Three mathematical problems of linear equation with one variable 1. It costs 4.7 yuan to buy 4 exercise books and 3 pencils. It is known that each pencil costs 0.5 yuan. How much is each exercise book? 2. A thin and long cylinder with a bottom radius of 5cm and a height of 36cm is forged into a short and fat cylinder with a bottom radius of 10cm. What is the height of the cylinder? 3. Use a 60cm long wire to form a rectangle (1) Let the width of the rectangle be 2 / 3 of the length, and find the length and width of the rectangle (2) Make the length of the rectangle 4cm more than the width, and find the area of the rectangle

Let each exercise book be x yuan. 4x + 3x0.5 = 4.74x = 4.7-1.5x = 3.2 ／ 4x = 0.8, the current height be xcm.3.14x10 ＆ 178; X = 3.14x5 ＆ 178; x36x = 5 ＆ 178; X36 ／ 10 ＆ 178; X = 93; (1) if the length is xcm, the width is 2 / 3xcm. 2 (x + 2 / 3x) = 605 / 3x = 30x = 182 / 3x = 18x2 / 3 = 12, the width is xcm

A junior one's one variable equation mathematical application problem!
Charge price of water
2 yuan / M & sup3 for water consumption not exceeding 6m & sup3; per month;
The monthly water consumption exceeding 6m & sup3 and not exceeding 10m & sup3 is 4 yuan / M & sup3;
The monthly water consumption exceeding 10m & sup3; is 8 yuan / M & sup3;
Q:
If the water supply of the household is 15m & sup3 in March and April; (the water consumption in April is more than that in March), and the bus water fee is 44 yuan, how many cubic meters of water will the household use in March and April?
It's better to explain how to solve the equation of first degree with one variable,

Because it is known that the total water consumption is 15m & sup3;, and the water consumption in April is more than that in March, so it can be divided into two cases: the first case: suppose that the water consumption in March is not more than 6m & sup3;, and that in April is more than 6m & sup3; but not more than 10m & sup3;. Then: suppose that the water consumption in March is XM & sup3;, and that in April is (

Taking the known circle x-4x + y + 2Y = 0 as an example, the tangent equation passing through a point P (0, - 2) on the circle is obtained

(x-2)^+(y+1)^=5
Take (2, - 1) as the center and root 5 as the radius
Then a straight line y = KX + B can be obtained from the center of the circle to P
x1=0,y1=-2,x2=2,y2=-1
The line is y = 1 / 2x-2
Then the product of the tangent of point P and its slope is - 1
The tangent slope of point P is - 2
Bring in (0, - 2)
The tangent equation is
y=-2x-2

If the line ax + 2Y + 6 = 0 and the line x + a (a + 1) y + (A & # 178; - 1) = 0 are perpendicular, find the value of a and the value of a when parallel. Please be more detailed

A:
1) When two lines are vertical:
The straight line ax + 2Y + 6 = 0 and the straight line x + a (a + 1) y + (A & # 178; - 1) = 0 are perpendicular
A = 0, the straight line is y + 3 = 0 and the straight line X-1 = 0, the two straight lines are vertical, which is in accordance with
A = - 1, the straight line is - x + 2Y + 6 = 0 and the straight line x = 0, which is not vertical and does not conform to the requirements
When a ≠ 0 and a ≠ - 1, the slope product is - 1: (- A / 2) * {- 1 / [a (a + 1)]} = - 1
So: 2 (a + 1) = - 1
The solution is a = - 3 / 2
To sum up, a = - 3 / 2 or a = 0
2) When two lines are parallel:
The line ax + 2Y + 6 = 0 is parallel to the line x + a (a + 1) y + (A & # 178; - 1) = 0
When a = 0 or a = - 1, the two lines are not parallel
Then the slopes are equal: k = - A / 2 = - 1 / [a (a + 1)]
(a+1)a^2=2
a^3+a^2-2=0
(a^3-1)+(a^2-1)=0
(a-1)(a^2+a+1)+(a-1)(a+1)=0
(a-1)(a^2+2a+2)=0
Because: A ^ 2 + 2A + 2 = (a + 1) ^ 2 + 1 > 0
So: A-1 = 0
The solution is: a = 1
The straight lines are x + 2Y + 6 = 0 and X + 2Y = 0, which are parallel to each other
To sum up, a = 1

Solving equation x ^ 2 + XY + y ^ 2 = 7 y ^ 2-5xy + 6y ^ 2 = 1
x^2+xy+y^2=7
y^2-5xy+6y^2=1

I guess it's wrong. Of course it's better now. I don't understand
I guess the title is
x^2+xy+y^2=7 x^2-5xy+6y^2=1

The ratio of number a to number B is 3 to 4, the ratio of number B to number C is 5 to 6, and the ratio of number a to number C is 1

5:8

① Y = x-5x to find the zeros of functions on X ∈ [- 2,2]; ② y = X-6 / X-1 to find the zeros of functions on X ∈ [- 2,2]

Solution 1: y = x (X-5) gives zero point x = 0, x = ± √ 5, so the zero point on X ∈ [- 2,2] is (0,0) solution 2: y = (x-x-6) / x = (x-3) (x + 2) / x, gives zero point x = 3, x = - 2, so the zero point on X ∈ [- 2,2] is (- 2,0)

Given a ^ 2 + B ^ 2 = 7, a + B = - 3, find the value of ab

a^2+b^2=7,
a+b=-3,
Square on both sides
a^2+b^2+2ab=9
So 7 + 2Ab = 9
2ab=2
ab=1

If the function FX is equal to x ^ 3 + ax ^ 2 + BX, a and B can be obtained symmetrically about 1 and 1 points

Let P (x, y) be any point in the image, then the symmetric point of P with respect to (1,1) is Q (2-x, 2-y),
It is known that Q is also on its image,
So, (2-x) ^ 3 + a (2-x) ^ 2 + B (2-x) = 2-y = 2-x ^ 3-ax ^ 2-bx
Expand and merge to (2a + 6) x ^ 2 - (4a + 12) x + (4a + 2B + 6) = 0,
The above formula holds for any real number X,
So 2A + 6 = 0, 4A + 12 = 0, 4A + 2B + 6 = 0,
The solution is a = - 3, B = 3

If the vertex of the parabola x ^ 2 = 2Y is the closest point on the parabola to the point a (0, a) (a > 0), find the value range of A
Take (0, a) as the center and a as the radius: X ＾ + (Y-A) ＾ = a ＾
The simultaneous equations are: 2Y + y ＾ - 2ay = 0
y＾+（2-2a）y=0
△=（2-2a）＾≥0
a≤1
∴0＜a≤1
But why △≥ 0?
Shouldn't a circle tangent to a parabola be △ 0?
^There's a "2" missing in the back
It's Square

Take (0, a) as the center and a as the radius: X ＾ 2 + (Y-A) ＾ 2 = a ＾ 2
The simultaneous equations are: 2Y + y ＾ 2-2ay = 0
y＾2+（2-2a）y=0
y(y-(2a-2))=0
So: y = 0 and y = 2a-2
If 2a-2 > 0, the parabola and the circle will have three intersections (because y = 2a-2 corresponds to positive and negative x)
In this case, the vertex is not the closest point to point a (0, a) on the parabola
Such as 2a-2