The 11th square number is 121. The sum of which consecutive 11 odd numbers
is 121

Answer:

a) 14.85

b) 0.1136

c) 74.7

Answer:

Step-by-step explanation:

Question 1

System of equations is,

y = -6x -------(1)

y = -4x - 2 ---------(2)

Substitute the value of y from equation (1) to equation (2)

-6x = -4x - 2

-6x + 4x = -2

-2x = -2

x = 1

From equation (1)

y = -6(1)

y = -6

Question (2)

System of equations is,

-y = x --------(1)

3x + 5y = 20 ---------(2)

Substitute the value of x from equation (1) to equation (2)

3(-y) + 5y = 20

-3y + 5y = 20

2y = 20

y = 10

From equation (1)

-y = x

x = -10

Question (3)

System of the equations is,

y = -2x - 7 --------(1)

9x - 10y = 12 ----------(2)

Substitute the value of y from equation (1) to equation (2)

9x - 10(-2x - 7) = 12

9x + 20x + 70 = 12

29x = 12 - 70

29x = -58

x = -2

From equation (1)

y = -2(-2) - 7

y = 4 - 7

y = -3

Answer:

S, BB, J

Step-by-step explanation:

Because you are going greatest (positive) to least (negative)

Highest to lowest

We are given the conversion factors from kilograms to grams and from pounds to ounces. We are asked to convert 360 kilograms into pounds. To do that, let's remember that 1 kg = 2.2 pounds. Therefore, to convert the measurement of 360 kilograms we multiply by the conversion factor in such a way that the unit we want to find is in the denominator, like this:

Solving the operation we get:

Answer:

isosceles triangle

Step-by-step explanation:

isosceles triangles have two equal sides

Maclaurin series of `f(x)` is given by:f(x) = `f(0)` + `f'(0)x` + `(f''(0)/2!) x²` + `(f'''(0)/3!) x³` + `(f⁴(0)/4!) x⁴` + `(f⁵(0)/5!) x⁵` + `(f⁶(0)/6!) x⁶` = `0 + 0x + 0x² + 0x³ + 0x⁴ + 0x⁵ + (7200/6!)x⁶` = `10x⁶`

Answer: `10x⁶`.

The given function is `f(x) = x⁶ sin(10x⁵)`. We need to find the Taylor series of `f` centered at `0` (Maclaurin series of `f`).

Formula used: The Maclaurin series for `f(x)` is given by `f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...... + (f^n(0)/n!)x^n`.

Here, `f(0) = 0` because `sin(0) = 0`.

Differentiating `f(x)` and its derivatives at `x = 0`:`f(x) = x⁶ sin(10x⁵)`

First derivative: `f'(x) = 6x⁵ sin(10x⁵) + 50x¹⁰ cos(10x⁵)`

Differentiate `f'(x)`

Second derivative: `f''(x) = 30x⁴ sin(10x⁵) + 200x⁹ cos(10x⁵) - 250x¹⁰ sin(10x⁵)`

Differentiate `f''(x)`

Third derivative: `f'''(x) = 120x³ sin(10x⁵) + 1800x⁸ cos(10x⁵) - 2500x⁹ sin(10x⁵) - 5000x²⁰ cos(10x⁵)`

Differentiate `f'''(x)`

Fourth derivative: `f⁴(x) = 360x² sin(10x⁵) + 7200x⁷ cos(10x⁵) - 22500x⁸ sin(10x⁵) - 100000x¹⁹ cos(10x⁵) + 100000x²⁰ sin(10x⁵)`

Differentiate `f⁴(x)`

Fifth derivative: `f⁵(x) = 720x sin(10x⁵) + 36000x⁶ cos(10x⁵) - 112500x⁷ sin(10x⁵) - 1900000x¹⁸ cos(10x⁵) + 2000000x¹⁹ sin(10x⁵)`

Differentiate `f⁵(x)`

Sixth derivative: `f⁶(x) = 7200 cos(10x⁵) - 562500x⁶ cos(10x⁵) + 13300000x¹⁷ sin(10x⁵)`

Evaluate at `x = 0`:

The derivatives of `f(x)` evaluated at `x = 0` are:f(0) = 0f'(0) = 0f''(0) = 0f'''(0) = 0f⁴(0) = 0f⁵(0) = 0f⁶(0) = 7200

Maclaurin series of `f(x)` is given by:f(x) = `f(0)` + `f'(0)x` + `(f''(0)/2!) x²` + `(f'''(0)/3!) x³` + `(f⁴(0)/4!) x⁴` + `(f⁵(0)/5!) x⁵` + `(f⁶(0)/6!) x⁶` = `0 + 0x + 0x² + 0x³ + 0x⁴ + 0x⁵ + (7200/6!)x⁶` = `10x⁶`

Answer: `10x⁶`.

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Answer:

lesser x =-4

greater x =1/2

Answer:

The first one is 5.83 or -5 5/6

The second one is -414 1/3

The last one is 290

Answer:

Standard Form:

720.058

Expanded forms can be written like a sentence or stacked for readability as they are here.

Expanded Notation Form:

 720.058 =

 700  

+ 20  

+ 0  

+   0.0

+   0.05

+   0.008

Expanded Factors Form:

   720.058 =

 7 × 100  

+ 2 × 10  

+ 0 × 1  

+ 0 ×   0.1

+ 5 ×   0.01

+ 8 ×   0.001

Expanded Exponential Form:

 720.058 =

7 × 102

+ 2 × 101

+ 0 × 100

+ 0 × 10-1

+ 5 × 10-2

+ 8 × 10-3

Word Form:

720.058 =

seven hundred twenty and fifty-eight thousandths

Step-by-step explanation:

Answer:

700 + 20 + 0 + 0.050 + 0.008

Step-by-step explanation:

The dimension of the cardboard is 12in by 5in

So when the cardboard is curved into a cylinder

there are two ways to do that

If diagram 'a' is folded into a cylinder

the circunference of the circular base form is 5 inches

then the radius is 0.8 in

since circuference = 2x pi x r

If diagram 'b' is folded into a cylinder

the circumference of the circular base is 12 inches

then the radius is 1.9 in

the volume

So diagram b gives the larger volume

Hence, when the cardboard in curved in such a way that the 12 inches side forms the base, the volume is larger than when the cardboard is curved such that the 5 inches side forms the base

Answer:

-13.04.

Step-by-step explanation:

The orthogonal trajectories of the family of circles x^2 + y^2 = a are given by the differential equation y = -x(dy/dx), where (x, y) represents a point on the orthogonal trajectory.

To find the orthogonal trajectories of the family of circles given by the equation x^2 + y^2 = a, where "a" is a constant, we can follow these steps:

Step a: Find the differential equation representing the family of circles.

Differentiating the equation x^2 + y^2 = a implicitly with respect to x, we get:

2x + 2y(dy/dx) = 0.

Step b: Find the slope of the tangent line to the circles.

Rearranging the equation obtained in Step a, we have:

dy/dx = -x/y.

Step c: Find the differential equation representing the orthogonal trajectories.

Since orthogonal trajectories have perpendicular slopes to the family of circles, the slope of the tangent line to the orthogonal trajectories will be the negative reciprocal of the slope of the circles. Therefore, the slope of the tangent line to the orthogonal trajectories is y/x.

Differentiating this expression implicitly with respect to x, we get:

(dy/dx) = -y/x^2.

Rearranging the equation obtained in Step c, we have:

y = -x(dy/dx).

This equation represents the differential equation for the orthogonal trajectories of the family of circles x^2 + y^2 = a.

In summary, the orthogonal trajectories of the family of circles x^2 + y^2 = a are given by the differential equation y = -x(dy/dx), where (x, y) represents a point on the orthogonal trajectory.

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The probability that a randomly chosen 10-year-old child in the country of United States of Heightlandia has a height of less than 54.6 inches is approximately 0.2776 or 27.76%.

To answer this question, we can use the standard normal distribution formula:

z = (x - μ) / σ

where:

x = the height value we want to find the probability for (in this case, 54.6 inches)

μ = the mean height of the population (55.9 inches)

σ = the standard deviation of the population (2.2 inches)

z = the corresponding z-score

Substituting the given values, we get:

z = (54.6 - 55.9) / 2.2 = -0.59

Now we need to find the probability that a randomly chosen child has a height less than 54.6 inches, which is equivalent to finding the area under the standard normal distribution curve to the left of z = -0.59.

Using a standard normal distribution table or calculator, we find that the area to the left of z = -0.59 is approximately 0.2776.

Therefore, the probability that a randomly chosen child has a height of less than 54.6 inches is approximately 0.2776 or 27.76%.

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Answer:

1/b

Step-by-step explanation:

b=b/1. 1/b would be a reciprical because it switches the numberator and denominator.

Answer:

The equation of the line that passes through the given points is ;

7y = -x + 12

Step-by-step explanation:

Here, we want to get the equation of the line that passes through the given points

The general equation form

is;

y = mx + b

where m is the slope and b is the y-intercept

Now, let us substitute the x and y coordinate values of each of the points;

for (-2,2); we have

2 = -2m + b

b = 2m + 2 •••••(i)

for F;

1 = 5m + b

b = 1-5m ••••••(ii)

Equate both b

1-5m = 2m + 2

1-2 = 2m + 5m

7m = -1

m = -1/7

Recall;

b = 2m + 2

b = 2(-1/7) + 2

b = -2/7 + 2

b = (-2 + 14)/7 = 12/7

The equation of the line is thus;

y = -1/7x + 12/7

Multiply through by 7

7y = -x + 12

such pair of equation that have only one pair of solutions which satisfy both equation is linear equation

Answer:

13.7

Step-by-step explanation:

Let us imagine that 0 on the numberline. -9.2 is to the left, and 4.5 is to the right of zero.

Since distance as positive, we can simply add the absolute values of -9.2 and 4.5 to obtain the distance:

|-9.2|+|4.5|=

9.2+4.5=

13.7

I hope this helps! :)

By using the concepts of Application of derivatives(A.O.D.) we get that in order to reach the cabin point P is to be 25  feet away from the cabin.

The concept of derivatives has been used on a small scale and a large scales. The concept of derivatives is used in many ways such as change of temperature or rate of change of shapes and sizes of an object depending on the conditions etc.,

The concept is to find that point such that the derivative of a given function will be zero, as we need to find the local minima.

So after drawing the picture according to the given question

Naming image as from left side APBCD in anticlockwise sense

Assume ,P is at x feet from the cabin, therefore ,AP=(100-x) feet

Using Pythogoras Theorem

\(AP^{2}+AD^{2}=PD^{2}\)

=>\(PD^{2}=AP^{2}+AD^{2}\)

=>PD=\(\sqrt{AP^{2}+AD^{2} }\)

=>PD=\(\sqrt{(100-x)^{2}+(100)^{2} }\)

Now, we have rate in terms of time, so we applying differentiation on both sides of equation with time

We also know

differentiation of \(\sqrt{x}\) is 1/2×\(\sqrt{x}\)

Differentiating both sides with time t

d(PD)/dt= \(\frac{1}{2\sqrt{(100-x)^{2}+(100)^{2} } }2\frac{x-100}\)×dx/dt

We have given d(PD)/dt as 3ft/sec and dx/dt as 5 ft/sec

After putting the values

=>3=\(\frac{(x-100)}{\sqrt{(100)^{2}+(100-x)^{2} } }\)     ×5

After squaring on both sides and moving x on one side of equation and constant term on the other side of equation we get

100-x=\(\sqrt{9/16 }\)×100

100-x=75

=>x=25feet

Hence Point p should be at 25 feet from the cabin

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Company A has a higher risk percentage (55%) compared to Company B (3%).

To compute the Coefficient of Variation (CV) for each company, we need to use the formula:

CV = (Standard Deviation / Mean) * 100

Let's calculate the CV for each company:

For Company A:

Risk Percentage = 55%

Return = 14%

For Company B:

Risk Percentage = 3%

Return = 14%

Since we don't have the standard deviation values for each company, we cannot calculate the exact CV. However, we can still compare the riskiness of the two companies based on the provided information.

The Coefficient of Variation measures the risk relative to the return. A higher CV indicates higher risk relative to the return, while a lower CV indicates lower risk relative to the return.

In this case, Company A has a higher risk percentage (55%) compared to Company B (3%), which suggests that Company A is riskier. However, without the standard deviation values, we cannot make a definitive conclusion about the riskiness based solely on the provided information. The CV would provide a more accurate measure for comparison if we had the standard deviation values for both companies.

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An identity equation is an equation that is always true for any value substituted into the variable. The last equation is called a trigonometric identity.

In its simplest form in algebra, the definition of an equation is a mathematical statement that shows that two mathematical expressions are equal.

Trigonometric Identities are the equalities that involve trigonometry functions and holds true for all the values of variables given in the equation. There are various distinct trigonometric identities involving the side length as well as the angle of a triangle.

Trigonometric identities are the equations that include the trigonometric functions such as sine, cosine, tangent, etc., and are true for all values of angle θ. Here, θ is the reference angle taken for a right-angled triangle.

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The statement that describes the relationship between the larger figure and the smaller figure, its image after a dilation is given as follows:

C. The edge lengths of the smaller figure are less than half than those of the larger figure.

The scale factor is given as follows:

k = 1/4 = 0.25.

A dilation can be defined as a transformation that multiplies the distance between every point in an object and a fixed point, called the center of dilation, by a constant factor called the scale factor.

The side lengths are given as follows:

Hence the scale factor is given as follows:

k = 3/12

k = 1/4

k = 0.25.

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Answer:

300/6=50 cups(1 pitcher contains 50 cups of lemonade)

convert into quarts:

50/4=12.5 quarts

Answer:

Step-by-step explanation:

The base and height will be the two equal sides of the isosceles right triangle.

⇒ b = h = x cm

\(Area \ of \ triangle = \dfrac{1}{2}bh\)

\(\dfrac{1}{2}bh = 72 \ cm^{2}\)

\(\dfrac{1}{2}*x*x=72\\\\\\\dfrac{1}{2}*x^{2}=72\\\\\\x^{2}=72*2\\\\x^{2}=144\\\\Take \ square \ root,\\\\\sqrt{x^{2}}=\sqrt{144}\\\\x = \sqrt{12*12}\\\\x = 12 cm\)

Hypotenuse² = b² + h²

                      = 12² + 12²

                       = 144 + 144

                     = 288

hypotenuse = √288

                     = 16.97 cm

Answer:

\(\large{\boxed{\sf Hypotenuse = 16.97\ cm }}\)

Step-by-step explanation:

Here it is given that the area of a right isosceles ∆ is 72 cm² . Let us assume that each equal side is x . Therefore the height and the base of the ∆ will be same that is x .

\(\sf\qquad\longrightarrow Area =\dfrac{1}{2}(base)(height)\\ \)

\(\sf\qquad\longrightarrow 72cm^2=\dfrac{1}{2}(x)(x)\\\)

\(\sf\qquad\longrightarrow x^2= 144cm^2\\ \)

\(\sf\qquad\longrightarrow x =\sqrt{144cm^2}\\ \)

\(\sf\qquad\longrightarrow \pink{x = 12cm }\)

Hence we may find hypotenuse using Pythagoras Theorem as ,

\(\sf\qquad\longrightarrow h =\sqrt{ p^2+b^2} \)

\(\sf\qquad\longrightarrow h =\sqrt{ (12cm)^2+(12cm)^2}\\\)

\(\sf\qquad\longrightarrow h =\sqrt{144cm^2+144cm^2}\\\)

\(\sf\qquad\longrightarrow h =\sqrt{288cm^2}\\\)

\(\sf\qquad\longrightarrow \pink{ hypotenuse= 16.97cm }\)

Hence the hypotenuse is 16.97 cm .

The rule of similarity that is not included in the given list is "RHS," which stands for "Right Angle, Hypotenuse, Side."

This rule states that if a right triangle has a hypotenuse and one side that are congruent to the hypotenuse and one side of another right triangle, respectively, then the two triangles are similar.

The other three rules of similarity that are listed in your question are:

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The measure of angle A in the isosceles triangle ABC is 62.5 degrees.

In an isosceles triangle ABC, where AB = AC, we are given that angle B (denoted as ∠B) measures 55 degrees. We need to calculate the measure of angle A (denoted as ∠A).

Since AB = AC, we know that angles A and C are congruent (denoted as ∠A ≅ ∠C). In an isosceles triangle, the base angles (the angles opposite the equal sides) are congruent.

Therefore, we have:

∠A ≅ ∠C

Also, the sum of the angles in a triangle is always 180 degrees. Hence, we can write:

∠A + ∠B + ∠C = 180

Substituting the given values:

∠A + 55 + ∠A = 180

Combining like terms:

2∠A + 55 = 180

Subtracting 55 from both sides:

2∠A = 180 - 55

2∠A = 125

Dividing by 2:

∠A = 125 / 2

∠A = 62.5

Therefore, the measure of angle A in the isosceles triangle ABC is 62.5 degrees.

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Answer:

56 dude

Step-by-step explanation:

Answer:

56!!!!!!

Step-by-step explanation:

By using the method of differentiation on the given MGF, we can find the covariance Cov(X1, X2) of the random vector [X1, X2]. The differentiation process involves calculating the expected values and variances of X1 and X2, enabling us to determine the relationship between the two variables and how they vary together.

To find the covariance Cov(X1, X2), we utilize the method of differentiation applied to the given MGF. The covariance is obtained by taking the second partial derivatives of the MGF with respect to t1 and t2. Specifically, we differentiate the MGF twice with respect to each of the variables and evaluate it at t1 = 0 and t2 = 0.

By taking the first partial derivative with respect to t1 and evaluating at t1 = 0 and t2 = 0, we obtain the expected value E(X1). Similarly, by taking the first partial derivative with respect to t2 and evaluating at t1 = 0 and t2 = 0, we get the expected value E(X2). These values represent the means of X1 and X2, denoted by μ1 and μ2, respectively.

Next, we proceed to take the second partial derivatives with respect to t1 and t2. Evaluating them at t1 = 0 and t2 = 0 gives us the variances Var(X1) and Var(X2), denoted by σ12 and σ22, respectively.

Additionally, the cross-partial derivative evaluated at t1 = 0 and t2 = 0 provides us with the covariance term Cov(X1, X2), which is the desired result.

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Answer:

The end points of line segment are:

\((x_1,y_1) = (5,4)\\(x_2,y_2) = (-4,-3)\)

Using distance formula,

Distance = \(\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}\) = \(\sqrt{(-4-5)^{2}+(-3-4)^{2}} = \sqrt{(-9)^{2}+(-7)^{2}} = \sqrt{81+49}=\sqrt{130} units\)

Answer: the price of one senior citizen ticket is $10, and the price of one student ticket is $5.

Step-by-step explanation:

Let's assume the price of one senior citizen ticket is 's' dollars and the price of one student ticket is 't' dollars.

According to the given information, on the first day, the school sold 7 senior citizen tickets and 11 student tickets, totaling $125. This can be expressed as the equation:

7s + 11t = 125 ---(1)

On the second day, the school sold 14 senior citizen tickets and 8 student tickets, totaling $180. This can be expressed as the equation:

14s + 8t = 180 ---(2)

We now have a system of two equations with two variables. We can solve this system to find the values of 's' and 't'.

Multiplying equation (1) by 8 and equation (2) by 11, we get:

56s + 88t = 1000 ---(3)

154s + 88t = 1980 ---(4)

Subtracting equation (3) from equation (4) eliminates 't':

(154s + 88t) - (56s + 88t) = 1980 - 1000

98s = 980

s = 980 / 98

s = 10

Substituting the value of 's' back into equation (1), we can solve for 't':

7s + 11t = 125

7(10) + 11t = 125

70 + 11t = 125

11t = 125 - 70

11t = 55

t = 55 / 11

t = 5

Therefore, the price of one senior citizen ticket is $10, and the price of one student ticket is $5.