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2 years ago

Solve this inequality and show the result on a number line.

Mathematics

2 answers:

MariettaO [177]2 years ago

8 0

$\\ \bull\tt\longmapsto 5(x-2)\geqslant 4(x-2)$

$\\ \bull\tt\longmapsto 5x-10\geqslant 4x-8$

$\\ \bull\tt\longmapsto 5x-4x\geqslant -8+10$

$\\ \bull\tt\longmapsto x\geqslant 2$

x is either 2 or greater than 2.

andrew11 [14]2 years ago

6 0

∙⟼5(x−2)⩾4(x−2)

∙⟼5x−10⩾4x−8

∙⟼5x−4x⩾−8+10

∙⟼x⩾2

x is either 2 or greater than 2.

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A biologist has been observing a tree's height. This type of tree typically grows by 0.22 feet each month. Fifteen months into t

Lyrx [107]

Data:

x: number of months

y: tree's height

Tipical grow: 0.22

Fifteen months into the observation, the tree was 20.5 feet tall: x=15 y=20.5ft (15,20.5)

In this case the slope (m) or rate of change is the tipical grow.

m=0.22

To find the line's slope-intercep equation you use the slope (m) and the given values of x and y (15 , 20.5) in the next formula to find the y-intercept (b):

$\begin{gathered} y=mx+b \\ 20.5=0.22(15)+b \\ 20.5=3.3+b \\ 20.5-3.3=b \\ 17.2=b \end{gathered}$

Use the slope(m) and y-intercept (b) to write the equation:

$\begin{gathered} y=mx+b \\ \\ y=0.22x+17.2 \end{gathered}$ A) This line's slope-intercept equation is: y=0.22x+17.2

B) To find the height of the tree after 29 months you substitute in the equation the x for 29 and evaluate to find the y:

$\begin{gathered} y=0.22(29)+17.2 \\ y=6.38+17.2 \\ \\ y=23.58 \end{gathered}$ Then, after 29 months the tree would be 23.58 feet in height

C) In this case as you have the height and need to find the number of moths you substitute the y for 29.96feet and solve the equation for x, as follow:

$\begin{gathered} 29.96=0.22x+17.2 \\ 29.96-17.2=0.22x \\ 12.76=0.22x \\ \frac{12.76}{0.22}=x \\ \\ 58=x \end{gathered}$ Then, after 58 months the tree would be 29.96feet tall

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1 year ago

How do you find percent decrease?

snow_lady [41]

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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

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3 years ago

What is the exact circumference of a circle with a radius of 45 cm?<br><br> HELP ASAP!!!

Sergeeva-Olga [200]

The answer is 282.7433

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3 years ago

Read 2 more answers

Toby buys packs of 12 racing cars for R49,99 and sells each racing car for R10 at school

Ratling [72]

Answer:

1.1 R5.83

1.2 R70.01

1.3 58.34%

Step-by-step explanation:

Given that he buys a pack of 12 racing cars for R49,99 then the unit cost for a car which is the total cost divided by the number of cars in a pack

= R49,99/12

= R4.17

If each car is sold for R10 then the profit made on each car which is the difference between the cost and the selling price

= R10 - R4.17

= R5.83

The profit margin on one pack of 12 racing cars

Profit margin is the ratio of the profit to the total selling price

The total profit (profit on one pack of 12 racing cars)

= R5.83 * 12

=R70.01

Total selling price = R10 * 12

=R120

Margin = R70.01 / R120

= 58.34%

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3 years ago