<h2>Answer:</h2>
A. Lowering the discount rate to provide more loans to banks
<em>P</em><em>l</em><em>s</em><em> </em><em>c</em><em>o</em><em>r</em><em>r</em><em>e</em><em>c</em><em>t</em><em> </em><em>m</em><em>y</em><em> </em><em>a</em><em>n</em><em>s</em><em>w</em><em>e</em><em>r</em><em> </em><em>i</em><em>f</em><em> </em><em>i</em><em>t</em><em>s</em><em> </em><em>w</em><em>r</em><em>o</em><em>n</em><em>g</em><em> </em>^_^
N = x-7
2x = 5n-7
2x = 5(x-7)-7
2x=5x-35-7
-3x=-42
x=14
n=7
7 is the answer
So first we want to find how far away they were after one hour.
Assuming they were going at a steady rate:
Since:
320 miles = distance traveled in 2 hours
We can divide both sides by 2 to get:
160 miles = distance traveled in 1 hour.
So we now know that the trains were 160 miles apart after 1 hour.
So we can assign the eastbound train a variable for its speed right now, or just x. And then the westbound train would be x + 18.
So, since the rate is in miles per hour, and they traveled 160 miles in one hour, we can say:
x + x + 18 = 160 miles
Then just solve.
2x + 18 = 160
-18 -18
2x = 142
/2 /2
x = 71
Since x equals the rate of the eastbound train, and we're looking for the westbound train, just add 18 to 71 because the westbound train was going 18 mph faster than the eastbound train.
18 + 71 = 89
So the rate of the westbound train is 89 mph.
Answer:
The commutative, associative, and distributive properties help you rewrite a complicated algebraic expression into one that is easier to deal with. When you rewrite an expression by a commutative property, you change the order of the numbers being added or multiplied.
Step-by-step explanation:
