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I don't think this teacher has taught junior high school math problems, so would you please help me (1)2a-2b-3a+5b+1 (2)x^2-x+2-x^2+6x-3 (3)3t-2p^2-1/2t+1/3p^2 (4)-5a^2b+6ab^2-4ab-2a^2b+4ab+3 Need process
Junior high school math problem master into ah, to design an isosceles reminder of the flower bed, bottom length of 100m, bottom length of 180m, bottom distance 80m, in the middle of the two waist line there is a horizontal channel, between the bottom there are two longitudinal channels, the width of each channel is equal, the area of the channel is one sixth of the body area How many meters should the width of the channel be (the result retains two decimal places) how to calculate this? We are learning the application problem of one yuan twice. Let's talk about the idea of solving the problem in the third year of junior high school. I can't do this similar problem in this unit
If the width of the channel is x, the sum of the areas of the two vertical channels is 80x + 80x = 160x; the median line of the trapezoid is (100 + 180) / 2 = 140, so the sum of the upper and lower bottoms of the transverse channel is 140x2 = 280. That is to say, the area of the transverse trapezoid is 280x / 2 = 140x
If BC = 4, Sina = 2 / 3, then the length of AC is?
This is simple. If you draw a graph, you can see that BC / AB = Sina and ab = 6. Because in ABC, C is 90, then ac * AC + AB * BC = AB * AB, so AC is 20 square root
Let a = (2x-6,5x), B = (1 / 2x, 3 + X / 5), then the minimum value of F (x) = a times B is?
This problem a = (x, y), B = (P, q), f (x) = a * B, should be x * P + y * q, get: 2x square - 3x + 15, such an equation, and then use the complete square formula to form a complete square + A constant form. This constant is the maximum value. Mobile phone typing is very tired
Calculation formula for radius of inscribed circle and circumscribed circle of right triangle
(it is known that the length of the three sides is a B C, where C is the hypotenuse)
The radius of circumcircle is half of side C
If u = R, a = {x ∧ 2 ＞ 4}, B = {x ∧ 2 X-6 ≤ 0}, then Cu (a ∪ b) =?
∵ B = {x | x2 + X-6 ≤ 0}, then x ≤ 6-x2,
And ∵ a = {x | x2 ＞ 4}, ∵ 6-x2 ＜ 2, B = {x | x ＜ 2}
∴a∪b={x|x＞4 or x＜2}
∴Cu(a∪b)={x|2≤x≤4}
In the known isosceles triangle △ ABC, ∠ a = 90 °, ab = AC, be bisects ∠ ABC, the proof is BC = AB + AE
Glad to answer for you!
Proof: do EF ⊥ BC, cross BC to F
∵ be bisection ∠ ABC
∴AE=EF
∵ RT △ ABC is an isosceles triangle
∴∠C=45°
The ∧ CEF is an isosceles right triangle
∴EF=CF
‖ AE = CF (equivalent substitution)
In △ Abe and △ bef
∠A=∠BEF
AE=EF
BE=BE
The congruence of △ Abe and △ bef
So AB = BF
In conclusion, AB + AE = BF + CF = BC
Let a = {x | X & # 178; - 5x + 6}, B = {x | X & # 178; - (2a + 1) x + A & # 178; + a = 0}, if B &; a, find the value of A
X & # 178; - 5x + 6 = 0 (X-2) (x-3) = 0, x = 2 or x = 3, a = {2,3}
X & # 178; - (2a + 1) x + A & # 178; + a = 0 (x-a) [x - (a + 1)] = 0, X1 = a or x2 = a + 1
∵x1≠x2 ∴a=2,a+1=3
∴a=2
Known: isosceles triangle ABC, angle a = 100 degrees, extend AB to D, make ad = BC, find the angle of BCD
Extend Ba to e, connect CE, make CE = AC ∠ BEC = ∠ CAE = 80
Make ∠ CEF = 60, cross BC to F, connect DF, then ∠ ECF = 60
ECF is an equilateral triangle
AB = AC = CE = EF = CF, BD = BF, BDF is isosceles triangle, ∠ ABC = 40
∠DFB=∠BDF=∠DEF=20
DF = EF = CF, DFC is isosceles triangle
∠BCD=10
There are six days in August that are the birthdays of 12 students. One of these six days is the birthday of many students. How many students may have birthdays at most on this day?
This should belong to the drawer principle. Six days is the birthday of 12 students, and one day is the birthday of many students. Because there must be at least one student's birthday every day, so the maximum number of students' birthdays may be 12-5 = 7. In addition to this day, there is only one student's birthday every other five days
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