The answer is: No
Becuase the equation itself is not correct. The equation is c = 2p.
Good Luck!
You know where the glacier is now, and how far it moves in
one year. The question is asking how close to the sea it will be
after many years.
Step-1 ... you have to find out how many years
Step-2 ... you have to figure out how far it moves in that many years
Step-3 ... you have to figure out where it is after it moves that far
The first time I worked this problem, I left out the most important
step ... READ the problem carefully and make SURE you know
the real question. The first time I worked the problem, I thought
I was done after Step-2.
============================
Step-1: How many years is it from 2010 to 2030 ?
(2030 - 2010) = 20 years .
Step-2: How far will the glacier move in 20 years ?
It moves 0.004 mile in 1 year.
In 20 years, it moves 0.004 mile 20 times
0.004 x 20 = 0.08 mile
Step-3: How far will it be from the sea after all those years ?
In 2010, when we started watching it, it was 6.9 miles
from the sea.
The glacier moves toward the sea.
In 20 years, it will be 0.08 mile closer to the sea.
How close will it be ?
6.9 miles - 0.08 mile = 6.82 miles (if it doesn't melt)
The company picnic cost $ 1300 for 80 employees
<em><u>Solution:</u></em>
Given that cost of a company picnic varies directly as the number of employees attending the picnic
Let "c" be the company picnic cost
Let "n" be the number of employees attending the picnic
Therefore,
Where "k" is the constant of proportionality
c = kn ---------- eqn 1
<em><u>Given that company picnic costs $487.50 for 30 employees</u></em>
Therefore substitute c = 487.50 and n = 30
<em><u>How much does a company picnic cost for 80 employees?</u></em>
Substitute n = 80 and k = 16.25 in eqn 1
Thus $ 1300 is the cost for 80 employees
The answer is 25%.
60/240=x/100
60*100=6000
6000/240=25
60/240=25/100
