The answer is A
Hopes this helps
Answer:
im lost good luck
Step-by-step explanation:
Answer:
9900
Step-by-step explanation:
Round 900 down because 47 is closer to 900 than 1000.
Answer:
28 gallons
Step-by-step explanation:
We are given that a sports car gets 14mpg in city driving and 19mpg for highway.
The model G=![\frac{1}{14} c+\frac{1}{19 }h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B14%7D%20c%2B%5Cfrac%7B1%7D%7B19%20%7Dh)
Where G=Amount of gasoline used (in gal) for c miles driven in the city and h miles driven on the high way.
Amount of gas used in 14 miles car in the city driving=1 gal
Amount of gas used in 1 mile car in the city driving=![\frac{1}{14} gal](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B14%7D%20gal)
Amount of gas used in c miles car in the city driving =
gal
Similarly, for car driving on highway
Amount of gas used in h miles for car driving on highway=![\frac{1}{19}h](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B19%7Dh)
c=98 miles in the city
h=399 miles on the highway
We have to find the amount of gas used required to derive 98 miles in the city and 399 miles on the highway.
Substitute the value of c and h in the given expression
Then, G=![\frac{1}{14}\times 98+\frac{1}{19}\times 399=7+21=28 gal](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B14%7D%5Ctimes%2098%2B%5Cfrac%7B1%7D%7B19%7D%5Ctimes%20399%3D7%2B21%3D28%20gal)
Hence, the amount of gas required to derive 98 miles in the city and 399 miles on the highway=28 gallons
Answer:
(2, 1 )
Step-by-step explanation:
Assuming the centre of dilatation is the origin.
Then multiply each of the coordinates of B by 0.5 , that is
B(4, 2 ) → (4(0.5), 2(0.5)) → (2, 1 )