The answer is 157 degrees because the line segment is 180 and if <AFB is 23, you take 23 away from 180 and you should get 157. To check, add 157 to 23 or take away 157 from 180.
Answer:
852
6×10(6)+7×10(5)+3×10(2)+8×10+2= 852
Step-by-step explanation:
ORDER OF OPERATIONS Will really help you solve this long problem.
Order of operations is PEMDAS.
P-parenthesis
E-exponents
M-multiplication
D-division
A-addition
S-subtraction
6)(10)(6)+(7)(10)(5)+(3)(10)(2)+(8)(10)+2
=(60)(6)+(7)(10)(5)+(3)(10)(2)+(8)(10)+2
=360+(7)(10)(5)+(3)(10)(2)+(8)(10)+2
=360+(70)(5)+(3)(10)(2)+(8)(10)+2
=360+350+(3)(10)(2)+(8)(10)+2
=710+(3)(10)(2)+(8)(10)+2
=710+(30)(2)+(8)(10)+2
=710+60+(8)(10)+2
=770+(8)(10)+2
=770+80+2
=850+2
=852
From the research I just did, it seems that 6 lines will be drawn when constructing an inscribed square.
Hope this helped :)
<h2>
Answer:</h2>
B. |y| = 1.6
<h2>
Step-by-step explanation:</h2>
A sketch of the vector is attached to this response.
As shown, the vector lies on the second quadrant and its horizontal and vertical components are given by x and y respectively.
Where;
|x| = v cos θ -----------------(i)
|y| = v sin θ -----------------(ii)
and
v = magnitude of the vector = 2.5km/hr
θ = angle that the vector makes with the +x axis = 90° + 50° = 140°
<em>To get the magnitude of the vector's vertical component</em>
<em>Substitute these values into equation(ii) as follows;</em>
|y| = 2.5 sin 140°
|y| = 2.5 x 0.6428
|y| = 1.607
|y| ≅ 1.6 km/hr
Therefore, the vertical component of the vector has a magnitude of 1.6 km/hr.