Solve for y 11x+y= 18x

Answer:

X=19

Step-by-step explanation:

2x + 10 = -28

     -10     -10         Subtract 10 from both sides

divide both sides by 2

That gives you

X= 19

Answer:

p ≈ 7.3 m

Step-by-step explanation:

using the Sine Rule in the triangle

\(\frac{p}{sinP}\) = \(\frac{r}{sinR}\) ( substitute values )

\(\frac{p}{sin40}\) = \(\frac{10}{sin62}\) ( cross- multiply )

p × sin62° = 10 × sin40° ( divide both sides by sin62° )

p = \(\frac{10sin40}{sin62}\) ≈ 7.3 m ( to the nearest tenth )

Answer:

\(\fbox {p = 7.3 m}\)

Step-by-step explanation:

Applying the Law of Sines :

\(\boxed {\frac{sin R}{PS} = \frac{sin P}{RS}}\)

Substitute the values :

Answer:

(3x - 2) (x - 2)

Step-by-step explanation:

im smart

Answer:

A

Step-by-step explanation:

i stands for square root of negative 1. If you multiple i and square root of 23 you will end up with the original

Answer:  .50x+15=25  

I hope this helps.

Answer:

x ≥20

Step-by-step explanation:

Answer:

watxha doing

Step-by-step explanation:

Answer:

C) The additive inverse is positive, while the multiplicative inverse is negative.

Step-by-step explanation:

The answer is C on Edge.   (VERIFIED Edge Answer)

Make sure to click "thanks" and rate this answer to let others know this is the correct answer for this question.  

Edge Tip: Answers do not mix or change.

Answer:

Step-by-step explanation:

V = l * w * h

l = 2

h = 2

w = 7/3

Volume = 2 * 2 * 7/3 = 28/3 = 9 1/3 or 9.3333

Answer:

The height is 5 cm.

Step-by-step explanation:

a = 1/2(\(base_{1}\) + \(base_{2}\))h

27.5 = 1/2(7+ 4) h

27.5 = 1/2 (11) h

27.5 = 5.5 h  Divide both sides by 5.5

5 = h

Answer:

5cm

Step-by-step explanation:

Given ,

To Find : The height of the Trapezoid

In order to find the height of the trapezoid we

\(27.5 = \frac{1}{2} X ( 7+4)h\)

\(27.5X2 = 11 X h\)

\(55 = 11h\\\frac{55}{11}= h\\ 5 = h\)

Hence, the height of the trapezoid is 5cm

Answer:

\( \frac{88}{7} units\)

Step-by-step explanation:

Circumference of semi-circle=

\(\pi \: r\)

\( \frac{22}{7} \times 4 \\ \frac{88}{7} \)

To lock in the highest amount of profit, you would perform the following trades: In Market #1: Buy the 20 Call option for $5.50. Sell the 20 Put option for $3.50.The underlying asset is priced at $22. In Market #2: Sell the 110 Put option for $9.50. Buy the 115 Put option for $2.50. The underlying asset is priced at $110.

By executing these trades, you would create a risk-free strategy known as an arbitrage opportunity. Here's how it works:

In Market #1, you buy the 20 Call option and sell the 20 Put option. Since the options have the same strike price (20), you effectively create a synthetic long position in the underlying asset. This means that if the price of the underlying asset increases, the value of the Call option will also rise, resulting in a profit. Conversely, if the price decreases, the Put option will offset the losses.

In Market #2, you sell the 110 Put option and buy the 115 Put option. This strategy allows you to create a synthetic short position in the underlying asset. If the price of the underlying asset drops, the value of the 110 Put option will increase, generating a profit. On the other hand, if the price increases, the losses will be limited by the 115 Put option.

The profit from this strategy is determined by the difference in premiums paid and received for the options. Since the option prices provided in the question do not specify the unit size, it is not possible to calculate the exact profit without that information.

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Answer:

108 inches cubed

Step-by-step explanation:

d is the same as double the radius so by dividing d in half you get the radius, with the radius you can solve the equation given, 4/3*3*3^3=108inches cubed

Answer:

y = 234t

Step-by-step explanation:

Divide the equation by t on both sides. See image.

The probability of a random sample of 80 households having a sample means a number of at least 2.55 televisions per household is 0.48173.

Given: Mean(μ) = 2.24, Standard deviation(σ) = 1.38, Sample size(n) = 1.28 and a random variable for X = 2.55

The formula to calculate the z-score from the mean. The standard deviation is:

z = (X - μ) / ( σ / √(n)) where,

z = z-score,

X = random variable,

μ = mean,

σ = standard deviation, and

n = sample size

Let's solve the given question:

We have,

Mean(μ) = 2.24,

Standard deviation(σ) = 1.38,

Sample size(n) = 80, and

a random variable for X = 2.55

As it is given sample means at least 2.55 so we have to find the Z(x ≥ 2.55).

Therefore, Z(x ≥ 2.55) = (2.55 - μ) / ( σ / √(n))

Substituting all values we get:

Z(x ≥ 2.55) = (2.55 - 2.24) / ( 1.38 / √(80))

Z(x ≥ 2.55) = 0.01 / 0.21819

Z(x ≥ 2.55) = 0.48173

Hence the probability of a random sample of 40 households having a sample means a number of at least 2.55 televisions per household is 0.48173.

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The purpose of a variable is to hold and represent a value Option C.

The purpose of a variable in programming or mathematics is to hold and represent a value that can be assigned, changed, and used in various operations or calculations. Variables are fundamental components of programming languages and mathematical equations, enabling flexibility and dynamic behavior in computational tasks.

Option (c) "to hold a value" is the most accurate answer, as variables are used to store data or information in memory locations. This value can be of different types, such as integers, floating-point numbers, characters, or even more complex data structures like arrays or objects.

Variables allow programmers to work with and manipulate data efficiently. By assigning values to variables, we can reference and modify them throughout the program, making it easier to manage and organize information.

Variables also play a crucial role in performing calculations, as mentioned in option (b). We can use variables in mathematical expressions and algorithms to perform arithmetic operations, comparisons, and other computations. By storing values in variables, we can reuse them in multiple calculations and update them as needed.

While option (a) "to assign values" is a specific use case of variables, it is not the sole purpose. Variables not only store values but also facilitate data manipulation, control flow, and the implementation of algorithms and logic.

Option (d) "to hold a constant value" is incorrect because variables, by definition, can hold varying values. Constants, on the other hand, are fixed values that do not change during the execution of a program. Option C is correct.

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Answer:   The point in the interior of the circle that is equidistant from every point on the circle is called the center of the circle.

Step-by-step explanation:

Answer:

5. D

6. I think D. Because is y=mx+b and b is y the intercept in y. Y= 0-4 and replacing the -4 would remain at the end

Answer:

183.079764 if you round it would be around 183

Step-by-step explanation:

Have amazing day!

Answer:

Step-by-step explanation:

Answer:

hi

Step-by-step explanation:

Answer:

Part A: two lines, one line will be across $72 all the way. the second line will be a slope.

part b: 5 or less

Step-by-step explanation:

part A: time is x in the equation y=12x and y is the cost. so, plug in the x values and see what y values are. then plot the points and then draw a straight line.

part b: the rider charges $72 dollars for the whole day. if you wanted a bike from bikerama for 6 hours, it would be the same price. so, go for less hours and the price will be better

This is the UML diagram for the development of the program, in which Shapes 2D is the superclass and Circle, Square, Triangle, Rectangle, Rhombus, and Parallelogram are subclasses.

​​​​​​​This is the program in C++ demonstrating the above classes.

#include<iostream>

using namespace std;

class shapes

{

public:

   string name;

   string color;

   virtual void getAttributes()=0;

};

class Circle: public shapes

{

public:

   float radius;

   Circle(string n,string c, float r)

   {

       name=n;

       color=c;

       radius=r;

   }

   float getPerimeter()

   {

       return(2*(3.142)*radius);

   }

   float getArea()

   {

       return((3.142)*(radius*radius));

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Square:public shapes

{

public:

   float side;

   Square(string n,string c, float s)

   {

       name=n;

       color=c;

       side=s;

   }

   float getPerimeter()

   {

       return(4*side);

   }

   float getArea()

   {

       return(side*side);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Triangle:public shapes

{

public:

   float base;

   float height;

   float side1;

   float side2;

   float side3;

   Triangle(string n,string c)

   {

       name=n;

       color=c;

   }

   float getPerimeter()

   {

       cout<<"Enter side1\n";

       cin>>side1;

       cout<<"Enter side2\n";

       cin>>side2;

       cout<<"Enter side3\n";

       cin>>side3;

       return(side1+side2+side3);

   }

   float getArea()

   {

       cout<<"Enter base\n";

       cin>>base;

       cout<<"Enter height\n";

       cin>>height;

       return((0.5)*base*height);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Rectangle:public shapes

{

public:

   float length;

   float breadth;

   Rectangle(string n,string c, float l,float b)

   {

       name=n;

       color=c;

       length=l;

       breadth=b;

   }

   float getPerimeter()

   {

       return(2*(length+breadth));

   }

   float getArea()

   {

       return(length*breadth);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Rhombus:public shapes

{

public:

   float diagonal1;

   float diagonal2;

   float side;

   Rhombus(string n,string c)

   {

       name=n;

       color=c;

   }

   float getPerimeter()

   {

       cout<<"Enter Side\n";

       cin>>side;

       return(4*side);

   }

   float getArea()

   {

       cout<<"Enter diagonal 1\n";

       cin>>diagonal1;

       cout<<"Enter diagonal 2\n";

       cin>>diagonal2;

       return((0.5)*diagonal1*diagonal2);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

class Parallelogram:public shapes

{

public:

   float base;

   float height;

   Parallelogram(string n,string c, float b,float h)

   {

       name=n;

       color=c;

       base=b;

       height=h;

   }

   float getPerimeter()

   {

       return(2*(base+height));

   }

   float getArea()

   {

       return(base*height);

   }

   void getAttributes()

   {

       cout<<"Name  :"<<name<<endl;

       cout<<"Color :"<<color<<endl;

   }

};

int main()

{

   int choice;

   while(1)

   {

       cout<<"\n\nEnter your choice :";

       cout<<"\n1 for Circle\n";

       cout<<"2 for Square\n";

       cout<<"3 for Triangle\n";

       cout<<"4 for Rectangle\n";

       cout<<"5 for Rhombus\n";

       cout<<"6 for Parallelogram\n";

       cin>>choice;

       system("cls");

       switch(choice)

       {

       case 1:

           {

               float r;

               cout<<"Enter radius\n";

               cin>>r;

               Circle c("Circle","Yellow",r);

               c.getAttributes();

               cout<<"Perimeter : "<<c.getPerimeter()<<endl;

               cout<<"Area : "<<c.getArea()<<endl;

           }break;

       case 2:

           {

               float side;

               cout<<"Enter side\n";

               cin>>side;

               Square s("Square","Red",side);

               s.getAttributes();

               cout<<"Perimeter : "<<s.getPerimeter()<<endl;

               cout<<"Area : "<<s.getArea()<<endl;

           }break;

       case 3:

           {

               Triangle t("Triangle","Green");

               t.getAttributes();

               cout<<"Perimeter : "<<t.getPerimeter()<<endl;

               cout<<"Area : "<<t.getArea()<<endl;

           }break;

       case 4:

           {

               float l,b;

               cout<<"Enter Length and breadth\n";

               cin>>l>>b;

               Rectangle r("Rectangle","Blue",l,b);

               r.getAttributes();

               cout<<"Perimeter : "<<r.getPerimeter()<<endl;

               cout<<"Area : "<<r.getArea()<<endl;

           }break;

       case 5:

           {

               Rhombus r("Rhombus","Purple");

               r.getAttributes();

               cout<<"Perimeter : "<<r.getPerimeter()<<endl;

               cout<<"Area : "<<r.getArea()<<endl;

           }break;

       case 6:

           {

               float b,h;

               cout<<"Enter base\n";

               cin>>b;

               cout<<"Enter height\n";

               cin>>h;

               Parallelogram p("Parallelogram","Pink",b,h);

               p.getAttributes();

               cout<<"Perimeter : "<<p.getPerimeter()<<endl;

               cout<<"Area : "<<p.getArea()<<endl;

           }break;

       }

   }

}

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Answer:

51 is 6% of 850

Step-by-step explanation:

0.06*850

(a) The real discrete Fourier transform (DFT) is calculated for the given data set to analyze the helicopter's acoustic signature.

(b) To obtain the ck values and illustrate the relative size of each term in the Fourier series, we calculate the magnitude of each coefficient and plot their amplitudes on a bar graph against the corresponding frequency component, k.

To analyze the helicopter's acoustic signature, the real DFT is computed for the provided data set. The DFT transforms the time-domain measurements of acoustic pressure into the frequency domain, revealing the different frequencies present and their corresponding amplitudes. This analysis helps in understanding the spectral characteristics of the helicopter's acoustic signature and identifying prominent frequency components.

Using the Fourier series representation, the amplitudes (ck's) of the different frequency components in the Fourier series are determined. These amplitudes represent the relative sizes of each term in the series, indicating the contribution of each frequency component to the overall acoustic signature. By plotting the amplitudes on a bar graph, the relative strengths of different frequency components become visually apparent, enabling a clear comparison of their importance in characterizing the helicopter's acoustic signature.

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A sequence of transformation that would move ΔABC onto ΔDEF is: D. a dilation by a scale factor of 1/2, centered at the origin, followed by a 90° clockwise rotation about the origin.

In Geometry, a dilation is a type of transformation which typically changes the size of a geometric object, but not its shape.

In this scenario an exercise, we would dilate the coordinates of the pre-image by applying a scale factor of 1/2 that is centered at the origin as follows:

Ordered pair B (-4, 2) → Ordered pair B' (-4 × 1/2, 2 × 1/2) = Ordered pair B' (-2, 1).

In Mathematics and Geometry, a rotation can be defined as a type of transformation which moves every point of the object through a number of degrees around a given point, which can either be clockwise or counterclockwise (anticlockwise) direction;

(x, y)                               →            (y, -x)

Ordered pair B' (-2, 1)   →          E (1, 2)

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Answer:

C

Step-by-step explanation:

you make my head hurt by reading this

Bueno, el valor de las 8 motos es de 8 x 7.560 = S/ 60.480

Los productos que recibe por pago tienen un valor de:

TV 7 x 620 = S/4.340

PC 8 x 3620 = S/28.960

IM 2 x 13.528 = S/27.056

MINI no sabemos, lo llamamos X

Sumemos los valores de los precios que recibió por la venta y lo igualamos al precio de las 8 motos vendidas:

4.340 + 28.960 + 27.056 + X = 60.480

60.356 + X = 60.480

X = 60.480 - 60-356

X = S/124

El mini cuesta S/124

Espero haberte ayudado!

The population after 11 years will be  56,181.

The rate of increase of the population would be represented with an exponential equation.

An exponential equation can be described as an equation with exponents. The exponent is usually a variable.

The general form of exponential equation is f(x) = \(e^{x}\)

Where:

Population after t years = \(p(1 + r)^{t}\)

Where:

= \(32,000(1 + 0.0525)^{11}\)

= \(32,000(1.0525)^{11}\)

= 56,181

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