to tell which is base is yeah
((B1+B2)/2)*h=A B1 and B2 are "Base 1" and "Base 2" (the parallel sides) h is the height A is, of course, the area
area= 1/2 h (b1+b2) h=heigth b1 and b2=base 1 and base 2
A = 1/2 (b1 + b2) h b1 = base 1 (usually the bottom) b2 = base 2 (usually the top) h = height
if the bases are b1 and b2, then: area = ½ × (b1 + b2) × h → 2 × area ÷ h = b1 + b2 → b2 = 2 × area ÷ h - b1 = 2 × 156.6 m² ÷ 18 m - 6 m = 11.4 m
What is the valid class declaration header for the derived class d with base classes b1 and b2?A. class d : public b1, public b2 {/*...*/};B. class d : class b1, class b2 {/*...*/};C. class d : public b1, b2 {/*...*/};D. class d : b1, b2 {/*...*/};The answer is A, C and D.B is not valid because "class" is not a valid access specifier.All the others are valid because private access is the default when the access specifier is omitted. Note that if class D were declared using the struct prefix, inheritance would default to public access rather than private.
Answer: No.Explanation: Area of trapezoid = 1/2(b1 + b2) * h where b1 is length of base one, b2 is length of base 2 and h is height.this equation = 1/2*b1*h + 1/2*b2*hdouble one base:1/2(2*b1+b2) *h = b1*h+1/2*b2*h = (b1+1/2*b2)*hin order for the area to double, both bases would have to double which would cancel out both 1/2's. Only one is cancelled out so the area would increase but not double
The formula for the area of a trapezoid is A = 1/2 (b1 + b2)/h where b1 is the upper base of the trapezoid , b2 is the lower base of the trapezoid and h is the height of the trapezoid. Since a triangle has only one base , replace either b1 or b2 with zero. Thus the area of a triangle is A = 1/2bh
No, you cannot work on a B1/B2 visa.
Yes, you can work on a B1/B2 visa.
A B1/B2 visa allows for both business (B1) and tourism (B2) purposes, while a B2 visa is specifically for tourism and visiting friends or family in the United States.
In a trapezium (trapezoid), the upper base (b1) and lower base (b2) can be found using the formula for the area (A) if the height (h) is known: [ A = \frac{1}{2} \times (b1 + b2) \times h ] Rearranging this formula allows you to solve for one base if you have the area, the other base, and the height. For example, if you know the area and one base, you can isolate the other base: [ b1 + b2 = \frac{2A}{h} ] Thus, you can find either base by substituting the known values.
To show all possible interleavings of two processes, let’s assume we have two atomic statements from Process A (A1, A2) and Process B (B1, B2). The possible interleavings could be: A1, A2, B1, B2 A1, B1, A2, B2 B1, A1, A2, B2 A1, B2, A2, B1 B1, B2, A1, A2 B2, A1, A2, B1 A2, A1, B1, B2 B2, B1, A1, A2 These interleavings illustrate the various ways the two processes can be executed in a concurrent setting.