Find the Greatest Common Factor (GCF)
<u>GCF = 6y^6</u>
Factor out the GCF. (Write the GCF first. Then, in parenthesis divide each term by the GCF.)
6y^6(24y^8/6y^6 + 6y^6/6y^6)
Simplify each term in parenthesis
<u>6y^6(4y^2 + 1)</u>
Answer:
d
Step-by-step explanation:
hchfjtjjgjhkgiyggb bbhhu
Answer:
A
Step-by-step explanation:
44 - 15 =29
Answer:
the no of liters required is 50.55 liters
Step-by-step explanation:
Given that
The car gas efficiency is 23.0 mi/gal
distance is 495 km
1 gal = 3.78 L
1 km = 0.6214 mi
We need to find out the no of liters of gas that would needed to complete a trip
So,
= (495 × 3.78 × 0.6214) ÷ (23)
= 50.55 Liters
hence, the no of liters required is 50.55 liters