There are "broadband network" and "dial-up network" for computer Internet access. The cost of dial-up network access consists of telephone fee and Internet fee. The former charging standard is: telephone fee 0.18 yuan / 3 minutes, Internet fee 7.2 yuan / hour. Now it is adjusted to: telephone fee 0.09 yuan / 3 minutes, Internet fee 4 yuan / hour. (1) before the tariff adjustment, the net name Zhang Yong was in his family budget, There has been a monthly expenditure of 70 hours on the Internet. How much is the budget? (2) after the tariff adjustment, the budget remains unchanged. How many hours can Zhang Yong surf the Internet every month at least?

There are "broadband network" and "dial-up network" for computer Internet access. The cost of dial-up network access consists of telephone fee and Internet fee. The former charging standard is: telephone fee 0.18 yuan / 3 minutes, Internet fee 7.2 yuan / hour. Now it is adjusted to: telephone fee 0.09 yuan / 3 minutes, Internet fee 4 yuan / hour. (1) before the tariff adjustment, the net name Zhang Yong was in his family budget, There has been a monthly expenditure of 70 hours on the Internet. How much is the budget? (2) after the tariff adjustment, the budget remains unchanged. How many hours can Zhang Yong surf the Internet every month at least?

1.70 * 7.2 = 504 yuan
2.504/4 = 126 hours

Xiao Ming has five cards with different numbers. Please draw the cards and complete the following questions
Number on card: - 3 - 50 + 3 + 4
1. Take out two cards to make the number product of the two cards maximum. How to extract? What is the maximum value?
2. Take two cards out of them to minimize the quotient of division of numbers on the two cards. How to extract? What is the minimum value?
3. Take out 4 cards and use the learned operation method, the result is 24. How to extract? Write the operation formula (each number can only be used once, at least write 3 kinds of operation formula)
Write in more detail

1,-3*-5=15
2,4/-5=-20
3,3*4*(-3--5) -3*(-5-3+0) (-5-3)*-3-0

1. We fill the second container with water, and then pour it into the first container. Q: after pouring, how many centimeters is the water surface in the first container from the bottle mouth
2. The area of one test field is 100 square meters more than three times that of the second one. The total area of the two test fields is 2900 square meters. How many square meters are the areas of the two test fields? (solved by equation)
3. Xiao Ming cut a strip of 4 cm wide from a square piece of paper, and then cut a strip of 5 cm wide from the remaining rectangular piece of paper. If the area of the strips cut twice is exactly the same, how much is the area of each strip? (use equation solution)

1. Volume of the second container filled with water: V1 = Π (8 / 2) square × 10 = 160 Π
Pour V1 volume of water into the first container, H = 160 Π / Π (4 / 2) square = 40 cm
As H > 39cm, the water will overflow after pouring,
Answer: the water surface in the first container is 0 cm from the bottle mouth
2. Set up the second plot with an area of x square meters
X + (3x + 100) = 2900, the solution is x = 700, 3x + 100 = 2200
Answer: the area of the first test field is 2200 square meters, and the area of the second test field is 700 square meters
3. Set the side length of the original square paper as x cm
4X = 5 (x-4), the solution is x = 4, then the area of the cut strip is 4x = 16
Answer: the area of each strip is 16 square centimeters

Do me a favor, this is the first volume of mathematics homework (2) 2.1 rational number addition (2) in the topic
Eight students took a one minute turn to fitness circle test. According to the standard of 32 turns, the number of times exceeded was positive, and the number of times insufficient was negative. The records are as follows
+2,-1,+3,0,-3,-3,+4,+2
I'll make it up tomorrow. Please use a simple method to find out the total number of times these 8 students turn to the fitness circle

2+（-1）+3+0+（-3）+（-3）+4+2
=[2+3+0+4+2]+[(-1)+(-3)+(-3)]
=11+(-7)
=4
32 × 8 + 4 = 260 (Times)

In △ ABC, SINB, cosa / 2, sinc are in equal proportion, then what shape must this triangle have?

SINB, cosa / 2, sinc are proportional sequence SINB * sinc = cos ^ 2 (A / 2) = (COSA + 1) / 2 (1) By integrating sum difference sin α sin β = - [cos (α + β) - cos (α - β)] / 2, cos (B + C) + cos (B-C) = cosa + 1, because in △ ABC, cos (B + C) = - cos [pi - (B + C)] = - cosa, so cosa + Co

All multiples of 11 are () A. prime B. composite C. prime or composite

Choose C
Among the multiples of 11, 11 is prime and the others are composite numbers

As shown in the figure, the image of two inverse scale functions y = K1X and y = k2x (where K1 > 0 > K2) in the first quadrant is C1, and the image in the second and fourth quadrants is C2. Set point P on C1, PC ⊥ X axis at point m, intersection C2 at point C, PA ⊥ Y axis at point n, intersection C2 at point a, ab ∥ PC, CB ∥ AP at point B, then the area of quadrilateral odbe is ()
A. |k1-k2|B. k1|k2|C. |k1•k2|D. k22k1

Let P (x, K1X), then a (k2xk1, K1X), C (x, k2x), s rectangle apcb = AP · PC = (x-k2xk1) (k1x-k2x) = (K1 − K2) 2K1, the area of quadrangle odbe = s rectangle apcb-s rectangle pnom-s rectangle mcdp-s rectangle AEON = (K1 − K2) 2k1-k1 - | K2 | - | K2 | = k22k1

Change the following sentences into synonyms
1、Rose is one of the best students in our class
Rose is_____ _____ _____ _____ _____ in our class
2、Sally often calls her parents on saturday
Sally often _____ _____ _____ _____ her parents on saturday

better than any other student

It is known that the analytic formula of parabola is y = a (X-2) + K. there are two different intersections A and B between the image and x-axis. Let its vertex be C and d be a point on the parabola symmetry axis. The side length of the quadrilateral ABCD is a diamond of 4 and has an inner angle of 60 °. Find the analytic formula of this function

If a ＞ 0, if a ＞ 0, CAD = 60, and then, if a ＞ 0, CAD = 60, and then 60 °, and then, if a (2-2, CAD = 60, and then, BAC = 30 ° in RT △ ACO, OC = 4 × sin30, so C (2, - 2). OA = 4cos, 60 ° = 4cos, 60, 60 = 2, and so, a (2-2, and so, a (2-2-2, so, a) so, y = 1 / 6 (X-2) & \\\\\35\; - 2 = 1 / 66x / 6x \\\\\\35\\\\\\\\\0,0), substituting y = a (X-2) &# 178; +Similarly, when a < 0, y = - 1 / 6 (X-2) & # 178; + 2 = - 1 / 6x & # 178; + 2 / 3x + 5 / 3;, and y = - root 3 / 2 (X-2) & # 178; + 2 times root 3 = - root 3 / 2x & # 178; + 2 times root 3x

When the values of a and B are, the value of the square of polynomial a + the square of polynomial b-4a + 6B + 13 is zero
And the square of a + the square of B - 4A + 6B + 18 has a minimum? What is the minimum?

a²+b²-4a+6b+13=(a-2)²+(b+3)²=0
So a = 2, B = - 3
a²+b²-4a+6b+18=(a-2)²+(b+3)²+5
When a = 2, B = - 3, there is a minimum, and the minimum is 5