C expression types (assignment expression, arithmetic expression, relational expression, logical expression, conditional expression, comma expression) and evaluation rules

C expression types (assignment expression, arithmetic expression, relational expression, logical expression, conditional expression, comma expression) and evaluation rules


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In C + +, the value of () is not a logical value in the following expressions: a arithmetic expression; B relational expression; c logical expression; D comma expression
Why do I think a and D can be chosen


A comma expression has a value, which cannot be replaced by a statement. The value of a comma expression is the value of the nth subexpression, that is, the value of the expression n
int a,b,c,d;
d=(a=1,b=a+2,c=b+3);
cout



The base of a triangle is twelve centimeters long. If the height remains the same, the area will be increased by sixteen square centimeters after the base is extended by four centimeters


Solution: the area ratio of triangles with the same height is equal to the ratio of the bottom
That is, the ratio of the original triangle area to the newly added triangle area is 12:4 = 3:1
Therefore, the original triangle area is: 16 * 3 = 48 (square centimeter)
(Note: you can also use the area of the new triangle to calculate the height of 8, and then calculate the area of the original triangle to be 48 square centimeters.)



Solve equation 4.2-0.2x = 0.5 (24-x)


4.2-0.2x=0.5(24-x)
4.2-0.2x=12-0.5x
0.5x-0.2x=12-4.2
0.3x=7.8
x=26



The circumference of a semicircle is 20.56 cm. What is the area of the semicircle


20.56÷(3.14÷2+1)
=20.56÷(1.57+1)
=20.56÷2.57
=8(cm)
8÷2=4(cm)
3.14×4²÷2
=3.14×16÷2
=50.24÷2
=25.12(cm²)
A: the area of this semicircle is 25.12 square centimeters
I'm glad to answer for you



25×3.2×1.25.


25×3.2×1.25,=25×4×(0.8×1.25),=100×1,=100.



As shown in the figure, △ ABC, the angular bisectors ad, be and CF intersect at point h, Hg ⊥ AC is made through point h, and the perpendicular foot is g, then ∠ ahe = ∠ CHG? Why?


Reasons: ∵ ad, be and CF are bisectors of △ ABC. Suppose ∵ bad = ∵ CAD = x, ∵ Abe = ∵ CBE = y, ∵ BCF = ∵ ACF = Z, then 2x + 2Y + 2Z = 180 ° i.e. x + y + Z = 90 ° in △ AHB, ∵ ∵ ahe is the outer angle of △ AHB, ∵ ahe = ∵ bad + ∵ Abe = x + y = 90 ° - Z in △ AHB



Let A1, A2 ,an,b1,b2,… , BN are real numbers and B1 ^ 2-b2 ^ 2 - -bn^2>0
Verification (A1 ^ 2-A2 ^ 2 -...) -an^2)(b1^2-b2^2-… -bn^2)
I know I can't - that's why I asked. I can't prove this.


A1 ^ 2-A2 ^ 2... - an ^ 20 let y = (a1x + B1) ^ 2 - (a2x + B2) ^ 2... - (anx + BN) ^ 2 = (A1 ^ 2-A2 ^ 2... - an ^ 2) x ^ 2 + 2 (a1b1-a2b2... - anbn) x + (B1 ^ 2-b2 ^ 2... - BN ^ 2) Cauchy inequality is proved by the discriminant less than or equal to zero because it is always greater than or equal to zero



If the similar area ratio of the two triangles is 1:4, then the ratio of the corresponding median line is


1:2



It is proved that the function f (x) = x + lgx has zeros on (0,1)
To complete the process


Certification:
Firstly, a derivation formula is proved
y = lgx = [lnx]/ln10
∴ dy/dx = 1/xln10
Back to the theme:
f(x) = x + lgx
df/dx = 1 + 1/xln10
When x ∈ (0,1), DF / DX > 0
F (x) is an increasing function
∵ when X -- > 0, lgx -- > - ∞
When X -- > 0, f (x) -- ∞
∵ when X -- > 1, f (x) -- > 1 + 0 = 1 > 0
On (0,1), the continuous smooth differentiable function f (x) has one end below the x-axis and one end above the x-axis
F (x) has zero on (0,1)