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In-s [12.5K]
3 years ago
7

Graph the line with the equation y= 2/5x+3 ? ​

Mathematics
1 answer:
saw5 [17]3 years ago
5 0
Below is the graph :)

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Someone took the elevator down 2 floors to the basement. would it be positive/negative?
kirill [66]

Answer:

Negative

Step-by-step explanation:

8 0
2 years ago
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Day 4: As a landscaper, one of your clients wants a rectangular garden in which the length is five feet
liberstina [14]

Answer:

The Area of Rectangular Garden is 1044  feet²    

Step-by-step explanation:

According to question

The perimeter of the garden = 82 ft

Let the length be L ft

The width be W ft

Now as per question

L = 5 + ( 2× W )

∵ Perimeter of Rectangle = 2 × ( Length + Width )

Or , Perimeter of Rectangle = 2 × ( L+ W )

Or, 82 = 2 × ( L+ W )

Or,  82 = 2 × [ 5 + ( 2 ×W ) + W ) ]

Or,  82 = 2 × ( 5 +3W )

Or,  41 = 5 + 3W

Or, 41 - 5 = 3W

So, 3W= 36

∴    W = \frac{36}{3} = 12 feet

I.e Width = 12 feet

And L = 5 + ( 2× W )

Or, Length = 5 + 24 = 29 feet

Now The Area of Rectangle = Length × width

So, The Area of Rectangle = 29 ft × 36 ft

The Area of Rectangle is 1044 feet²

Hence The Area of Rectangular Garden is 1044  feet²        Answer

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3 years ago
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Answer:

Hola no se chino.......

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What are the solutions of the quadratic equation? 2 x 2 – 16 x + 32 = 0
Evgen [1.6K]

Answer:

4 - 16 x + 32 = 0

- 16 x = - 4 - 32

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X = 2.25

3 0
3 years ago
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Suppose you have a light bulb that emits 50 watts of visible light. (Note: This is not the case for a standard 50-watt light bul
natali 33 [55]

Answer:

0.049168726 light-years

Step-by-step explanation:

The apparent brightness of a star is

\bf B=\displaystyle\frac{L}{4\pi d^2}

where

<em>L = luminosity of the star (related to the Sun) </em>

<em>d = distance in ly (light-years) </em>

The luminosity of Alpha Centauri A is 1.519 and its distance is 4.37 ly.

Hence the apparent brightness of  Alpha Centauri A is

\bf B=\displaystyle\frac{1.519}{4\pi (4.37)^2}=0.006329728

According to the inverse square law for light intensity

\bf \displaystyle\frac{I_1}{I_2}=\displaystyle\frac{d_2^2}{d_1^2}

where

\bf I_1= light intensity at distance \bf d_1  

\bf I_2= light intensity at distance \bf d_2  

Let \bf d_2  be the distance we would have to place the 50-watt bulb, then replacing in the formula

\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.

7 0
3 years ago
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