It is known that the image of inverse scale function y = 3x-m intersects with the image of linear function y = K / X at (1,5) (1) Find the analytic expressions of these two functions (2) Find the intersection of the two other functions Wrong... The first function is y = 3x + M

It is known that the image of inverse scale function y = 3x-m intersects with the image of linear function y = K / X at (1,5) (1) Find the analytic expressions of these two functions (2) Find the intersection of the two other functions Wrong... The first function is y = 3x + M

If y = 3x-m and y = K / X intersect at (1,5), then (1,5) substitute y = K / X5 = K / 1, k = 5Y = 5 / x into y = 3x-m5 = 3 * 1-m, M = - 2, so y = 3x + 2, y = 5 / X intersection ordinates are equal, so y = 3x + 2 = 5 / x3x ^ 2 + 2x-5 = 0 (x-1) (3x + 5) = 0x = 1 is (1,5), x = - 5 /
The answers are as follows
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If x = 1, then y = 5
The solution is m = 2
K=5
Analytic formula y = 3x + 2, y = 5 / X
Let 3x + 2 = 5 / X
(x-1)(3x+5)=0
x=-5/3
Coordinates (- 5 / 3, - 3)
As shown in the figure, the image of the inverse scale function y = KX (K ≠ 0) passes through the point (- 3,1) and intersects with the straight line y = − 23x + m at two points a (x1, Y1) and B (X2, Y2), and X1 and X2 satisfy 1x1 + 1x2 + 13 = 0. (1) find the analytic expression of the inverse scale function; (2) find the value of M and the area of △ AOB
(1) By substituting (- 3, 1) into y = KX, we can get the analytic expression of k = - 3 × 1 = - 3, and the inverse proportion function is y = − 3x; (2) the intersection of the inverse proportion function y = − 3x and the straight line y = − 23x + m at two points a (x1, Y1), B (X2, Y2), and the intersection of the inverse proportion function y = − 3x and the straight line y = − 23x + M
It is known that the image of inverse scale function y = K / X intersects with the image of primary function y = 3x + m at points (1,5)?
Because (1,5) is the intersection of inverse proportional function and linear function
So k = 1 × 5 = 5, so y = 5 / X
And 3 + M = 5, M = 2, then y = 3x + 2
So 3x + 2 = 5 / X
The solution is X1 = 1, X2 = - 5 / 3
So the coordinates of the other intersection point are (- 5 / 3, - 3)
（-5/3，-3）
Given the function y = - 3x + 7, [1] when x is greater than 2, find the value range of function value y, [2] when y is greater than 2, find the value range of independent variable x
1、
X>2
So - 3x
【1】 When x is greater than 2, y is less than 1.
【2】 When y is greater than 2, X is less than 5 / 3
1. X = (Y-7) / - 3 is greater than 2
Y-7 is greater than - 6
Y is greater than 1
Y is greater than 2
7-3x greater than 2
3x less than 5
Less than 3 / 5 x
(1) X = (7-y) / 3 is obtained from the original function
Because x > 2
So (7-y) / 3 > 2
7-Y>6
Y2
So 7-3x > 2
X
Given the set a = {x | y = LG (2x & # 178; + 3x-2)}, the set B = {y | y = 2 & # 178; - 1 / 2 & # 178; + 1}, find (CRA) U B
The union of a's complement on R and B
A = {x | 2x & # 178; + 3x-2 > 0}, so a = (- 2,1 / 2) complement is (- ∞, - 2] ∪ [1 / 2, + infinity)
Is set B unit set y = 4-1 / 4 + 1 = 19 / 4? Or is the title copied wrong
If M and N are opposite numbers, X and y are reciprocal, and m and N are not zero, then XY (M + n) - Mn + 5xy=______ .
∵ m and N are opposite to each other, X and y are reciprocal to each other, ∵ M = - N, xy = 1, ∵ m + n = 0, Mn − 1, ∵ original formula = 1 × 0 - (- 1) + 5 × 1 = 0 + 1 + 5 = 6
Who has all the formulas for math from grade one to grade six?
Mathematics formula of grade 1-6 in primary school
Number of copies × number of copies = total number of copies / number of copies = total number of copies / number of copies = number of copies
2. 1 times × times = several times △ 1 times = several times △ 1 times
3. Speed × time = distance △ speed = time distance △ time = speed
4. Unit price × quantity = total price △ unit price = total quantity △ quantity = unit price
5. Work efficiency × work time = total amount of work △ work efficiency = total amount of work time △ work time = work efficiency
6. Addend + addend = sum - one addend = another addend
7. Subtracted - subtracted = difference subtracted - difference = subtracted difference + subtracted = subtracted
8. Factor × factor = product △ one factor = another factor
9. Divisor / divisor = quotient divisor / quotient = divisor × divisor = divisor
Primary school mathematics figure calculation formula
1. Square C perimeter s area a side length perimeter = side length × 4 C = 4A area = side length × side length s = a × a
2. Cube V: Volume A: edge length surface area = edge length × edge length × 6 s surface = a × a × 6 volume = edge length × edge length × edge length v = a × a × a
3. Rectangle
C perimeter s area a side length
Perimeter = (length + width) × 2
　　C=2(a+b)
Area = length × width
　　S=ab
4. Cuboid
V: Volume s: Area A: length B: width H: height
(1) surface area (L × W + L × H + W × h) × 2
　　S=2(ab+ah+bh)
(2) volume = length × width × height
　　V=abh
5 triangles
S area a bottom h height
Area = bottom × height △ 2
　　s=ah÷2
Triangle height = area × 2 △ bottom
Triangle bottom = area × 2 △ height
6 parallelogram
S area a bottom h height
Area = bottom × height
　　s=ah
7 trapezoid
S area a upper bottom B lower bottom h height
Area = (upper bottom + lower bottom) × height △ 2
　　s=(a+b)× h÷2
8 round
S area C perimeter Π d = diameter r = radius
(1) perimeter = diameter ×Π = 2 ×Π × radius
　　C=∏d=2∏r
(2) radius
9 cylinder
V: Volume H: height s; bottom area R: bottom radius C: bottom perimeter
(1) side area = perimeter of bottom surface × height
(2) surface area = side area + bottom area × 2
(3) volume = bottom area × height
(4) volume = side area △ 2 × radius
10 cone
V: Volume H: height s; bottom area R: bottom radius
Volume = bottom area × height △ 3
Total number △ total number of copies = average number
The formula of sum difference problem
(sum + difference) △ 2 = large number
(sum difference) △ 2 = decimal
The problem of sum times
Sum (multiple-1) = decimal
Decimals × multiples = large numbers
(or sum - decimal = large)
Differential multiple problem
Difference (multiple-1) = decimal
Decimals × multiples = large numbers
(or decimal + difference = large)
The problem of tree planting
1. The tree planting problem on non closed lines can be divided into the following three cases
(1) if trees are to be planted at both ends of the non closed line, then:
Number of plants = number of segments + 1 = total length △ plant spacing-1
Total length = plant spacing × (number of plants - 1)
Plant spacing = total length (number of plants - 1)
(2) if trees are to be planted at one end of the non closed line and not at the other end, then:
Number of plants = number of segments = total length △ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length △ number of plants
(3) if trees are not planted at both ends of the non closed line, then:
Number of plants = number of segments-1 = total length △ spacing-1
Total length = plant spacing × (number of plants + 1)
Plant spacing = total length (number of plants + 1)
The number of closed trees on the line is as follows
Number of plants = number of segments = total length △ plant spacing
Total length = plant spacing × number of plants
Plant spacing = total length △ number of plants
Profit and loss
(profit + loss) △ the difference between the two distributions = the number of shares participating in the distribution
(big profit - small profit) △ the difference between the two distributions = the number of shares participating in the distribution
(big loss - small loss) △ the difference between the two distributions = the number of shares participating in the distribution
Encounter problem
Encounter distance = speed and X encounter time
Encounter time = encounter distance △ speed and
Speed sum = encounter distance △ encounter time
Follow up questions
Pursuit distance = speed difference × pursuit time
Pursuit time = pursuit distance △ speed difference
Speed difference = pursuit distance △ pursuit time
Flow problem
Downstream velocity = hydrostatic velocity + water velocity
Countercurrent velocity = still water velocity - water velocity
Hydrostatic velocity = (downstream velocity + countercurrent velocity) △ 2
Water flow velocity = (downstream velocity countercurrent velocity) △ 2
Concentration problem
Weight of solute + weight of solvent = weight of solution
Weight of solute / weight of solution × 100% = concentration
Weight of solution x concentration of solute
Weight of solute △ concentration = weight of solution
Profit and discount
Profit = selling price cost
Profit margin = profit / cost × 100% = (selling price / cost-1) × 100%
Up and down amount = principal × up and down percentage
Discount = actual selling price △ original selling price × 100% (discount < 1)
Interest = principal × interest rate × time
After tax interest = principal × interest rate × time × (1-20%)
Length unit conversion
1 km = 1 000 m 1 m = 10 decimeters
1 decimeter = 10 cm 1 meter = 100 cm
1 cm = 10 mm
Conversion of area units
1 sq km = 100 ha
1 ha = 10000 M2
1 square meter = 100 square decimeter
1 square decimeter = 100 square centimeter
1 sq cm = 100 sq mm
Volume (volume) product unit conversion
1 cubic meter = 1000 cubic decimeter
1 cubic decimeter = 1000 cubic centimeter
1 cubic decimeter = 1 liter
Ml = 1 cm3
1 cubic meter = 1000 liters
Conversion of weight unit
1 ton = 1000 kg
1kg = 1000g
1kg = 1kg
Conversion of RMB units
1 yuan = 10 Jiao
1 jiao = 10 points
1 yuan = 100 points
Time unit conversion
1 century = 100 years 1 year = December
Big month (31 days): January, March, may, July, August, October, December
Small month (30 days): April, June, September and November
The average year is 28 days in February and leap year is 29 days in February
There are 365 days in a normal year and 366 days in a leap year
1 day = 24 hours, 1 hour = 60 minutes
1 minute = 60 seconds 1 hour = 3600 seconds
Calculation formula of perimeter area volume of primary school mathematics geometry
1. Perimeter of rectangle = (length + width) × 2, C = (a + b) × 2
2. Perimeter of square = side length × 4 C = 4A
3. Area of rectangle = length × width s = ab
4. Square area = side length × side length s = A.A = a
5. Area of triangle = bottom × height △ 2 s = ah △ 2
6. Area of parallelogram = bottom × height s = ah
7. Trapezoid area = (upper bottom + lower bottom) × height △ 2 s = (a + b) H △ 2
8. Diameter = radius × 2 D = 2R radius = diameter △ 2 r = D △ 2
9. Circumference of circle = circumference × diameter = circumference × radius × 2 C = π d = 2 π R
10. Area of circle = circumference × radius × radius
4X plus 3x minus 5 equals KX minus 20x plus 20K, which is a one variable linear equation about X, k equals?
4X ^ 2 + 3x-5 = KX ^ 2-20x + 20K is a linear equation of one variable about X, so k = 4
3X-5=-20X+80
23X=85
X=85/23
4X ^ 2 + 3x-5 = KX ^ 2-20x + 20K is a linear equation of one variable about X, so k = 4
3X-5=-20X+80
23X=85
X=85/23
4X ^ 2 + 3x-5 = KX ^ 2-20x + 20K is a linear equation of one variable about X, so k = 4
3X-5=-20X+80
23X=85
X=85/23
If the image vertex of quadratic function y = - x ^ 2-bx + C is in the third quadrant, then the value range of B and C is
b---------,c------------
b>0,c
The vertex is in the third quadrant, so the axis of symmetry is in the third quadrant, - B / 2A
A = {x | 5-x > 2x-1}, B = {x | - 7 ≤ 3-2x ≤ 5}. Find ① a ∩ B ② a ∪ B ③ (CRA) ∪ (CRB) ④ a ∩ (CRB)
First, the set is reduced to a = {x | X