NO. They are not.
We can prove this by turning those numbers into fractions:
16 / 14 = 1.1428 ; 64 / 60 = 1.0667
Or simply:
64 / 16 = 4
60 / 14 = 4.29
To get the equivalent of 16 to 14; we must multiply both numbers by 4.
16 * 4 = 64
14 * 4 = 56
The equivalent of 16 to 14 is 64 to 56.
Answer:
C≈18.85
Step-by-step explanation:
Answer:
-66
Step-by-step explanation:
-(-2)^2=-4
-3(-2)(-3)+2(-3)^3= -62
-4-62= -66
Answer:
0.049168726 light-years
Step-by-step explanation:
The apparent brightness of a star is
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
According to the inverse square law for light intensity
where
light intensity at distance
light intensity at distance
Let
be the distance we would have to place the 50-watt bulb, then replacing in the formula
Remark: It is worth noticing that Alpha Centauri A, though is the nearest star to the Sun, is not visible to the naked eye.
So 75 points is your 100%. You can do 60 over 75 is equal to X over 100. You divide your numerator(60) by your denominator(75) to get 85.7142857%, or to round to your nearest percent, your answer would be roughly 86%.