A worker originally planned to process a batch of parts within a limited time. If he processes 10 parts per hour, he can overfulfil 3; if he processes 11 parts per hour, he can complete 1 hour in advance. How many parts are there in this batch? How many hours will it take to finish as planned?

The total number of parts is: 10x-3 = 10 × 8-3 = 77. A: there are 77 parts in this batch, which will take 8 hours to complete according to the original plan

A workshop can produce 180 A-type parts or 120 B-type parts every day. If three or two A-type parts and two B-type parts can form a set, how to arrange the number of days to produce the most complete sets of products in 30 days?

Set X days to produce type a parts and (30-x) days to produce type B parts. According to the meaning of the question, we can get the following solution: x = 15.30-x = 30-15 = 15. Answer: 15 days to produce type a parts and 15 days to produce type B parts

A workshop produces 180 a-parts or 120 b-parts every day. If 3 a-parts and 2 b-parts form a set, it will take 30 days
How to arrange the number of days to produce a and B parts?

Solution: let a x days, then B (30-x) days
180x:120(30-x)=3:2
The solution is x = 15
A: 15 days for each party

If the image of the function y = AX2 + 1 is tangent to the line y = x, then a = ()
A. 18B. 14C. 12D. 1

When a ≠ 0, a = 14 is obtained from △ = 0, so the answer is B

In this paper, three integers a & # 178;, B & # 178;, 2Ab are given,
Write the formula and the process of factorization

a²-b²=(a+b)(a-b)

How to calculate the range of y = x ^ 2-2x + 60 ≤ x ≤ 4 by derivative method?

y=x^2-2x+6
y‘=2x-2=0
x=1
When x ≤ 1, y'1, y '> - 0, y = f (x) increases monotonically on [1,4]
So the minimum value Ymin = f (1) = 1-2 + 6 = 5
The maximum value at the endpoint is ymax1 = f (0) = 6
ymax2 = f(4)=16-8+6=14
So the range of Y is [5,14]

Score calculation
By the way, 2000

1) 4/14+8/14=6/7 2) 7/10+2/10=9/10 3) 4/7-3/7=1/7 4) 3/14+7/14=5/7 5) 4/11+5/11=9/11 6) 4/15+2/15=2/5 7) 9/10-3/10=6/10 8) 7/13+5/13=12/13 9) 6/13+5/13=11/13 10) 2/8+3/8=5/8 11) 9/13-9/13=0 12) 9/13+3...

If the solutions of the equations {Χ + 2Y = 6, X-Y = 9-3a are a pair of identical numbers, find the value of a?

X + 2Y = 6 --- Formula 1
X-Y = 9-3a --- formula 2
Equation 1-2
Y=A-1
Substituting Formula 1, we get
X=8-2A
It is known that x = y
So A-1 = 8-2a
A=3

Put the nine numbers 1-9 into the box of the following formula to make the equation hold

3+9-7=5,6*8/12=4
Of course, you can also change the order appropriately
When solving problems, you must know that 5 and 7 are impossible in multiplication and division, so only in addition and subtraction, so the scope is narrowed! In fact, you can think that odd numbers are difficult to set up in multiplication and division, in addition, you can regard the original formula as? +? =? +?,? *? =? *?, so it will be easier to think about it!

It is known that the asymptote equation of hyperbola is x ± y = 0, and the distance between two vertices is 2

When the focus of hyperbola is on the x-axis, let the analytic expression be
x²/a²-y²/b²=1
b/a=1 ;2a=2
The solution is a = b = 1
The analytic formula is X & sup2; - Y & sup2; = 1
When the focus of the hyperbola is on the y-axis, let the analytic expression be
y²/b²-x²/a²=1
b/a=1 ;2b=2
The solution is a = b = 1
The analytic formula is Y & sup2; - X & sup2; = 1

A worker originally planned to process a batch of parts within a limited time. If he processes 10 parts per hour, he can overfulfil 3; if he processes 11 parts per hour, he can complete 1 hour in advance. How many parts are there in this batch? How many hours will it take to finish as planned?