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frutty [35]
2 years ago
13

Which arrangement shows275, 5.36, 535, and 336 in order from least to greatest?

Mathematics
1 answer:
Vadim26 [7]2 years ago
7 0

Answer:

5.36,275,336,535

Step-by-step explanation:

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30mL:45mL simplify the ratio
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30ml:45ml\\\\\dfrac{30}{45}=\dfrac{30:5}{45:5}=\dfrac{6}{9}=\dfrac{6:3}{9:3}=\dfrac{2}{3}\\\\\boxed{30ml:45ml=2ml:3ml}
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Find the area of a triangle that has a base of 16 inches and a height of 5 inches.
sertanlavr [38]
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In this excerpt the servant is a comic figure because he unknowingly invites his master
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Yes he would be

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Triangle PQR hAS vertices P (4,-1),Q(-2,7), and R (9,9). find an equation of the median from R
strojnjashka [21]
A median intersects at the midpoint of the opposite length. The midpoint is:

x,m = (4+-2)/2 = 1
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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

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

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

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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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