How many kilometers is the distance between a and B? How many kilometers does the second car travel?

How many kilometers is the distance between a and B? How many kilometers does the second car travel?

Distance between the two places: 160 △ 4 / 7 = 280km
The second car goes: 280 × 1 / 4 = 70dkm

Application problems of solving equations in the third year of junior high school
1. In the supermarket, if you sell the goods with the purchase price of 40 by 50, you can sell 500 pieces, but the price increases by 1 yuan, and the sales volume is less than 10 pieces. If you are a manager, in order to earn 8000, how much should you set the price? How much should you buy at this time?
(the equation is not like this: stock (500-10 (x-50))
Then: (x-40) (180-10 (x-50)) = 8000 ask everyone to correct: how can I always find the small tree? How can I do the equation?
2. The farm needs to build a rectangular chicken farm. One side of the chicken farm is against the wall (the wall is 10 meters wide), and the other three sides are surrounded by iron wires. There is a wooden fence in the middle, with a total length of 23 meters. Please design a chicken farm, so that the area can reach 40 square meters
3. Cut a 20cm long iron wire into two sections, and make a square with the length of each section of iron wire as the perimeter. In order to make the sum of the areas of the two squares 17cm square, what are the lengths of the two sections? Let's say: for one section, X is: quarter x square + (quarter times (20-x)) square = 17‘
Guide me a few questions, I slowly skilled myself to deliberate
--------------------------------(wall) sketch: for example, the middle one on the left is a wooden fence, with a total length of 23m
1 1 1
1 1 1（x）
1 1 1
1 1 1
-----------------------------------
2. The farm needs to build a rectangular chicken farm. One side of the chicken farm is against the wall (the wall is 10 meters wide), and the other three sides are surrounded by wooden railings. There is a wooden railings in the middle, with a total length of 23 meters. Please design a chicken farm, so that the area can reach 40 square meters.

The first question: the answer is pricing, 60 yuan, or 80 yuan. Set the price increase as X, the profit is 50 + x-40, the quantity is 500-10x, and then the profit x quantity = 8000. It's a very simple one yuan quadratic equation, the solution is 10 or 30. Children, you check it again, good habit, and find that 10 and 30 are the answers. So, when the price is 60 yuan, you can buy 400

The square of (a + B-C)

(a+b-c)^2=[(a+b)-c]^2=(a+b)^2-2c(a+b)+c^2
=a^2+2ab+b^2-2ac-2bc+c^2
=a^2+b^2+c^2+2ab-2ac-2bc

A mathematical problem is as follows: to make x multiply by (the second power of x plus a plus 3)=
To make x times (the second power of x plus a plus 3) = x times (the second power of x plus 5) plus 2 times (B plus 2), what are the values of a and B respectively?
It's a process

x*(x^2+a+3)=x*(x^2+5)+2*(b+2)
-> x^3+(a+3)x=x^3+5x+2(b+2)
-> (a-2)x=2(b+2)
To make it true for any X
Then the coefficient of X A-2 = 0 → a = 2
2(b+2)=0 ->b=-2
So a = 2, B = - 2

Is the '/' sign a division sign or a multiplication sign in mathematics?

The division sign is equivalent to X / x = x / X of X

A vector problem: triangle ABC is inscribed in a circle with o as the center and 1 as the radius, and 3oa + 4ob + 5oC = 0
①OA*OB,OB*OC,OC*OA
② Area of triangle ABC
(I gave up the arrow)

That is, 3oa + 4ob = 5cO, because 345 is just a group of Pythagorean numbers, so OA is perpendicular to ob, so OA.OB=O It can also be obtained by using the angle of 345 OB.OC= -4/5, OC.OA= -3/5.
So the sine value of AOC is 3 / 5, the sine value of BOC is 4 / 5, so the area of AOC is 3 / 5, the area of BOC is 4 / 5, the area of AOB is 1 / 2, so the area of ABC of triangle is 6 / 5

In the triangle ABC, if high AD and high be intersect at point h, and BH = AC, then the degree of angle ABC is equal to

∠ABC=45°.
It is proved that: ∵ ad ⊥ BC, ∵ DAC + C = 90 °,
∵BE⊥AC,∴DBH+∠C=90°,
∴∠DAC=∠DBH,
∵∠BDH=∠ADC=90°,BH=AC,
∴ΔDBH≌ΔDAC,
∴BD=AD,
That is, Δ abd is an isosceles triangle,
∴∠ABC=45°.

X-1 of 2x-5 = x + 2 of 3-2

2x-x/5-1=3-x/2+2
2x-x/5-1+1=3-x/2+2+1
2x-x/5=6-x/2
2x-x/5+x/2=6-2/x+x/2
2x+0.3x=6
2.3x/2.3=6/2.3
x=60/23

The right focus of the known ellipse C: X Λ 2 / a Λ 2 + y Λ 2 / b Λ 2 = 1 (a > b > C) is f (1,0)
It is known that the right focus of ellipse C: X Λ 2 / a Λ 2 + y Λ 2 / b Λ 2 = 1 (a > b > C) is f (1,0)
Description: given that the right focus of ellipse C: X ∧ 2 / a ∧ 2 + y ∧ 2 / b ∧ 2 = 1 (a > b > C) is f (1,0) and the point (- 1, √ 2 / 2) is on ellipse C. (1) find the standard equation of ellipse C. (2) given that the moving line L passes through point F and intersects with ellipse C at two points a and B, ask: is there a fixed point Q on the x-axis such that the multiplication of vector QA by vector QB = - 7 / 16 holds? If so, find out the coordinates of point Q. if not, please explain the reason

C = 1; point (- 1, √ 2 / 2) on the ellipse has 1 / A ^ 2 + 1 / 2B ^ 2 = 1.1#
Then a ^ 2-C ^ 2 = B ^ 2 is brought into 1 # to get a ^ 2 = 2 or a ^ 2 = 1 / 2 (rounding off, because a > C in the ellipse)
Standard equation: x ^ 2 / 2 + y ^ 2 = 1
The elliptic equation is x ^ 2 + 2Y ^ 2 = 2
Let the line l be my + 1 = x (slope k = 1 / m, in fact y = K (x-1)) and substitute it into the elliptic equation
We get (my + 1) ^ 2 + 2Y ^ 2-2 = 0, that is, (m ^ 2 + 2) y ^ 2 + 2my-1 = 0.1#
Let Q coordinate (x3,0) a (x1, Y1) and B (X2, Y2) vector qaqb = - 7 / 16 be equivalent to x1x2-x3 (x1 + x2) + X3 ^ 2 + y1y2 = - 7 / 16 (2 #)
y1y2 = -1/(m^2+2); x1+x2 = 2+m(y1+y2) = 2-2m^2/(m^2+2)=4/ (m^2+2)
x1x2 = m(y1+y2)+1+y1y2m^2 = (-2m^2+m^2+2-m^2)/(m^2+2)=6/(m^2+2)-2
Let n = m ^ 2 + 2
6/N-x3*4/N+x3^2-1/N=25/16
(1 / N) (5-4x3) = 25 / 16-x3 ^ 2
1 @ when X3 = 5 / 4, the left side = 0, the right side = 0, constant hold! Test when the line is parallel to the X axis, also hold (think about why it is not parallel to the Y axis)
2 @ when x3 ≠ 5 / 4, 1 / N = 5 / 4-x3, the right side is constant, but the left side is variable
So there is a fixed point Q (5 / 4,0) which makes the proposition hold

Known: triangle ABC, angle a90 degrees, ad vertical BC and D, be bisector angle B, intersection AB and E, EF vertical BC and F, FM vertical AC and M. find FM equal to fd
Tips!

Let's make et parallel to BC through e, and let's make ad parallel to t, because ∠ 1 = ∠ 2, so AE = ef (the distance from the point on the bisector to both sides of the angle is equal) because ad is perpendicular to BC, EF is perpendicular to BC, so ad is parallel to EF, so ∠ Tae = ∠ FEM in triangle AET and triangle EFM, AE = ef (proved) ∠ Tae = ∠ FEM (proved)