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How to calculate the slope k in the primary function? How about constructing a right triangle and then two right sides?
I don't know what math stage you are? Now the answer is as follows: if you are at the junior middle school level: make a vertical line along a point on the primary function (straight line) and intersect with the x-axis to form a right triangle (maybe this is the solution of the right angle side). Use the opposite side next to the adjacent side, that is, the Tg of the included angle α The value of is the slope k
Is the slope of the primary function negative? For example, the book says that in the line y = KX + B, K is called the slope of the line. If K is negative, is the slope also negative?
Of course, the slope can be negative. When the inclination angle of the straight line (the angle between the part above the X axis and the positive direction of the X axis) is greater than 90 degrees, the tangent of the inclination angle, that is, the slope is negative
A function of degree, with respect to the slope It's a higher-order function. I forgot a knowledge point about the slope of the opening. The two straight lines K1 and K2 on the image K1 is greater than 0, K2 is greater than 0, what do you mean? how? For what? I remember this on the knowledge point, which reminds me of it
K > 0 inclination angle, i.e. the included angle with the right end of X axis is an acute angle
k = y2- y1/ x2 -x1
Solve the equations ay + BX = C, CX + AZ = B, BZ + CY = A. It's best to take photos and have a detailed process. Thank you Solve the equations ay + BX = C, CX + AZ = B, BZ + CY = a It's best to take photos. There's a detailed process. Thank you
ay+bx=c------acy+bcx=cc(1)
cx+az=b------bcx+abz=bb(2)
bz+cy=a------abz+acy=aa(3)
(1) + (2) + (3) gets ABZ + ACE + BCX = 1 / 2 (AA + BB + CC) (4)
therefore
(4) - (1)
ABZ = 1 / 2 (AA + BB CC), then Z can be found
(4) - (2)
Acy = 1 / 2 (AA BB + CC), then y can be found
(4) - (3)
BCX = 1 / 2 (- AA + BB CC), then x can be found
a^3(bz-cy)^3+b^3(cx-az)^3+c^3(ay-bx)^3 Factorization
Is it a factorization? The result is
-3abc(ay-bx)(az-cx)(bz-cy) .
The partial derivative of binary function is known. The equation f (Y / x, Z / x) = 0 determines the function z = Z (Z, y), and its f (U, V) is differentiable. Find AZ / ax, AZ / ay az/ax=[(y/x)f'1+(z/x)f'2]/f'2 az/ay=-f'1/f'2
[Jun Lang lieying] the team will answer the question for you. It should be that z = Z (x, y) is actually a simple answer. Just find the partial derivative on both sides of F (Y / x, Z / x) = 0. In fact, it is the transformation of implicit function derivation. First find the partial derivative of X to get f'1 * (- Y / x ^ 2) + F'2 * (AZ / xax-z / x ^ 2) = 0, and get AZ / AX = [(Y / x) f'1 + (Z / x)
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