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How to solve 8x-5x = 0.39
3x=0.39
x=0.39/3
x=0.13
8x-5x = 18 (solving equation)
8x-5x=18
(8-5)x=18
3x=18
x=18÷3
x=6
Fill in the appropriate units in the brackets to make the equation hold. 1 [] - 9 [] = 11 [] - 99 [] = 11 [] plus 999 [] = 1 []
1 (decimeter) - 9 (CM) = 1 cm 1 (meter) - 99 (CM) = 1 cm 1 (g) + 999 (g) = 1 kg
It is known that the left vertex of the ellipse x ^ 2 / 4 + y ^ 2 = 1 is a. two mutually perpendicular chords am and an intersect the ellipse at M and N. do Mn always pass the fixed point on the X axis?
Elliptic equation: X & sup2; / 4 + Y & sup2; = 1, that is, X & sup2; + 4Y & sup2; = 4A & sup2; = 4, a = 2, point a (- 2,0) when the slope of AM changes, let the slope of am be K, then the slope of an is - 1 / K. linear am equation: y = K (x + 2) linear an equation: y = - 1 / K (x + 2)
The perimeter of a rectangle is 20 decimeters, and the length is three times the width. How many square decimeters is the area of a rectangle?
If the width is x decimeters, the length is 3x decimeters
So (3x + x) * 2 = 20
8x=.20
x=20÷8=2.5
3x=7.5
So the area is 7.5 * 2.5 = 18.75 square decimeters
Solving the quadratic equation 2x + 1 = 3Y ① 4x-9y = 8 ②
By substituting 4x = 6y-2 into ②, 3Y = - 10 is obtained
The solution is y = - 10 / 3
Substituting y into 1 gives x = - 11 / 2
So the solutions of the equations are as follows
x=-11/2,y=-10/3
If a focus of the ellipse x ^ 2 / A ^ 2 + y ^ 2 = 1 is f, the point P is on the ellipse, and | vector op | = | vector of |, then the area s of △ OPF
1/2
Easy to get root of abscissa of focus (A2-1)
Then there is a joint solution of circular equation x2 + y2 = A2-1 and ellipse x ^ 2 / A ^ 2 + y ^ 2 = 1
I y i = 1 / radical (A2-1)
So s = 1 / 2 * x * y = 1 / 2
It is proved that if P is prime, then there are at least p prime numbers between P and P square
P is not equal to 2
P prime numbers (including prime number P)
This is just a program to find out whether a number is prime or not
CLS
INPUT N
F=1
FOR I=2 TO SQR(N)
IF N MOD I=0 THEN F=0
NEXT I
IF F=1 THEN PRINT"YES" ELSE PRINT"NO"
END
It is known that the function f (x) = x ^ 3 + BX ^ 2 + CX is an odd function, and the function g (x) = x ^ 2 + CX + 3 is a decreasing function in the interval (- ∞, 3) and an increasing function in the interval (3, + ∞)
Finding the value of a and B
I calculate, B = 0, C = - 6?
F (x) = x ^ 3 + BX ^ 2 + CX is an odd function
So f (x) + F (- x) = 0
x^3+bx^2+cx+(-x)^3+b(-x)^2+c(-x)=0
x^3+bx^2+cx-x^3+bx^2-cx=0
2bx^2=0
b=0
g(x)=x^2+cx+3
=[x-(-c/2)]^2-c^2/4+3
Axis of symmetry x = - C / 2
Decrease on the interval (- ∞, 3), increase on (3, + ∞), and find the value of B and C
So x = 3 is the axis of symmetry
So - C / 2 = 3, C = - 6
So B = 0, C = - 6
If the function f (x) = LG ^ (AX + √ x ^ 2 + 1) is an odd function on R, find the value of A
If the function f (x) = LG ^ (x ^ 2 + 1 under the root of AX + is an odd function on R, find the value of A
^ ^
f(-x)=lg(-ax+√(x^2+1) )=-f(x)=-.=lg(1/ax+.)
therefore
-ax+√(x2+1)=1/(ax+√(x2+1))
Both sides multiply the same denominator
(x² +1)-a² x² =1
That is X & sup2; - A & sup2; X & sup2; = 0
x² =a² x²
A = 1 or a = - 1
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