Can you solve a math problem in your first year of high school?

1) The general formula of the equal difference series is: an=a1+(n-1)d is substituted into n=10, an=30, n=20, an=50, and we get:

30=a1+9d

50=a1+19d

Solution: d=2, a1=12

an=12+(n-1)*2

2) The summation formula of the equal difference series is: sn=n*a1+n(n-1)d 2 then there is: sn=12n+n(n-1)=242

Solution: n=11

Let the first term be a1 and the tolerance is d, then a1+9d=30, a1+19d=50, and the system of equations is solved a1=12 d=2, so an=10+2n, sn=(a1+an)*n 2, so 242=(12+10+2n)*n 2, then n=11

1) d = (a20-a10) 10 = 2, the difference between a10 and a20 is ten terms.

a1=a10-(10-1)d=30-18=12, so an=12+2(n-1)=2n+10

2) sn=n*a1+n(n-1)d 2=242, so n=22 or —11 (rounded, n is positive).

Summary. The translation and stretching problem of trigonometric functions, the difference and method of panning first and stretching first - Translation: There are up-and-down and left-right translations.

The "w" of y=sin[wx+b] is the scaling of the function, and the scaling depends on the value of wI don't understand how to find the translational telescopic transformation of trigonometric functions -- Analysis: The formula "add left and subtract right, add up and subtract" y=sinx move right h unit length, get:

y=sin(x-h)y=sinx shifts h unit length to the left, and obtains:y=sin(x+h).

Hello, can you provide me with the full question?

1) If there is a question to know a Qingxiang which Xun 0, then there is a = 2 3, and (-2, 0) and (6, 0) are substituted into the unanswered slow pulse knowledge function t=2 16, =8, =3 4

f（x）=2√3cos（πx/8+3π/4）

2）g（x）=√3cos（π（x-2）/8+3π/4）

As can be seen from the figure, g(x) has a pole if it is not monotonic in [-2,1], and the minimum value can be found.

2）g（x）=2√3cos（π（x-2）/4+3π/4）

In the interval of [-2,1], when x=-1, y has a maximum value of 2 3

When x = -2, y = 6

When x=1, y=0

Then the range is 0,2 3

The expansion and contraction of the abscissa is transformed into the period of the trigonometric function, that is, the coefficient of x changes, which becomes 2 times the original, that is, the ordinate is unchanged, and the abscissa is reduced to half of the original early, and becomes the original 1 2 is the source of the hail with the ordinate unchanged, and the abscissa expands to 2 times the original. y=sinx - the abscissa is unchanged, the ordinate becomes the original a times to y=asinx——— the ordinate is unchanged, the abscissa becomes one-tenth of the original to y=asin x - if it is positive, the resulting image is shifted to the right by one unit, if it is negative, the image is shifted to the left by one unit to obtain y=asin(x).

The translational transformation of a trigonometric image has 1Translate first and then expand. 2.Expand and contract first and then translate.

The translation and stretching problem of trigonometric functions, the difference and method of panning first and stretching first - Translation: There are up-and-down and left-right translations. The translation up and down is mainly based on the translation of x, remember the law of "left plus right minus"!

Wake up: the coefficient of x must be 1) to translate up and down is to add or subtract the whole equation! The "w" of y=sin[wx+b] is the scaling of the function, and the scaling depends on the value of w

I don't understand how to find the translational telescopic transformation of trigonometric functions -- Analysis: The formula "add and subtract left and right, add and subtract up" y=sinx move right h unit length, get: slag y=sin(x-h)y=sinx left move h unit length, get:

y=sin(x+h)

Solution: Ream-1 3 - x = x -1 3

x=0So, let x=1 3 (that is, let x=any real number other than 0, here for convenience, take x=1 3).

f(-1/3-1/3)=f(1/3-1/3)∴f(-2/3)=f(0)

f(-2 3)=7 3 - 2 3 bf(0)=1

7/3 - 2/3 b=1

b=2

Isn't it f(-1 3-x)=f(-1 3+x), so the axis of symmetry x=-1 3

Then -b (2*3)=-1 3

b = 2 for d

According to the nature of the quadratic function, it is a parabolic function with an axis of symmetry, then, according to the nature of symmetry, it is similar to f(-1 3-x) = f(-1 3+x) in the problem

The axis of symmetry can be obtained, so the axis of symmetry x=-1 3

Then -b (2*3)=-1 3

b = 2 for d

Solution: Let x=sina y=1+cosa

x+y+csina+1+cosa+c

2sin(a+π/4)+c+1≥0

2sin(a+π/4)≥-c-1

sin(a+π/4)≥-1

2sin(a+π/4)≥-2

When -c-1 - 2, the inequality is constant.

c≥√2-1

The value of c can be [ 2-1,+

I can't read your question clearly.

It is recommended that you use the first formula to deduce the relationship between x and y.

Then bring it to the second formula.

Move the part about x y to the other side of the great equal.

Re-evaluate the range. The c-value is obtained.

I'm a junior in high school.

Look at the picture first, and then take a closer look at it step by step.

For reference, please smile.

Parallel lines, the slope is the same, i.e. the coefficient ratio of x and y is the same, so the option of x and 2y should be selected.

After the point (1,-1), that is, the introduction of x 1, y -1 into the equation can be solved. So choose A

The correct answer is A.

The straight line parallel to the straight line l:x+2y-3=0 can be set to l1:x+2y+c=0

p(1,-1) on l1, substituting the equation of the straight line of l1 yields: 1-2+c=0, c=1, the linear equation of l1 is x+2y+1=0

Comparing the ABCD option, the correct answer is A.

x=1, y=-1 is substituted into the equation and the equation is correct. If more than one is satisfied, the slope is revalidated.

A correct.

1.Are you so sleepy that you have given the wrong conditions? b=（sinα，cosβ）?

a b, then 4 3=cos sin

then tan = 3 4

tan2α=2tanα/(1-tan²α)2*3/4÷（1-9/16）=24/7

Guess [1+tan( +tan( -45)]=2 5-1 4) (1+1 10)=3 22

tan(α+45°)=tanα+tan45°)/1-tanαtan45°)=tanα+1)/(1-tanα)

So (tan +1) (1-tan ) = 3 22

3.There is a problem with the problem, the equation x (the square of x) + 3 times the root number 3 + 4, this is not an equation!

Here's how it works. Tan and Tan are known to be the two roots of the equation x (the square of x) + 3 times the root number 3+4.

The excavation type is tan + tan = -b 2a, tan tan = c a

tan(α+tanα+tanβ)/1-tanαtanβ)=

f(0+0)=f(0)*f(0) f(0)≠0 f(0)=1>0x>0 f(x)>1>0

x<0 -x>0 f(-x)>0

f(x-x)=f(x)*(x)=f(0)=1f(x)=1/f(-x)>0

In summary, for any x r, there is always f(x)>0

Because for any a,b r, there is f(a+b)=f(a)*f(b), so a=0,b=0 there is f(0)=f(0)*f(0) f(0)≠0 f(0)=1

For any x less than 0, -x>0

f(x+(-x))=f(x)*f(-x)

f(0)=f(x)*f(-x)

1=f(x)*f(-x)

So f(-x)=1 f(x)>0

In summary, for any x r, there is always f(x)>0

f(0)=f(0)*f(0) at x=0 Because f(0) is not equal to 1, f(0)=1

x>0 f(x)>1.

x<0 f(a-a)=f(a)*f(-a)=f(0)=1, let a>0 f(-a)=1 f(a) is a positive number, let a<0 Same as above.

So for any x r, there is always f(x)>0

Solution: From the coordinates of point A (0,4), we can know that f(0)=4

Then the parabola of the straight line where BC is located is obtained from the coordinates of B and C.

Let the function of the line where bc is located be y=kx+b

Bringing b(2,0) and c(6,4) into the above equation respectively yields the following equations:

0=2k+b

4=6k+b

The solution is k=1 b= -2

So the analytic expression of this function is y=x-2

Then bring x=4 to get y=4-2

y=2, so f(f(0)))=2

First inside and then outside, the inner function f(0) can be seen from the figure as 4

Then find the outer function f(4).

As can be seen from the figure, it is 2