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2 years ago

Which is likely to have a mass close to 2 kilograms? (1 point)

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

7 0

A pineapple they way about 2 killograms

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Graph the system of inequalities y> 2x-3 and 3y-x≤6

Veronika [31]

Y > 2x - 3
3y - x ≤ 6
 ↓
y = 2x - 3
3y - x = 6

 3(2x - 3) - x = 6
3(2x) - 3(3) - x = 6
 6x - 9 - x = 6
 6x - x - 9 = 6
 5x - 9 = 6
 + 9 + 9
 5x = 15
 5 5
 x = 3
y = 2x - 3
y = 2(3) - 3
y = 6 - 3
y = 3
(x, y) = (3, 3)

4 0

3 years ago

out of a senior class of 100 students the.class offers (president vice president treasurer needs to be arranged.com how many dif

kondaur [170]

Answer: 33

explanation: 100/3 = 33.333, and since it isn't possible for 0.333 to be three different students being picked, there are only 33 different ways.

8 0

2 years ago

Evaluate each expression for the given value(s)

Butoxors [25]

Answer:

-16

Step-by-step explanation:

Substitute the values into the equation

$= \frac{ ({ - 12 + 4}^{2}) }{ - 4} \\ = \frac{ {( - 8)}^{2} }{4} \\ = \frac{64}{4} \\ = - 16$

<h2>OladipoSeun♡˖꒰ᵕ༚ᵕ⑅꒱</h2>

4 0

3 years ago

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

L = luminosity of the star (related to the Sun)

d = distance in ly (light-years)

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

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