Let a = 3, B = 4, C = 5. (1) a + b > C & & B = = C (2) a | B + C & & B-C (3)! (a > b) & amp; (4)! (x = a) & amp; (Y = b) Write the values of the following logical expressions. Let a = 3, B = 4, C = 5 (1)a+b>c&&b==c (2)a||b+c&&b-c (3)!(a>b)&& ( 4)!(x=a)&&(y=b)

Let a = 3, B = 4, C = 5. (1) a + b > C & & B = = C (2) a | B + C & & B-C (3)! (a > b) & amp; (4)! (x = a) & amp; (Y = b) Write the values of the following logical expressions. Let a = 3, B = 4, C = 5 (1)a+b>c&&b==c (2)a||b+c&&b-c (3)!(a>b)&& ( 4)!(x=a)&&(y=b)

I learned this before, but I forgot it

226. If there is a statement: int a = 5; a + +; here the value of expression a + + is a) 7 b) 6 C) 5 d) 4

Choose C 5
In the current sentence a has not added 1, you need to complete this sentence to add 1

Given that the function y = f (x) is a periodic function, its minimum positive period is 2, and f (x) = x square when x ∈ [- 1,1], then the value of F (- 3) is

Because f (x) is a periodic function and the minimum positive period is 2,
So f (x + 4) = f (x + 2) = f (x),
So f (- 3) = f (- 3 + 4) = f (1) = 1

ABCD multiplied by 9 equals DCBA, and ABVD equals how much?

1089
If B = 1, then C = 0 (because the four digits are less than 1112), then ABCD = 1109, which is not suitable for the topic. If B = 0, then c * 9 =? 2, then C = 8, which is suitable for the topic
So ABCD = 1089

Find the area of the triangle formed by the image and two coordinate axes of the first-order function y = 0.5x + 1

The graph of this linear function is a straight line, intersecting X-axis at (- 2,0) and y-axis at (1,0)
It is easy to know that the area of the triangle surrounded by the two coordinate axes is:
s=0.5*2*1=1

One is 4.5dm high and its volume is 81 cubic decimeters. The other is 3 decimeters high. What is its volume?

The bottom area of the first cylinder = volume △ height = 81 △ 4.5 = 18 decimeters & sup2;
=The bottom area of the second cylinder
The volume of the cylinder = bottom area × height
=18×3
=54 cubic decimeter

Call for help, count 24 - 4, 6, 7, 9

1:6 ÷ 4 ×(7 + 9)2:(6 ÷ 4) × (7 + 9)3:6 ÷ (4 ÷ (7 + 9))4:6 ÷ 4 ×(9 + 7)5:(6 ÷ 4) × (9 + 7)6:6 ÷ (4 ÷ (9 + 7))7:(6 × (7 + 9)) ÷ 48:6 × (7 + 9) ÷ 49:6 × ((7 + 9) ÷ 4)10:(6 × (9 + 7)) ...

How does ρ = root 2 * cos (Θ + π / 4) become a rectangular coordinate equation,

ρ = radical 2 * cos (Θ + π / 4) ρ ^ 2 = 2cos (θ + π / 4) = 2 (√ 2 / 2cos θ - √ 2 / 2Sin θ) = √ 2 (COS θ - sin θ) i.e. ρ ^ 2 = √ 2 (COS θ - sin θ) i.e. ρ ^ 3 = √ 2 (ρ cos θ - ρ sin θ) x = ρ cos θ, y = ρ sin θ, ρ ^ 2 = x ^ 2 + y ^ 2, then (x ^ 2 + y ^ 2) √ 2 (X-Y)

One pound =? Kgft =? Cm and other conversion formulas

0.4536 kg = 1 lb
0.3048 M = 1 ft

If the perimeter of a diamond is 20cm and the ratio of two diagonals is 4:3, the area of the diamond is ()
A. 12cm2B. 24cm2C. 48cm2D. 96cm2

Let the diagonals of the diamond be respectively 8x and 6x. It is known that the perimeter of the diamond is 20cm, so the side length of the diamond is 5cm. According to the properties of the diamond, the diagonals of the diamond are divided vertically, that is, (4x) 2 + (3x) 2 = 25. The solution is x = 1, so the diagonals of the diamond are respectively 8cm and 6cm, so the area of the diamond is 12 × 8 × 6 = 24cm2, so B