Cp=7512.75×[100÷(100+77÷4)]
=7512.75×(100÷119.25)
=6300
c.p=6300
The answer would be 3. {0,3}.
Remember: {x,y}
2(0)+3=3
0+3=3
0+3=3
Answer:
7 square units
Step-by-step explanation:
As with many geometry problems, there are several ways you can work this.
Label the lower left and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.
The applicable formulas are ...
area of a trapezoid: A = (1/2)(b1 +b2)h
area of a triangle: A = (1/2)bh
So, our areas are ...
AQRS = AWSQE - AWSR - AEQR
= (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)
Factoring out 1/2, we have ...
= (1/2)((2+5)·4 -2·2 -5·2)
= (1/2)(28 -4 -10) = 7 . . . . square units
A ) The angle formed by 2 consecutive radii:
Ф= 360° : 8 = 45°
b ) The angle formed by a radius and a side of a polygon:
α = (180° - 45°): 2 = 135° : 2 = 67.5°
Answer:
B ) 45°, 67.5°
Answer:
So in two weighings, we can find a single light coin from a set of 3 × 3 = 9. By extension, it would take only three weighings to find the odd light coin among 27 coins, and four weighings to find it from 81 coins.
Step-by-step explanation:
