It is known that: as shown in Figure 1, M is a certain point on the fixed length line AB, C and D respectively start from m and B and move to the left along the straight line BA at the speed of 1cm / s and 3cm / s, and the direction of motion is shown by the arrow (C is on the line am, D is on the line BM) (1) If AB = 10cm, when points c and d move for 2S, find the value of AC + MD. (2) if points c and d move, there is always MD = 3aC, fill in the blank directly: am=______ Ab. (3) under the condition of (2), where n is a point on the line AB and an-bn = Mn, the value of mnab is obtained

It is known that: as shown in Figure 1, M is a certain point on the fixed length line AB, C and D respectively start from m and B and move to the left along the straight line BA at the speed of 1cm / s and 3cm / s, and the direction of motion is shown by the arrow (C is on the line am, D is on the line BM) (1) If AB = 10cm, when points c and d move for 2S, find the value of AC + MD. (2) if points c and d move, there is always MD = 3aC, fill in the blank directly: am=______ Ab. (3) under the condition of (2), where n is a point on the line AB and an-bn = Mn, the value of mnab is obtained

(1) When points c and d move for 2S, CM = 2cm, BD = 6cm ∵ AB = 10cm, CM = 2cm, BD = 6cm ∵ AC + MD = ab-cm-bd = 10-2-6 = 2cm (2) 14 (3) when point n is on line AB, as shown in the figure ∵ an-bn = Mn ∵ BN = am = 14ab, ∵ Mn = 12ab, that is, mnab = 12

It is known that: as shown in Figure 1, M is a certain point on the fixed length line AB, C and D respectively start from m and B and move to the left along the straight line BA at the speed of 1cm / s and 3cm / s, and the direction of motion is shown by the arrow (C is on the line am, D is on the line BM)
(1) If AB = 10cm, when points c and d move for 2S, find the value of AC + MD. (2) if points c and d move, there is always MD = 3aC, fill in the blank directly: am=______ Ab. (3) under the condition of (2), where n is a point on the line AB and an-bn = Mn, the value of mnab is obtained

(1) When points c and d move for 2S, CM = 2cm, BD = 6cm ∵ AB = 10cm, CM = 2cm, BD = 6cm ∵ AC + MD = ab-cm-bd = 10-2-6 = 2cm (2) 14 (3) when point n is on line AB, as shown in the figure ∵ an-bn = Mn ∵ BN = am = 14ab, ∵ Mn = 12ab, that is, mnab = 12

As shown in the figure, points a and B are on the straight line Mn, ab = 11cm, circle a and circle B
As shown in the figure, point a and B are on the straight line Mn, ab = 11cm, and the radius of circle a and circle B is 1cm. Circle O moves from left to right at the speed of 2cm per second. At the same time, the radius of circle B is also increasing. The relationship between radius R (CM) and time t (seconds) is r = 1 + T (T > = 0). Question: (1) try to write the functional expression between distance D (CM) and time t (seconds) between point ab
(2) How many seconds after point a starts, the two circles are tangent?

1.d=11-2t
2.d'=11-1-2t-(1+t)
When d '= 0, two circles are tangent
t1=3s
d''=11+1-2t+(1+t)=0
t2=13s

As shown in the figure, C is a point on the line AB, AC; BC = 2; 3, e and F are the midpoint of AB and BC respectively, and EF = 3cm, so the length of AB can be obtained

AB:BC=5:3
(1/2AB):(1/2BC)=5:3
EB: FB = 5 / 3
EB=EF+FB=3+FB
(3+FB)/FB=5/3
FB=9/2
EB=3+9/2=15/2
AB=2EB=15

As shown in the figure, points E and F are the midpoint of line segments AC and BC respectively. If EF = 2.5cm, calculate the length of line segment ab

∵ E and F are the midpoint of AC and BC, respectively, ∵ EC = 12ac, FC = 12bc, ∵ EF = ec-fc = 12ac-12bc = 12 (ac-bc) = 12ab = 2.5cm, ∵ AB = 5cm

Given that the common Parts BD = 1 / 3, ab = 1 / 7, CD, e and F are the midpoint of AB and CD respectively, and the distance between E and F is 8cm, the length of AB and CD can be obtained
The graph points are a, e, D, B, F, C in turn!

A——E——D——B——F——C
∵BD＝1/3AB
∴AB＝3BD
∵ e is the midpoint of ab
∴BE＝AB/2＝3/2BD
∵BD＝1/7CD
∴CD＝7BD
∵ f is the midpoint of CD
∴DF＝7/2BD
∴BF＝DF-BD＝7/2BD-BD＝5/2BD
∴EF＝BE+BF＝3/2BD+5/2BD＝4BD
∵EF＝8
∴4BD＝8
∴BD＝2
∴AB＝3BD＝6（cm）
∴CD＝7BD＝14（cm）

As shown in the figure, C and D are two points on the line AB, e is the midpoint of AC, and F is the midpoint of BD. if EF = 20 and CD = 8, then the length of AB is（

A——E——C————D—F—B
∵ e is the midpoint of AC
∴CE＝AC/2
∵ f is the midpoint of BD
∴DF＝BD/2
∴EF＝CE+DF+CD＝（AC+BD）/2+8
∴（AC+BD）/2+8＝20
∴AC+BD＝24
∴AB＝AC+BD+CD＝24+8＝32
The math group answered your question,

As shown in the figure, it is known that the line segment ad = 6cm, the line segment AC = BD = 4cm, e and F are the midpoint of the line segments AB and CD respectively, and EF is calculated

∵AD=6cm，AC=BD=4cm，∴BC=AC+BD-AD=2cm；∴EF=BC+12（AB+CD）=2+12×4=4cm．

As shown in the figure, the common parts of AB and CD are BD = 1 / 4AB = 1 / 5CD, e, f are the midpoint of AB and CD respectively, EF = 20cm, and the length of BD is calculated
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A E D B F C

Let BD = a, then AB = 4A, CD = 5a, be = 2A, DF = 2.5A
And EF = be + df-bd = 3.5A to get BD

As shown in the figure, given that ab: BC: CD = 2:3:4, e and F are the midpoint of AB and CD respectively, and ad = 45cm, find the length of line EF
——————————————
A E B C F D

Let AB = 2A, then BC = 3A, CD = 4A, AB + BC + CD = 45, that is, 2A + 3A + 4A = 45, the solution is a = 5, and EF = 30cm