Solve a math problem in the first year of junior high school winter vacation life p23. 1

Solution: Set x two-point balls.

Column equation: 3*3+2x+(14-x-3)=289+2x+14-x-3=28

20+x=28

x=814-3-8=3 (pcs).

Answer: 8 two-point shots, 3 one-point shots.

Subtract the remaining points from the three-point shot.

Subtract the number of shots left on the three-point shot.

The number of one-point balls is 11*2-19=3

The two-point shot is 11-3 = 8.

Let x one-point balls be put into x and y two-point balls should be put in.

then x+y+3=14 (14 shots, including 3 three-pointers) x+2y+3 3=28 (28 points in total) get y=8

Substituting x=3

That is, a total of 3 one-point shots and 8 two-point shots.

It's a simple question.

Solution: Set a two-point shot to hit the X ball, and a one-point shot (free throw) to shoot the Y ball.

2x+y+3x3=28

x+y+3=14

Let's figure it out for yourself.

Answer: x is 8 y is 3

Suppose the two-point shot was made x times and the one-point shot was made y times.

A total of 28 points, three three-pointers are 9 points, and the remaining score is 28-3*3=19 columns of equations 2x+y=19 x+y=11

The result of the knot is: x=8 y=3

If there are x two-point balls, then one point (19-2x) balls.

x+(19-2x)+3=14

x=8 then 8 two-point shots and 3 one-point shots.

Solution: Set x two-point shots and y free throws.

There is 2x+y+3*3=28, x+y+3=14

The solution is x=8, y=3

8 on a two-point shot and 3 on a one-point (free throw)?

Eight of two points and three of one-pointers.

He shot 8-of-18 from two-point range and 3-of-3 from the free-throw line

8 two-pointers and 3 one-pointers.

46+12) divided by one-third, minus 14, and divided by two-thirds.

Let the total number of parts be x

Completed on the first day 1 3*x+14 pieces.

Completed on the second day (x-(1 3*x+14))*2 3-12 pieces.

On the 3rd day, 46 pieces were completed, then the rolling god x=1 3*x+14+(x-(1 3*x+14))*2 3-12+46

Solve the big loss x = 174 pieces.

Set the total to y! That is, y minus 3 points, 1y minus 12, minus 3 points, 2y plus 14 minus 46 ....All right.

If the selling price is x, then the purchase price is (x-2).

x=6, so the purchase price is 6-2=4 yuan.

It's simple.

Because the original selling price is profitable 2 yuan, and the current selling price is profitable yuan, so after the price change, there is less profit yuan, which is the result of a 7% discount, so this yuan is 30% of the original selling price =, so the original selling price is yuan, so the purchase price is 6-2 = 4 yuan.

Question: Which plan is the most cost-effective?

If you want 45 cars to be x, then 45x 270, solve x 6, x take 6, then 6 250 = 1500 yuan.

If you want to take 60 cars to y, then 60y 270, the solution is y, because the car has no whole number, so y takes 5, then it takes 5 300 = 1500 yuan.

Both plans are the same price, and you can do both.

Question: Which cars can I rent the cheapest?

Check it out for yourself, I can't guarantee that it's right.

The upstairs has already said that I can only push it up.

ABC.

acb=90°，∠cab=60°

ab = 2ac = 4 (cm) Pythagorean theorem.

pm⊥aby=1/2amxpm

x1tx (3 under the root number) t

3) t under the root number

When t=0, m coincides with a, nb=ab-an=ab-mn=4-1=3(cm).

At this point the amp does not exist (because am=0).

t>0 again. As cd ab in d, ad=1 2

ac=1(cm)

Pythagorean theorem. When AM>AD, APM does not exist. xtt

T assumes that mnqp is rectangular, then pq ab mn, pq=mn=1 cpq=cab=60°

The two straight lines are parallel and the isotope angles are equal.

cp=1/2pq

Pythagorean theorem. ∴ap=ac-pc=3/2

MNQP is rectangular.

pm⊥ab，∠pma=90°

a=60°

am=1/2

ap=3/4(cm)

am=1xt at this time.

t=3/4(s)

t=3/4，0

t There is a case where the mnqp is rectangular.

Suppose that cab is similar to cpq, and c is the common angle between cab and cpq.

cab=∠cpq

pq‖abpm⊥ab，qn⊥ab

MNQP is rectangular.

Same as , t=3 4