Let x be her initial distance from the building, then tan 37 = 130/x
x = 130/tan 37 = 130/0.7536 = 172.5 feet
Let y be her distance from the building after moving forward, then
tan 40 = 130/y
y = 130/tan 40 = 130/0.8391 = 154.9
After moving forward, she is 172.5 - 154.9 = 17.6 ft closer.
Answer:
120(0.96)^n
Step-by-step explanation:
6% = 0.06
1 - 0.06 = 0.94
p = 120(0.96)^n
Answer:
The sweat chloride reference value is less than 30 mmol/L. A value of more than 60 mmol/L of chloride in the sweat is consistent with a diagnosis of cystic fibrosis. The values of 30-60 mEq/L may represent heterozygous carriers, these carriers cannot be accurately identified with a sweat chloride test.
14 divided by 10.5 is 1.3333 that goes on and on never stops.
