Can anyone help me find the problem with the following functions: 10

dim a as integer, b as integer, c as integer, t as integer'Define 4 integer variables.

a = val('a is the numeric value in text box 1.

b = val('b is the numerical value in text box 2.

c = val('c is the numerical value of text box 3.

if a > b then'If a is larger than b, t = a: a = b: b = c, :

The colon is the separator of the statement written on the same line Note that b=c here should be b=t, compare the size of a and b, assign the large value to b, and the small value to a

t = a: a = b: b = c

end if

if a > c then'Ditto.

t = a: a = c: c = t

end if

if b > c then'Ditto.

t = b: b = c: c = t

end if complete a e.g. a=23, b=12, c=34, then = a &" " & b & " " & c'Let label label4 show the value of abc, & is the connector, and 12 23 34 appears

end sub

I used you to find a function that didn't go wrong and ran the result correctly.

In general, static variables should not be assigned to subfunctions.

Let the equation for the straight line after translation be y=2x+b

After the dot (2, -1), then b = -5

The equation is y=2x-5

That is, the original image of the rift brother is shifted 4 units to the right, or 8 units upwards.

In layman's terms, because y is inversely proportional to -3x, the larger y is, the smaller -3x is, that is, the larger y is, the larger x is also (because the coefficient is negative); Yes Because x is proportional to 4 z, the larger the x, the larger the 4 z, that is, the smaller z z: the larger y the combination, the smaller z is, so b! is chosen

Because y is inversely proportional to -3x, y=k1 (-3x) because x is proportional to 4 z, so x=k2 (4 z) so y=k1 (-3 (4k2 z)=k1z (-12k2) so y is proportional to z.

What you're trying to say is:

y = 1 - 1 2 * cos( x 3) x r right? When x 3 is an odd multiple of , y obtains a maximum value of 3 2, and when an even multiple of y, a minimum value of 1 2

That is, when x is 6k+3 (k is an integer), y obtains the maximum value of 3 2, and when x is 6k, y obtains the minimum value of 1 2

I don't think it's very difficult, but your formula is not very clear, and it is easy to cause ambiguity, so it is recommended to use the formula tool with word or use"Parentheses"Clearly express the rules of operation of the formula

cos x itself is a period of 2 periodic function, so, y should also be a period of 2 periodic function, just need to find the maximum and minimum value in a period, find the zero point of cos x, the minimum point, the maximum point, one by one substitution y, I think you should be able to get the result!

f(x)=lnx/x

f ′(x) = [1/x*x-lnx]/x² = (1-lnx)/x²

At x [1 2,1], f (x) = (1-lnx) x 0 is constant and f(x) is continuously and monotonically increasing.

f(1/2)=ln(1/2)/(1/2)=2ln(1/2)f(1)=0

f(x) is bounded at [1 2,1], and 2ln(1 2) f(x) 0

option a is correct.

First of all, we should pay attention to the condition in the question, we know that f(x) is an odd function, and we know that the image of the odd function is symmetrical about the origin, so f(-2)=-f(2); And because f(2)=0, so f(-2)=f(2)=0; The main thing here is the definition of odd functions.

The first gets a = -2 or 3, and the second gets a = -1 or 3

Proportional: y=k*x k is not equal to 0 (* is a multiplier sign) The image is a straight line through the origin (I won't draw it), property 1with respect to origin symmetry; 2.When k>0, the function is incrementing at (-oo, +oo), and vice versa when k<0;

Inverse proportionality: y=k x The image is hyperbolic (the left is the case where k is positive, and within each quadrant, the single subtraction is single, and the right image is opposite in nature) The image is symmetrical with respect to the origin.

Primary function: y=k*x+b k is not equal to 0 The image is a straight line (the proportional function is a special case when the primary function crosses the origin, i.e., b=0) When k>0, the function is an increasing function at (-oo, +oo), and vice versa when k<0; and with respect to point (0,b), point symmetry (i.e., the intersection of the image with the x-axis and y-axis).