Answer:
no
Step-by-step explanation:
38 minus 15
38 - 15
15 less than 38
15 < 38
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)
Answer:
222- 50+1=0? 225 - 50 = 175 175+1 = 176 176+0 = 176
Answer: Chuck's travel at a rate of 52mph
Step-by-step explanation:
For Chuck's trip:
D=RT
104= (R+4)T
T= 104 / (R+4)
For Dana's trip:
96 = RT
T= 96/R
Set both equation for Chuck's and Dana together
104/(R+4) =96/R
Then we cross multiply
96(R+4) = 104R
96R + 384 = 104R
104R - 96R = 384
8R = 384
To get R, divide both side by 8
8R/8 = 384/8
R= 48mph
This means Dana's speed is 48mph
Chuck's speed will be: 48mph+4mph = 52mph
Answer:
Slope = 1
Step-by-step explanation:
