A math problem 13 Math problem 13

Waypoint dialing: According to the title, sell according to the price, you can get a profit.

Yuan. Now at 80** of the price, 80 of the original profit should be obtained, ie.

Yuan. The actual situation is that not only did not get a profit, but lost 832 yuan. This resulted in a total loss of 1600

Yuan. This 1600

Yuan. That's exactly $20 of the list price. Solution 1].

Yuan. Solution 2].

According to the original price of 80 **, exactly 832 loss

Yuan. It can be solved according to this series of equations. Establish. Certain.

The cost of the commodity is x yuan. x+

80％=x-83280％x+

80％=x-832

80 x ten 768 = x-832

x-80％x=768+832

20％x=1600

x=8000

A: The purchase price of this product is 8000 yuan.

The definition of a prime number is that there is no divisor other than itself and 1, and proving a comfortable prime can of course be done in the same way, but we can follow the steps below to prove it.

1) Take the arithmetic square root of this number and take its integer part.

2) Use this number to remove 2 to see if it can be eliminated.

3) Use this number to remove 3 to see if it can be eliminated.

4) Divide continuously, all the way to the integer part of the arithmetic square root of this book. If none of these numbers are divisible, then the number is prime.

In other words, if a number cannot be divided by any number within the square root of its arithmetic from 2, it is prime.

There is no divisor other than 1 and itself, there is, not prime, no, is prime.

Example: 1 2 3 5 7 11 13 17 19....is a prime number.

The rest are not prime numbers [except 0].

Looking at this number, there is no divisor other than pull 1 and itself, there is, it is not a prime number, and it is not, it is a prime number.

There is no divisor other than 1 and itself, such as: 2 3 5 7 11 13 17 19 23 29 31 37 41 ...is a prime number.

See if it's divisible by all prime numbers less than it.

Not divisible by a number other than 1 and itself (e.g. ).

Write a program on your computer.

Simple oral arithmetic of numbers.

When x tends to be infinite:

e x is infinity, sinx and cosx are bounded**, so e x + cosx is equivalent to e x, e x -sinx is equivalent to e x, the original limit is e x e x, and the limit result is 1.

Another way of thinking, dividing up and down by e x:

cosx e x is the bounded quantity divided by infinity, and the result is 0;

In the same way, the sinx e x limit result is 0;

The original limit is: 1 1, and the limit result is 1.

The numerator and denominator are divided by e x, and because e (-x)sinx and e (-x)cosx are the product of infinitesimal and bounded functions, the original limit lim [1+e (-x)cosx] [1+e (-x)sinx]=[1+0] [1-0] 1.

Original formula = (e x-1) (e x+1) = 1 when x tends to infinity.

Original formula = (e x-1) (e x+1) = 1 when x tends to infinity.

So the original formula = 1

First understand the defined domains of the basic functions, and then solve them in turn.

The answer and the process are shown in the diagram.

That's right, is there something wrong with what the teacher said?

Can you draw an image of a straight line? The area of the triangle is half of the product of the horizontal and vertical coordinates where the line intersects the coordinate axis.

13, (1) solution; Because b(1,-3) c(-2,1) so d(3 2, -2).

2) Solution: Passing the point C as CE vertical AB in E

So the S triangle abc = 1 2ab*ce

CE is the perpendicular line of the triangular ABC.

Because a(2,4) b(1,-3) c(-2, 1) so ab = root number 50

ab = 5 times the root number 2

ac=5bc=5

So ac=bc

So the triangle ABC is an isosceles triangle.

So CE is the perpendicular line, midline of the isosceles triangle ABC.

So ae = be = 1 2ab = 5 times the root number 2 2 angles ceb = 90 degrees.

So the triangle ceb is a right triangle.

So BC 2 = be 2 + ce 2

So ce = 5 times the root number 2 2

So the s triangle abc = 25 2

So the area of the triangle abc is 25 2

Make use of the two-point distance formula:

ab=5√2,ac=5,bc=5

Obviously, there are: ab 2 = ac 2 + bc 2, ac = bc isosceles right angle, ab is hypotenuse.

The midline of the BC side is long l

l 2 = 5 2 + (5 2) 2, l = 5 5 2 Area: 1 2 * 5 * 5 = 25 2

Rotate the triangle APC so that AC and AB coincide with the point P, point P to point Q, and connect PQ. Then there are AQB= APC, PAQ=60 degrees, AQ=AP=PQ=3, and bq=P4. Bury wide trouble.

In the triangle APQ, BQ 2+PQ 2=16+9=25=PB 2, therefore, the triangle APQ is a right triangle, PQB=90 degrees, so, AQB= PQA+ PQB=150 degrees, that is, there are APC=150 degrees.