A math problem that asks for how and why

Set the cut part to 1 part. Then the rest is 6 servings.

So the full length of the rope can be divided into 6+1=7 parts.

So the cut part is 1 7

When cutting an extra 10 cm, let the rope cut part at this time is 1 part, then the remaining part is 5 parts.

So at this time, the full length of the rope can be divided into 5+1=6 parts.

So the cut part at this time is 1 6

Then the ratio of 10 cm to the rope is: 1 6 -1 7 = 1 42, so the total length of the rope is 10 (1 42) = 10*42 = 420 cm.

Solution: Think of the rest as 1, then.

In the first case, the subtraction is 1 6 (1+1 6) = 1 7 of the total length of the rope, and in the second case, the subtraction is 1 5 (1+1 5) = 1 6 of the total length of the rope, so the length of the rope is 10 (1 6-1 7) = 10 1 42 = 420 (cm).

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Welcome to the "Mathematics and Physics Infinity" team.

Original length = 10 (1 6-1 7).

If you cut off the remaining one-sixth, you will subtract the original length of 1 7

If you cut off the remaining fifth, you will subtract the original length of 1 6

The central angle of the circle o=arc length radius = l1 ao=l2 co, that is: l1 co=l2 ao=l2 (d+co)co=l2 d (l1-l2);

Then the area of the flower bed is:

s=½l1×oa-½l2×co=½[l1×(d+co)-l2×co]=½[l1×d+(l1-l2)×co]=½[l1×d+l2×d]=½d(l1+l2)

Sector area formula s=1 2 arc length radius shaded part = sector OAB - sector OCD

See if you can do it, I don't understand and then give you the detailed process.

Let oa=r1, oc=r2, and the central angle of the circle is x, then l1=r1*x, l2=r2*x, because r1-r2=d, so x=(l1-l2) d,, so as to obtain r1,r2, and then according to the area of the sector is equal to 1 2*x*r 2, the area of the two sectors can be obtained respectively, and finally subtracted.

Solution: Let oc=od=r, then oa=ob=r+d, and the cod angle is , then there is an ab arc length.

l1=2π（r+d）α/360

There is a CD arc length.

l2=2πrα/360

Combining the above 2 equations, the solution is r=l2d(l1-l2), = 180(l1-l2) d

Available flower bed area = OAB area - OCD area = R+D) 2 360 - R2 360= (2rd+d 2) 360=(2r+d)(l1-l2) 2=(l1+l2)d 2

Let A to C ship X units, and the total shipping cost is S

s=400x+200(20-x)+600(5-x)+300[10-(5-x)]

400x+4000-200x+3000-600x+3000-1500+300x

8500-100x

When x is larger, the value of s is smaller.

Because the maximum value of x is x=5

Therefore, the minimum value of S = 8500-100 * 5 = 8500-500 = 8000, that is, 5 units are transported from A to C, 15 units are transported from B to C, and 10 units are transported from B to D.

s=400x+600(5-x)+200(20-x)+300(10-5+x)=400x+3000-600x+4000-200x+1500+300x=8500-100x

More equations, we can see that the larger the x, the smaller the s, and because the maximum of x is 5, s=8500-500=8000

Replace the x in f(x) with x+1 and the answer is obtained. i.e. f(x+1) = (x+1) +2(x+1).

(1) Proof:

Even OC, OD

oc=od=r

OCD is an isosceles triangle.

E is the CD midpoint.

ob⊥cdbf//cd

ob bfb on round o.

BF is the tangent of the circle O.

(2) BCD=38 degrees.

BAF = 38 degrees.

r=5ab=2=10

abf = 90 degrees.

bf=tan∠baf*ab=

Even the AC ob on the AB.

ab cd with dot e

e is the midpoint of cd.

ac=adcab= baf=38 degrees.

ab is the diameter, acb = 90 degrees.

bc=sin∠cab*ab=

This should be a physics problem.

According to the formula: f=ma, f=-12000n (because f is the resistance, then add a negative sign), m=2x1000kg

So a=-6m s negative sign represents the gradual decrease in acceleration, and the initial speed of the car on a horizontal highway is 54km h, i.e., 54 according to the mechanical formula v (t)-v (0)=2as brackets represent the subscript.

s=[0-15²]/(2x(-6))=

This is a very simple topic for the application of the law of conservation of energy.

The force received by the car has the braking force, the ability gravity, and the support force of the ground to face it. On the horizontal highway, only the braking force does the work, and it does the negative work.

According to the energy conservation formula, there is: -fs = 0-2 mv squared, i.e., -12000*s= (54km h into SI units 15m s) solution s=

You need to glide for meters to stop.

I think we should use the law of conservation of energy, not the velocity formula, because the title does not say that it is a uniform deceleration of brakes, so a can not be determined, that is, the result is -fs=0-1 2mv squared, s=