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It is known that Y1 is positively proportional to X (scale coefficient is K1) and Y2 is inversely proportional to X (scale coefficient is K2). If the image of function y = Y1 + Y2 passes through points (1,2), (2,12), then the value of 8k1 + 5k2 is______ ．
Let Y1 = K1X, y2 = k2x, then y = Y1 + y2 = K1X + k2x, substituting (1, 2), (2, 12) into: K1 + K2 = 22k1 + K22 = 12, the solution is: K1 = − 13k2 = 73 ｛ 8k1 + 5k2 = − 83 + 353 = 9
Synthesis of inverse proportion function and Pythagorean theorem
Place the square piece of paper ABCD with area of 1 in the plane rectangular coordinate system, point B.C is on the X axis, A.D and B.C are symmetrical about the Y axis, fold point C to C 'on the Y axis, the crease is BP, and the image of the existing inverse scale function passes through point P, then the analytical expression of the inverse scale function is?
The straight line y = - x + 2, root sign 3 + 2 and x.y axis intersect at two points A.B and P point AB respectively, ∠ POA = 30 ° rotate OP 90 ° counterclockwise around o point to make point P rotate to point P1. If the hyperbola y = K / X passes through point P1, then k =?
1. We know that △ c'bc is an isosceles triangle, BC '= BC = 1
The ordinate of C 'is [1 ^ 2 - (1 / 2) ^ 2] = root 3 / 2
Δ c'bc is an equilateral triangle, so ∠ PBC = 30 degree
∴PC＝BCtan30°＝√3/3
∴P（1/2,√3/3）
Let the analytic expression of the inverse proportional function be y = K / x, then
√3/3＝k/(1/2),k＝√3/6
∴y＝(√3/6)x
2. The equation of OP is y = √ 3x, and the simultaneous solution of y = - x + 2 √ 3 + 2 gives x = 2, y = 2 √ 3
∴P（2,2√3）
After rotation, the coordinate of P1 is P1 (- 2 √ 3,2)
Substituting the hyperbola y = K / x, we get 2 = K / (- 2 √ 3)
The solution is k = - 4 √ 3
An inverse proportion function in junior high school
The annual electricity price of a place was 0.8 yuan and the annual electricity consumption was 100 million kwh. This year, it is planned to adjust the electricity price to 0.55-0.75. According to calculation, if the electricity price is adjusted to X Yuan, the new electricity consumption of this year is in a positive proportion of Y (100 million kwh) and (x-0.4) yuan, and when x = 0.65, y = 0.8
(1) Find the function expression between Y and X;
(2) If the cost price per kilowatt hour is 0.3 yuan, when the price is adjusted to what amount of yuan, the revenue of the power sector in this year will increase by 20% compared with the previous year?
result.
When y = 0, x = 0.8; when y = 0.8, x = 0.65
Substituting function y = K (x-0.4) + B
Solution
As shown in the figure, ad bisects ∠ BAC, de ⊥ AB in E, DF ⊥ AC in F, and DB = DC, proving: EB = FC
It is proved that: ∵ ad bisects ∠ BAC, de ⊥ AB in E, DF ⊥ AC in F, ∵ de = DF; ∵ de ⊥ AB in E, DF ⊥ AC in F. ∵ in RT △ DBE and RT △ DCF, de = dfdb = DC ≌ RT △ DBE ≌ RT △ DCF (HL); ≌ EB = FC
As shown in the figure, ab = AC, BD = CD
It is proved that in △ ADC and △ ADB, CD = DBAC = abad = ad ≌ ADC ≌ ADB (SSS) and ≌ B = C
As shown in the figure, BD = CD, angle abd = angle ACD, de and DF are perpendicular to AB and AC respectively, and the intersection line is at e and F
Verification: de = DF
Brother, there is no picture~~~
As shown in the figure, line segments BD and AC intersect at point O, bisectors PC and PD connecting AB, CD, ∠ ACD and ∠ abd intersect at point P. try to explain the reason why ∠ P = 1 / 2 (∠ a + ∠ d)
Correct your question first, the bisector of ACD and abd should be PC and Pb respectively
I believe you have drawn the picture, so I will not draw it, just say the steps
First of all, you connect B and C, with ∠ AOB as the outer angle of the triangle OBC, ∠ AOB = ∠ OBC + ∠ OCB
So ∠ AOB + PBD + PCA + P = 180 ° Formula 1
Because PC and Pb are bisectors of ∠ ACD and ∠ abd respectively, ∠ PBD = 1 / 2 ∠ abd ∠ PCA = 1 / 2 ∠ ACD
So equation 1 can be changed to ∠ AOB + 1 / 2 ∠ abd + 1 / 2 ∠ ACD + ∠ P = 180 degree
The two sides are multiplied by 2
2 ∠ AOB + abd + ACD + 2 ∠ P = 360 ° type 2
In triangle AOB and triangle cod,
The formula of ∠ AOB + ∠ abd + ∠ a = 180 ° is 3
The formula of ∠ AOB + ∠ ACD + ∠ d = 180 ° is 4
Formula 3 and formula 4 are as follows
2 ∠ AOB + ∠ abd + ∠ ACD + ∠ a + ∠ d = 360 ° type 5
From equation 2 to equation 5, it can be concluded that:
2∠P=∠A+∠D
Therefore:
∠P=1/2（∠A+∠D）
As shown in the figure, it is known that BD = CD, ∠ 1 = ∠ 2. State the reason for △ abd ≌ △ ACD
It is proved that in △ abd and △ ACD, ∵ BD = CD, ∠ 1 = ∠ 2 (known), ad = ad (known), and ≌ abd ≌ ACD (SAS)
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- 1. Junior high school function formula and Pythagorean theorem, the best picture, I am doing mold, thank you for your help I hope I can give you the picture above
- 2. On the Pythagorean theorem in junior high school It is known that the two right angle sides of the RT triangle ABC are a and B, the hypotenuse is C, and the height of the hypotenuse is h. try to judge the shape of the triangle with H, C + H, and a + B as sides Please explain why,
- 3. In the triangle B2 + C2 BC = A2, find a (ABC followed by square)
- 4. As shown in the figure, it is known that △ ABC and △ DEB are isosceles right triangles, ∠ ACB = ∠ EDB = 90 °, point E is on side AC, CB and ED intersect at point F
- 5. As shown in the picture, OD.OE They are the bisectors of ∠ AOC and ∠ BOC, ∠ AOD = 40 ° and ∠ BOE = 25 ° respectively. To find the degree of ∠ AOB is not only the answer, but also the process plus the answer
- 6. As shown in the figure, OE bisects ∠ BOC, OD bisects ∠ AOC, ∠ BOE = 20 ° and ∠ AOD = 40 ° to calculate the degree of ∠ doe
- 7. As shown in the figure, the bisectors of ∠ CBD and ∠ BCE at the two outer angles of △ ABC intersect at point O, ∠ a = 40 ° to calculate the degree of ∠ BOC
- 8. In the triangle ABC, a = 3, B = root 3, a = 60 degree solution triangle
- 9. (1) In triangle ABC, a = 2, B = 2 √ 3, B = 60 ° are known to solve triangle ABC (2) In triangle ABC, we know a = √ 3, B = √ 2, B = 45 ° to solve triangle ABC
- 10. In the triangle ABC, a = 18, B = 16, a = 150, solve the triangle
- 11. Help out on the inverse proportion function and Pythagorean theorem of the problem it! A little basic on the line, but not too simple!
- 12. As shown in the figure, in △ ABC, points D and E are on AB and AC respectively, de ‖ BC, ∠ CED = ∠ BDC. (1) prove: △ DCE ∽ CBD; (2) if BC = 2CD, s △ ade = 1, find the value of s △ ABC
- 13. In the geometry shown in the figure, EA vertical plane ABC, DB vertical plane ABC, AC vertical plane BC and AC = BC = BD = 2ae. M is the midpoint of ab Finding the tangent of the angle between de and plane EMC
- 14. In the triangle ABC, ab = BC, the point D E F is the midpoint on the side of BC AC AB respectively
- 15. As shown in Figure 1, in △ ABC, points e.d.f ab.bc.ca If ad bisects ∠ BAC, then the quadrilateral AEDF is a diamond?
- 16. As shown in the figure, in the quadrilateral ABCD, ∠ bad = ∠ ABC, ad = BC, AC intersects with point o at BD. let's say that the triangle OAB is an isosceles triangle
- 17. As shown in the figure, it is known that ad, be and CF are respectively the heights of the three sides of △ ABC, h is the perpendicular, and the extension line of ad intersects the circumscribed circle of △ ABC at point G
- 18. In △ ABC, ab = AC, be = CF, EF intersect BC with D
- 19. As shown in the figure, the triangle ABC is an equilateral triangle, and D is the midpoint on the side ab. it is known that the area of the triangle BDE is 5 square centimeters. Find the area of the equilateral triangle ABC
- 20. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons