Make a graph of the following region 1< x+y in the standard (X

The coordinates of rectangle ABCD are A(1, 2), B(4, 5),C(9, 0) and D(6, -3). What is the area of the rectangle?

puteri [66]

the answer is 30 Square Units.

5 0

3 years ago

If 1 cm represents 30 km on a map and colorado is shown by a 20cm by 15 cm rectangle on the map, what is the approximate actual

zubka84 [21]

Answer:

20*30km = 600km

15*30km = 450km

600km * 450km = 270000 km^2

Step-by-step explanation:

assuming 1cm --- 30km causes that 2cm --- 60km, 3cm --- 90km, 4cm --- 120km etc. Notice that number of km's is 30 times greater than the number of cm's on a map

using this logic, you calculate both real dimensions of Colorado by multiplying 20*30 = 600 km and 15*30 = 450km

Then we use a formula of an area of a rectangle $A=a\cdot b$ assuming $a=600km, \ b=450km$ and get $A=600\cdot 450=270000km^2$ which is an actual area of Colorado

3 0

3 years ago

Identify a pair of supplementary angles in the figure. answers : 2 and 4 1 and 6 1 and 5 5 and 6

Nuetrik [128]

Answer:

2 and 4

Step-by-step explanation:

This is the only one where the two angles add up to 180 degrees, aka, a straight line.

6 0

3 years ago

14.3p – 32.24 = 127.92 14.3p – 32.24 + 32.24 = 127.92 + 32.24 14.3p = 160.16 StartFraction 14.3 p Over 14.3 EndFraction equals S

DiKsa [7]

Answer:

p = 11.2

Step-by-step explanation:

The computation is shown below:

Data provided in the question

2.6(5.5p – 12.4) = 127.92

Now

Distributive Propertyis

14.3p - 32.24 = 127.92

Addition Property is

14.3p = 127.92 + 32.24

Division Property is

14.3p ÷ 14.3 = 160.16 ÷ 14.3

p = 11.2

We simply find the value of p by applying the distributive property, addition property, and the division property and the same is to be considered

8 0

3 years ago

The planets in our solar system do not travel in circular paths. Rather, their orbits are elliptical. The Sun is located at a fo

qwelly [4]

1. The distance between the perihelion and the aphelion is 116 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Step-by-step explanation:

Let us revise the equation of the ellipse is

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , where the major axis is parallel to the x-axis

The length of the major axis is 2a
The coordinates of the vertices are (± a , 0)
The coordinates of the foci are (± c , 0) , where c² = a² - b²

∵ The Sun is located at a focus of the ellipse

∴ The sun located ate c

∵ The perihelion is the point in a planet’s orbit that is closest to the

Sun ( it is the endpoint of the major axis that is closest to the Sun )

∴ The perihelion is located at the vertex (a , 0)

∵ The closest Mercury comes to the Sun is about 46 million miles

∴ The distance between a and c is 46 million miles

∵ The aphelion is the point in the planet’s orbit that is furthest from

the Sun ( it is the endpoint of the major axis that is furthest from

the Sun )

∴ The aphelion is located at the vertex (-a , 0)

∵ The farthest Mercury travels from the Sun is about 70 million miles

∴ The distance from -a to c is 70 million miles

∴ The distance between the perihelion and the aphelion =

70 + 46 = 116 million miles

1. The distance between the perihelion and the aphelion is 116 million miles

∵ The distance between the perihelion and the aphelion is the

length of the major axis of the ellipse

∵ The length of the major axis is 2 a

∴ 2 a = 116

- Divide both sides by 2

∴ a = 58

∴ The distance from the center of Mercury’s elliptical orbit to the

closest end point to the sun is 58 million miles

∵ The distance between the sun and the closest endpoint is

46 million miles

∴ The distance from the center of Mercury’s elliptical orbit and

the Sun = 58 - 46 = 12 million miles

2. The distance from the center of Mercury’s elliptical orbit and the Sun is 12 million miles

∵ The major axis runs horizontally

∴ The equation is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

∵ a = 58

∵ c is the distance from the center to the focus of the ellipse

∴ c = 12

∵ c² = a² - b²

∴ (12)² = (58)² - b²

- Add b² to both sides

∴ (12)² + b² = (58)²

- Subtract (12)² from both sides

∴ b² = (58)² - (12)² = 3220

- Substitute these values in the equation

∴ $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

3. The equation of the elliptical orbit of Mercury is $\frac{x^{2}}{3364}}+\frac{y^{2}}{3220}=1$

The eccentricity (e) of an ellipse is the ratio of the distance from the

center to the foci (c) and the distance from the center to the

vertices (a) ⇒ $e=\frac{c}{a}$

∵ c = 12

∵ a = 58

∴ $e=\frac{12}{58}$ = 0.207

4. The eccentricity of the ellipse is 0.207 to the nearest thousandth

If the eccentricity is zero, it is not squashed at all and so remains a circle.

If it is 1, it is completely squashed and looks like a line

∵ The eccentricity of the ellipse is 0.207

∵ This number is closed to zero than 1

∴ The shape of the ellipse is near to the shape of the circle

5. The value of the eccentricity tell you that the shape of the ellipse is near to the shape of the circle

Learn more:

You can learn more about conics section in brainly.com/question/4054269

#LearnwithBrainly

5 0

3 years ago