Answer:
Step-by-step explanation:
v = {[(20sin36°)i + (20cos36°)j] + 10i} mi/h
vE = 20sin36º + 10 = 21.76 mi/h
vN = 20cos36° = 16.18 mi/h
v = √(vE2 + vN2) = √(21.762 + 16.182) mi/h = 27.12 mi/h
θ = tan-1(vN/vE) = tan-1(16.18/21.76) = 36.6º north of east
<u>Q</u><u>uest</u><u>ion</u><u>:</u>
To Simplify:
118 {121÷(11×11)-(-4)-(3-7)}
<u>Solu</u><u>tion</u>:
↠118 {121÷121+4-(-4)}
↠118 {1+4+4}
↠118 {5+4}
↠118{9}
↠118×9
↠1062
▬▬▬▬▬▬▬▬▬▬▬▬
The upper Solution is done by applying BODMAS
<u>Abou</u><u>t</u><u> </u><u>BODMAS</u><u>:</u>
B→ Bracket
O→ Of
D→ Division
M→ Multiplication
A→ Addition
S→ Subtraction
A and D i think.(sry this needed to be 20 characters long)
Answer:
idk if it's right and i'm really sorry if it isn't
∴ x=
Step-by-step explanation:
f(x)=3
fx=3
÷f on both sides
x=
