It is known that the image of the first-order function y = (M-3) x + 2m + 4 passes through the intersection m of the straight line y = - 13X + 4 and the Y axis, and the analytic expression of the first-order function is obtained

It is known that the image of the first-order function y = (M-3) x + 2m + 4 passes through the intersection m of the straight line y = - 13X + 4 and the Y axis, and the analytic expression of the first-order function is obtained

If the intersection of the straight line y = - 13X + 4 and the Y axis is m, then the coordinates of M are (0,4). Substituting it into y = (M-3) x + 2m + 4, we can get: 4 = (M-3) × 0 + 2m + 4, and the solution can get: M = 0, so the analytic formula of the first-order function is: y = - 3x + 4
Let a = {x | x = a + B radical 2, a, B ∈ Q}, if x1 ∈ a, X2 ∈ a
(1) Do x1x2 and X1 / x2 belong to a
(2) If B = = {x | x = a + B radical 2, a, B ∈ Z}, do x1x2, X1 / x2 belong to B and why?
The sum of the analytic functions (y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y = - 2, y =
By the way, teach me how to do this kind of topic,
Two points decide a straight line
The image and the line y = - 3x + 2 intersect at the same point on the Y axis
First, find out the intersection of this line and Y axis, that is (0,2)
So the straight line passing through points (0,2) and (2, - 3)
And then we use the two-point formula, OK?
Given the set M = {X - 1 ≤ x ≤ 2} n = {x y = root x}, then Mun=
A、R B、[-1,+∞） C、（-∞,2） D、（-1,0）∪（0,+∞）
X > = 0 is required in n set
So {- MUX}
Select b
B,{-1,+∞)
N is x ≥ 0, so Mun = [- 1, + ∞)
Given that the image of the first-order function passes through the point (2, - 3) and intersects with the line y = - 3x + 2 at the same point on the Y axis, the analytic expression of the first-order function is obtained
The line y = - 3x + 2 has only one intersection point with the Y axis, that is, (0,2)
Once again, the image of the function passes through points (2, - 3) and intersects with the line y = - 3x + 2 at the same point on the Y axis
The first-order function passes through (2, - 3) and (0,2)
Let the analytic formula be y = KX + B
∴2k+b=-3
0×k+b=2
The solution is: B = 2, k = - 5 / 2
∴y=(-5/2)x+2
In the right triangle ABC, angle c is a right angle, AC is equal to BC, another point P is in the triangle ABC, and satisfies PA = 3, Pb = 1, PC = 2, so the degree of angle BPC can be obtained
To solve; connect
∵△ ABC and △ PCE are isosceles right triangles
∴CE=CP,AC=BC,∠ACB=PCE=90°.
∴∠ACP=∠ECB
∴△ACP≌△ECB
∴EB=AP=6
∵PE2=2PC2=32
∴PE2+PBE=BE2
∴∠BPE=90°,
Therefore, BPC = BPE + CPE = 90 ° + 45 ° = 135 °
If we know that the image of a linear function and the line y = - 3x + 2 intersect at the same point on the Y axis and pass through the point (2, - 3), then the linear function is
y=（-5/2x）+2
In the cone p-abc, PA = Pb = PC = bc. and the angle BAC = 90, what is the angle between PC and ABC on the bottom
Speed. Be more specific
Just saw ~ hehe, because PA = Pb = PC, and ∠ BAC is a right angle, because the projection of the vertex of the triangular pyramid on the bottom is the intersection of the three sides of the median line of the triangle, that is, the point with equal distance to the three vertices, and because △ ABC is a right triangle, so the projection of the vertex P of the triangular pyramid on the bottom is the midpoint of the line BC, and △ PBC is
Let P1, P2 ···, PN be 1,2, ··, the proof of arbitrary arrangement of N: 1 / (P1 + P2) + 1 / (P2 + P3) + ··· + 1 / (PN-1 + PN) > (n-1) / (n + 2)
Big hand to solve. Process to understand ah
Use Cauchy inequality
((P1+P2)+(P2+P3)+...+(P(n-1)+Pn))(1/(P1+P2)+1/(P2+P3)+...+1/(P(n-1)+Pn))
≥ (1+1+...+1)² = (n-1)².
And (P1 + P2) + (P2 + P3) +... + (P (n-1) + PN) = 2 (P1 + P2 +... + PN) - P1 PN
P1, P2,..., PN is an arrangement of 1,2,..., N, so P1 + P2 +... + PN = 1 + 2 +... + n = n (n + 1) / 2
And P1 + PN ≥ 1 + 2 > 2, so (P1 + P2) + (P2 + P3) +... + (P (n-1) + PN) < 2n (n + 1) / 2-2 = n & # 178; + n-2 = (n-1) (n + 2)
So 1 / (P1 + P2) + 1 / (P2 + P3) +... + 1 / (P (n-1) + PN)
≥ (n-1)²/((P1+P2)+(P2+P3)+...+(P(n-1)+Pn)) > (n-1)/(n+2).
In the triangular pyramid p-abc, the angle BAC = 90 degrees, PA = Pb = PC = BC = 2Ab = 2, the verification plane PBC, the vertical plane ABC 2, and the cosine value of the two faces b-ap-c
The key is the second question
(1) Taking the midpoint D of BC, connecting PD and ad, it is easy to get ad = BD = CD = 1, then there is PD vertical BC, Pythagorean theorem obtains PD = radical 3, PD ^ 2 + ad ^ 2 = PA ^ 2, Pythagorean theorem inverse theorem, PD vertical ad, then PD vertical plane ABC, plane PBC, vertical plane ABC (2) plane PAB make be vertical PA to e, plane PAC make ef vertical PA intersection AC