Let’s say it takes time t hours for the interception.
In that time the carrier travels 28t nm west and the helicopter 130t nm.
Now we use the sine rule to find the angle x which lies between the distance 175 and 130t.
sin(x)/28t=sin35/130t. So sin(x)=28sin35/130=14sin35/65=0.1235 and x=7.0964 degrees.
Therefore the helicopter should use a bearing of 35+7.0964=42.0964 degrees north of east or 47.9 degrees east of north approx.
340+.06(sales)
340+.06(660)
$379.60
Answer:
4) -4
5) -6x-20
Step-by-step explanation:
Add x with x's and numbers with numbers
4) -6x + 9 + 6x -13
-6x+6x+9-13
x's cancel
9-13 = -4
5) 8x - 9 -11 -14x
8x-14x-9-11
-6x-20
hope this helps
B I have done this before
Mid term :
Q1 = (88 + 85)/2 = 86.5
Q2 = (92 + 95)/2 = 93.5
Q3 = 100
IQR = Q3 - Q1 = 100 - 86.5 = 13.5
final exams :
Q1 = (65 + 78)/2 = 71.5
Q2 = (88 + 82)/2 = 85
Q3 = (95 + 93)/2 = 94
IQR = Q3 - Q1 = 94 - 71.5 = 22.5
so the final exams has the largest IQR
