Answer:
$5820.07
Step-by-step explanation:
Substitute 243 into the equation for n:
![c=117\sqrt[3]{243^{2} }+1264\\c=117\sqrt[3]{59049}+1264\\c=117*38.9407384+1264\\c=4556.06639+1264\\c=5820.06639](https://tex.z-dn.net/?f=c%3D117%5Csqrt%5B3%5D%7B243%5E%7B2%7D%20%7D%2B1264%5C%5Cc%3D117%5Csqrt%5B3%5D%7B59049%7D%2B1264%5C%5Cc%3D117%2A38.9407384%2B1264%5C%5Cc%3D4556.06639%2B1264%5C%5Cc%3D5820.06639)
Round 5820.06639 to the nearest hundredth:
5820.07
$5820.07
<span>More please help me solve the problem
The stars have different brightness. The brightest stars are Grade 1 and the least degree luimnoase stars 6. The brightness of the stars shrinks 2.5 times with the passing from one grade to another. Whenever the stars are brighter Grade 1, Grade 6 than the stars?</span>
x = 1
5(1)= -4y + 4
5= -4y + 4
-4. -4 1 = -4y
y = -¼
x = 2...
10 = -4y + 4 6 = -4y y = -3/2
x = 3..
15 = -4y + 4 11 = -4y y = -11/4
