The distance from kim's house to nyc is 100 miles more than the distance from her house to la. if the sum of the distances is 3,000 miles, how far is it from kim's house to nyc?

Distance from kims house to NYC is 1550 miles

Solution:

Let "a" be the distance from kims house to NYC

Let "b" be the distance from kims house to LA

Given that distance from Kim's house to NYC is 100 miles more than the distance from her house to LA

Therefore,

distance from Kim's house to NYC = 100 + distance from her house to LA

a = 100 + b eqn 1

The sum of the distances is 3,000 miles

distance from kims house to NYC + distance from kims house to LA = 3000

a + b = 3000 eqn 2

Let us solve eqn 1 and eqn 2

Substitute eqn 1 in eqn 2

100 + b + b = 3000

2b = 3000 - 100

2b = 2900

From eqn 1,

a = 100 + 1450

Thus distance from kims house to NYC is 1550 miles

you need to show the graph

step-by-step explanation:

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

so the change in y-value for function f is (6 )) = 9.

the table tells us

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).