A few math problems, please help, a few math problems, everyone helps

Too much - just know that the first one is to make the angle bigger and the chances of scoring higher, I've done - -

The first is to extend and then add two theorems that "one of the outer angles of a triangle is greater than any one of the inner angles that is not adjacent to it" to draw conclusions.

The second one has three sides of a, b, and c

s=4a/2=2a s=12b/2=6b

So a=3b

Then c is in the range of a-b, i.e., 2b, i.e., 3, so h=4 or 5

1) H=4, C=3B, then A=C, not in line with the topic, discarded.

2) H=5, C=12B5 holds.

So high is 5 third, if you're doing this kind of problem for the first time, or if your mind is not very clear, the enumeration method is definitely the best choice, although it looks clumsy, but it is very convenient to write it all out one by one.

Be careful not to miss the first example of the plan that uses all 15s advertising, and then replace the two 15s with one 30s, and then list them all, eliminate the non-qualified (require each type of advertising ** not less than twice) and it's OK, and then ask what to answer.

Fourth, there is a standard practice for this kind of problem, which is to use the basic solution of a one-dimensional equation and make it in the form of ax=b, because it is true regardless of the value of x, so the equation has an infinite number of solutions, a=0, b=0

The lazy way to do this is to take a few values of x, and to connect a few equations to get the value of a b.

The fifth - -

Please read the question carefully, carefully list the system of equations, and I believe you will do it--

The sixth one, the same as above, is a basic problem, pay attention to the uphill when you go, and the downhill when you come back, don't make a mistake.

In addition, uphill and downhill application problems equations (groups) can sometimes be difficult to solve, just be careful.

The seventh one is simpler than the first two, and I think lz is here to ask for the answer, and I'm sorry, I don't have time to do it, so I can only tell you some questions that need to be rationalized.

The eighth one is a little dizzy, and it's good to make up more.

One 50-cent coin, one 25-cent coin, 4 10-cent coins.

The ninth, let's go to the problem of the system of equations

However, it seems to be an indefinite equation.

And the numbers can also be made up.

However, the question requires the calculation ......

Set up a win, draw b, lose c.

3a+b=19 a+b+c=12

Solve this indefinite equation and discard the undesirable solution.

The tenth is the problem of inequality.

Set the planned daily water consumption x tons.

x+x then the total actual water consumption is 110x 2179....

The total water consumption is more than 2179t.

There must be something wrong with the question 10, either the conditions are too few or the question is wrong.

You can only find the lower limit here, and you don't have to keep it.

I don't know if it's about total water consumption or daily water consumption.

That's it, the time relationship, there is no answer, only the idea.

If you have any questions about the answer to a question, please add it and I can find time to help you do it.

1.2007 2009 + 1 = under the root number

Under the root number ((2008 + 1) * (2008-1) + 1) = under the root number (2008 * 2008-1 * 1 + 1) = under the root number (2008 * 2008) = 2008

2.n(n+2)+1=n*2+1=n+n*2+1) = n+1(n+1) = n+1

Root number 11 = 8....

b = 5 - root number 11 = 1....

3a-2b=22

4.Rational numbers (

1 3,46,0,216 cube roots).

Irrational numbers: (8 under the change sign, 1 2 under the change sign.)

2 out of 2. Positive real numbers: (

8 under 1 2

The cube root of 216) real numbers: (

8 under 1 2

The cube root of 216.

2 out of 2.

1. It is known that the parabola y=ax 2+(4 3+3a)x+4 intersects with the x-axis at two points a and b, and intersects with the y-axis at the point c, whether there is a real number a, so that abc is a right-angled triangle, if it exists, the value of a is requested, if it does not exist, please explain the reason.

Solution: Exists. y=ax 2+(4 3+3a)x+4 intersects with the y-axis at the point c, then c(0,4).

1) Assuming the parabola opening is upward, a>0, then the intersection of the parabola and the x-axis is on the same side of the y-axis.

then abc cannot be a right triangle.

2) Assuming that the parabola opening is downward, a<0, then the intersection of the parabola and the x-axis is on either side of the y-axis.

Then y=0, the coordinates of a and b are ((-3,0), (4 3a,0), and point c is a right-angle point.

According to the Pythagorean theorem, out a=-1 4<0

So when a=-1, abc is a right triangle.

Forget about it, find the formula, and set it one by one.