Step-by-step explanation:

congruent through  the lines

Answer:

f(x) = x × 0.12 if  15,000 ≤ x ≤ 30,000

        x × 0.18 if  x ≥ 30,000

Step-by-step explanation:

I. When 12% is 12/100 = 0.12

 and 18% is 18/100 = 0.18

The missing values are:

BC = 25, BN = 16, NC = 9, and AN = 12

Given:

AB = 20

AC = 15

Using the Pythagorean theorem in triangle ABC:

CB = √AC² + AB²

CB= √20² + 15²

CB= √625

CB= 25

Since, AN is altitude we have ΔNBA~ΔABC and ΔNAC~ΔABC

BN / AN = BA / CB

BN/ 20= 20/25

BN = 16

and, AN / BA = AC / CB

AN /20 = 15/ 25

AN = 12

Now, CN=  CB - BN

CN = 25- 16

CN= 9

BC^2 = 20^2 + 15^2

BC^2 = 400 + 225

Since ΔANC is a right triangle so

AN = √15²-12²

AN = √81

AN = 9

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¡Hola! Creo que tu respuesta es una expresión algebraica B, aunque no estoy 100% seguro. ¡Espero que esto te ayude! Buena suerte y que tengas un gran día. ❤️

The student answered 27 out of 30 questions correctly on the history test, achieving a 90% accuracy rate.

To calculate the number of questions the student answered correctly, we can multiply the total number of questions (30) by the percentage of questions answered correctly (90%). The calculation is as follows:

Number of questions answered correctly = Total number of questions × Percentage of questions answered correctly

= 30 × 0.90

= 27

Therefore, the student answered 27 questions correctly on the history test.

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Answer:

A. y = (x + 3)(x - 2)

Step-by-step explanation:

I used the points where the line crosses the x - axis. (0,-3 and 0,2).

Also, graph it. I use desmos.

Answer:

A)

Step-by-step explanation:

Graphed the equation

Answer: A. 60 inches

Step-by-step explanation: 18 x 12 is 216 (area) so 18 + 18 + 12 + 12 = perimeter aka 60 inches.

The function is neither an even function nor an odd function

The function is given as

g(x) = x + x^2

Calculate g(-x)

So, we have

g(-x) = -x + (-x)^2

Evaluate

g(-x) = -x + x^2

Calculate -g(x)

So, we have

-g(x) = -x - x^2

In the above computations;

g(x) does not equal g(-x) and -g(x)

Hence, the function is neither an even function nor an odd function

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The area of the region enclosed by the curve r = 9cos(3θ) is (27/2)π.

What is integration?

In mathematics, and notably in calculus, integration is a fundamental notion. It is a mathematical process that seeks to determine a function's integral. The accumulation or total of all infinitesimally small changes in a quantity is represented by the integral.

To find the area of the region enclosed by the curve r = 9cos(3θ), we can set up a double integral in polar coordinates.

In polar coordinates, the area element is given by dA = r dr dθ. To determine the limits of integration, we need to find the values of θ where the curve intersects itself and encloses a region.

The polar curve r = 9cos(3θ) completes one loop for every 2π/3 radians, so we can integrate over the range 0 ≤ θ ≤ 2π/3. The corresponding limits for r can be determined by setting r = 0 and solving for θ.

At r = 0, we have:

0 = 9cos(3θ)

cos(3θ) = 0

The equation cos(3θ) = 0 has solutions at θ = π/6, π/2, 5π/6. These values divide the interval [0, 2π/3] into three subintervals.

Now we can set up the double integral:

Area = ∬R dA

Using polar coordinates, we have:

dA = r dr dθ

The limits of integration are:

0 ≤ r ≤ 9cos(3θ)

0 ≤ θ ≤ 2π/3

Thus, the double integral becomes:

Area = ∫[0 to 2π/3] ∫[0 to 9cos(3θ)] r dr dθ

Now we can evaluate this double integral:

Area = ∫[0 to 2π/3] (1/2)r² ∣[0 to 9cos(3θ)] dθ

Area = (1/2) ∫[0 to 2π/3] (81cos²(3θ)) dθ

Using the trigonometric identity cos²(3θ) = (1 + cos(6θ))/2, we can simplify further:

Area = (1/2) ∫[0 to 2π/3] (81/2)(1 + cos(6θ)) dθ

Area = (81/4) ∫[0 to 2π/3] (1 + cos(6θ)) dθ

Now we can integrate term by term:

Area = (81/4) [(θ + (1/6)sin(6θ)) ∣[0 to 2π/3]]

Area = (81/4) [(2π/3 + (1/6)sin(4π) - (1/6)sin(0))]

Simplifying further:

Area = (81/4) [(2π/3 + 0 - 0)]

Area = (81/4) (2π/3)

Area = (27/2)π

Therefore, the area of the region enclosed by the curve r = 9cos(3θ) is (27/2)π.

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According to the solving the dimensions of the triangles are as follow:

a . 29

b. 58

c. 57

A mathematical space's dimension is defined informally as the smallest number of coordinates required to specify any point within it. Thus, a line has a dimension of one since only one coordinate is required to identify a point on it - for example, the point at 5 on a number line.

2-D shapes really had no thickness and they can only be measured on two sides. Two-dimensional objects include the square, circle, rectangle, and triangle.

x=length of the shortest side

2x=length of another side

x+28=length of the third side

144=x+2x+x+28

144=4x+28

144-28=4x

116=4x

x=116/4

x=29 feet (one side)

Putting the value of x to find second side

2x=58 feet (second side)

x+28= 57 feet (third side)

the perimeter of the triangle = sum of all the sides

                                                = 29+58+57

                                                =144

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The slope of the ideal vector, determined by examining the coefficients of the variables in the regression equation is:

Cleaning Ability: 2.3

Cost: 1.0

Disinfecting Ability: 0.6.

In this case, the regression equation is:

Overall Preference = -2.1 + 2.3 * Cleaning Ability + 1.0 * Cost + 0.6 * Disinfecting Ability

The coefficients of the variables represent the weights or importance assigned to each variable in determining the overall preference. Therefore, the slope of the ideal vector would be the coefficients of the variables in the regression equation.

Based on the given regression equation, the slope of the ideal vector would be:

Cleaning Ability: 2.3

Cost: 1.0

Disinfecting Ability: 0.6

These values indicate the relative importance or impact of each variable on the overall preference for Slippery Soap.

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Answer:

56

Step-by-step explanation:

just multiply 8 by 7. to check do 56 divided by 7 which equals 8.

Answer:

P=56

Step-by-step explanation:

Multiply both sides by 7 &p/7=8x7

Simplify P=56

The general formula of a polynomial is:

P(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_2x^2 + a_1x + a_0

A polynomial is a mathematical expression consisting of variables (such as x), coefficients, and exponents. The general formula of a polynomial is P(x) = a_nx^n + a_(n-1)x^(n-1) + ... + a_2x^2 + a_1x + a_0. Here, a_n, a_(n-1), ... , a_2, a_1, and a_0 are coefficients, and n is the degree of the polynomial. The highest exponent in the polynomial is the degree of the polynomial. For example, in the polynomial 3x^2 + 5x – 8, x^2 is the highest exponent, so the degree of the polynomial is 2. The coefficient of the highest exponent is known as the leading coefficient. In the example above, 3 is the leading coefficient. Polynomials can be used to represent a variety of functions, from simple linear equations to complex curves.

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Answer:

Heeeheeeeheee

Step-by-step explanation:

So right

The answer are: a. Total outflows: $2,007,901, b. Total inflows: $827,080, c. Net present value: $824,179, d. Should the old issue be refunded with new debt? Yes

To determine whether the old bond issue should be refunded with new debt, we need to calculate the present value of total outflows, the present value of total inflows, and the net present value (NPV). Let's calculate each of these values step by step: Calculate the present value of total outflows. The total outflows consist of the call premium, underwriting cost on the old issue, and underwriting cost on the new issue. Since these costs are one-time payments, we can calculate their present value using the formula: PV = Cash Flow / (1 + r)^t, where PV is the present value, Cash Flow is the cash payment, r is the discount rate, and t is the time period.

Call premium on the old issue: PV_call = (7% of $22 million) / (1 + 0.1)^16, Underwriting cost on the old issue: PV_underwriting_old = $530,000 / (1 + 0.1)^16, Underwriting cost on the new issue: PV_underwriting_new = $680,000 / (1 + 0.1)^16. Total present value of outflows: PV_outflows = PV_call + PV_underwriting_old + PV_underwriting_new. Calculate the present value of total inflows. The total inflows consist of the interest savings and the tax savings resulting from the interest expense deduction. Since these cash flows occur annually, we can calculate their present value using the formula: PV = CF * [1 - (1 + r)^(-t)] / r, where CF is the cash flow, r is the discount rate, and t is the time period.

Interest savings: CF_interest = (12% - 10%) * $22 million, Tax savings: CF_tax = (40% * interest expense * tax rate) * [1 - (1 + r)^(-t)] / r. Total present value of inflows: PV_inflows = CF_interest + CF_tax.  Calculate the net present value (NPV). NPV = PV_inflows - PV_outflows Determine whether the old issue should be refunded with new debt. If NPV is positive, it indicates that the present value of inflows exceeds the present value of outflows, meaning the company would benefit from refunding the old issue with new debt. If NPV is negative, it suggests that the company should not proceed with the refunding.

Now let's calculate these values: PV_call = (0.07 * $22,000,000) / (1 + 0.1)^16, PV_underwriting_old = $530,000 / (1 + 0.1)^16, PV_underwriting_new = $680,000 / (1 + 0.1)^16, PV_outflows = PV_call + PV_underwriting_old + PV_underwriting_new. CF_interest = (0.12 - 0.1) * $22,000,000, CF_tax = (0.4 * interest expense * 0.4) * [1 - (1 + 0.1)^(-16)] / 0.1, PV_inflows = CF_interest + CF_tax. NPV = PV_inflows - PV_outflows.  If NPV is positive, the old issue should be refunded with new debt. If NPV is negative, it should not.

Performing the calculations (rounded to the nearest whole dollar): PV_call ≈ $1,708,085, PV_underwriting_old ≈ $130,892, PV_underwriting_new ≈ $168,924, PV_outflows ≈ $2,007,901,

CF_interest ≈ $440,000, CF_tax ≈ $387,080, PV_inflows ≈ $827,080.  NPV ≈ $824,179.  Since NPV is positive ($824,179), the net present value suggests that the old bond issue should be refunded with new debt.

Therefore, the answers are:

a. Total outflows: $2,007,901

b. Total inflows: $827,080

c. Net present value: $824,179

d. Should the old issue be refunded with new debt? Yes

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Answer:

x=10 , x=13

Steps:

5.

4x+5x=180-90

9x=90

x=90/9

x=10

6.

(3x+7)+(2x+3)=180-105

5x+10=75

5x=65

x=65/5

x=13

Step-by-step explanation:

☯ \( \underline{ \underline{ \large{ \tt{S \: O \: L \: U \: T\: I \: O \: N}}}} : \)

✰ \( \underline{ \underline{ \large{ \text{For \: Question \: Number \: 5}}}} : \)

\( \large{ \tt{90 \degree + 4x \degree + 5x \degree = 180 \degree}}\) [ Sum of angle in a triangle ]

⤑ \( \large{ \tt{90 \degree + 9x \degree = 180 \degree}}\)

⤑ \( \large{ \tt{9x \degree = 180 \degree - 90 \degree}}\)

⤑ \( \large{ \tt{9x \degree = 90 \degree}}\)

⤑ \( \large{ \tt{ \frac{9x}{9} = \frac{90}{9} \degree}}\)

⤑ \( \boxed{ \large{ \tt{x = 10 \degree}}}\)

----------------------------------------------------------

✰ \( \underline{ \underline{ \text{For \: Question \: Number \: 6}}} : \)

\( \large{ \tt{75 \degree = 3x \degree + 7 \degree + 2x \degree + 3 \degree}}\) [ Exterior angle is equal to the sum of two non-adjacent interior angles ]

⟶ \( \large{ \tt{75 \degree = 5x \degree + 10 \degree}}\)

⟶ \( \large{ \tt{5x \degree \: + 10 \degree = 75 \degree}}\)

⟶ \( \large{ \tt{5x \degree = 75 \degree - 10 \degree}}\)

⟶ \( \large{ \tt{5x \degree = 65 \degree}}\)

⟶ \( \large{ \tt{ \frac{5x}{5} = \frac{65}{5} \degree}}\)

⟶ \( \boxed{\large{ \tt{x = 13 \degree}}}\)

Hope I helped ! ✧

Have a wonderful day / night ! ♪

Let me know if you have any questions regarding my answer!

\( \underline{ \underline{ \mathfrak{Carry \: On \: Learning}}}\) !! ♕

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