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What is the area of the triangle ABC when the line y = four-thirds + 4 intersects with X at a, intersects with y at B, and O is the origin There are a few more. Give me 20 more ① What is the image line that the image of a linear function y = - 2x + 3 does not pass through ② It is known that the image of positive scale function y = K1 x intersects with the image of y = k2-9 at point P (3, - 6) If the first-order function y = k2x-9 intersects the x-axis at point a, the coordinates of a can be obtained,
Where's C? If it's ABO, the area is 3 times 4 divided by 2 = 6
1. Without going through three, you can remember a formula: the same positive is no more than four, the same negative is no more than one, the positive and negative are no more than two, the negative and positive are no more than three, the former represents K, and the latter represents B
2. Substituting the coordinates of point P into two analytic expressions, the solution is K1 = - 2, K2 = 1
Because the image intersects with the Y axis, so X-9 = 0, the solution is x = 9, so the coordinate of point a is (9,0)
This question is wrong, y = 4 / 3 + 4... There are no Unknowns at all
In the image, the line intersects at points (a, b), and 0 is the origin
(1) Without passing through the third quadrant, that is, the range of impossible values of Y is [0, -∞) (x is all real numbers)
(2) The coordinates of point P are brought into y = K1X and y = k2-9 respectively, and K1 = - 2, K2 = 1
The function intersects the X axis, that is, y = 0,
Take y = 0, K2 = 1 into the function and get x = 9
That is: a... unfold
This question is wrong, y = 4 / 3 + 4... There are no Unknowns at all
In the image, the intersection point is at the origin (a)
(1) Without passing through the third quadrant, that is, the range of impossible values of Y is [0, -∞) (x is all real numbers)
(2) The coordinates of point P are brought into y = K1X and y = k2-9 respectively, and K1 = - 2, K2 = 1
The function intersects the X axis, that is, y = 0,
Take y = 0, K2 = 1 into the function and get x = 9
That is: the coordinate of a is (9,0) folded
Given that a is an acute angle, the logarithm of 1 + cosa with 10 as the base = m, and the logarithm of 1-cosa with 10 as the base = n, what is the logarithm of sina with 10 as the base
It is known that a, B and C are all positive real numbers, and satisfy that log takes 9 as the base (9a + b) as the pair = log takes 3 as the base root, AB as the logarithm, and find the value range of C that makes 4A + B ≥ C constant
1)log[10](1+cosA)=m,log[10]1/(1-cosA)=n,log[10]sinA=
1+cosA=10^m
1/(1-cosA)=10^n; 1-cosA=10^(-n)
Sina = 1-cosa ^ 2 = 10 ^ (m-n) under the root sign
log[10]sinA=(m-n)/2
2) Log with 9 as the base (9a + b) as the pair = log with 3 as the root, AB as the logarithm
Log [9] (9a + b) = AB under the root of log [3]
1 / 2log [3]] (9a + b) = AB under the root of log [3]
9a＋b=ab;b=9a/(a-1)
4a＋b≥c
4a+9a/(a-1)=4(a-1)+9/(a-1)+13>=19
C
As shown in the figure, a and B are any two points symmetrical about the origin on the image of the function y = 2x, BC ‖ X axis, AC ‖ Y axis, and the area of △ ABC is denoted as s, then ()
A. S=2B. S=4C. 2＜S＜4D. S＞4
Let the coordinates of point a be (x, y), then B (- x, - y), xy = 2. AC = 2Y, BC = 2x
Lg2x-2lg (x-4y) + lgY = 0 find the logarithm of log with 2 as the base, Y / X
How much is it?
lg2x-2lg(x-4y)+lgy=0
lg2x-lg(x-4y)²+lgy=0
lg(2x*y/(x-4y)²)=0
2x*y/(x-4y)²=1
That is, 2XY = (x-4y) & sup2;
x²-10xy+16y²=0
(x-2y)(x-8y)=0
X = 2Y or x = 8y
And because x, Y > 0 (to make lgx, lgY meaningful)
Therefore, if x = 2Y, then LG (x-4y) = LG (- 2Y) is meaningless
So x = 8y
Log is based on 2, the logarithm of X / y = log2 (8) = 3
As shown in the figure, a and B are any two points symmetrical about the origin on the image of the function y = 2x, BC ‖ X axis, AC ‖ Y axis, and the area of △ ABC is denoted as s, then s=______ .
As shown in the figure, connect OC, let AC and X axis intersect at point D, BC and Y axis intersect at point E. ∵ A and B are symmetrical about the origin, BC ∥ X axis, AC ∥ Y axis, ≁ AC ⊥ X axis, ad = CD, OA = ob, ∥ s △ cod = s △ AOD = 12 × 2 = 1, ∥ s △ AOC = 2, ∥ s △ BOC = s △ AOC = 2, ∥ s △ ABC = s △ BOC + s △ AOC = 4
The logarithm of 64 ^ 1 / 3 minus 1 + log with two as the bottom eight is? LG25 + 2lg2 =?
64 ^ 1 / 3 minus 1 + log
=4-1+3
=6
lg25+2lg2
=2lg5+2lg2
=2(lg5+lg2)
=2
As shown in Figure AB, the area of △ ABC can be obtained from any two points BC ‖ X axis AC ‖ Y axis symmetrical about the origin on the image with function y = 4 / X
Let the intersection of AC and X axis be d
Because the analytic expression of inverse proportion function is y = 4 / X
∴S△AOD=2
∵ A and B are symmetric about the origin
∴AO=OB
∴S△ABC=4S△AOD=8
Let a (m, n), because a and B are symmetrical about the origin, B (- m, - n), so in RT △ ABC (∠ C is a right angle), BC = 2m, AC = 2n, s △ ABC = 1 / 2BC × AC = 2Mn. Because a (m, n) is a point on y = 4 / x, so Mn = 4, so s △ ABC = 8
If 2lg (x-2y) = lgx + lgY, then log2xy=______ .
∵ 2lg (x-2y) = lgx + lgY, ∵ x − 2Y ＞ 0x ＞ 0, y ＞ 0 (x − 2Y) 2 = XY, the solution is xy = 4. ∵ log2xy = log24 = 2
As shown in the figure, a and B are any two points symmetrical about the origin on the graph of function y = 1 x, AC ∥ Y axis, BC ⊥ Y axis, then the area s of △ ABC=______ .
The area of △ ABC s = 4 × 12 ×| K | = 2
The solution equation is: ① the cubic power of - (x-3) = 27, ② the - 2nd power of | - 5 | + (& # 189;) + (the third root - 27) - the square of root (- 2) - (root 7-1)
solve equations:
① The third power of - (x-3) = 27
-（x-3)=3
x-3=-3
X=0
② - 2 power of | - 5 | + (& # 189;) + (cubic root - 27) - square of root (- 2) - (root 7-1)
=5+√2/2-3-2-√7+1
=1+√2/2-√7
RELATED INFORMATIONS
- 1. As shown in the figure, a and B are any two points symmetrical about the origin on the graph of function y = 1 x, AC ∥ Y axis, BC ⊥ Y axis, then the area s of △ ABC=______ .
- 2. If we know that a and B are on the image of quadratic function y = AX2, the abscissa of these two points are - 2, 1 respectively, and the triangle AOB is a right triangle (o is the origin of coordinates), we can find the value of A
- 3. If the distance between the vertex A and the positive half axis of the graph a and B is the maximum distance on the positive half axis of the graph A. 32a−12aB. 32a+12aC. 62a−12aD. 62a+12a
- 4. Given that the right angle vertex a of RT triangle ABC moves on the straight line ρ cos θ = 9, C is the origin, and the angle ACB is 30 °, the polar coordinate equation of the trajectory of vertex B is obtained
- 5. Given that the length of the line segment l passing through point P (2,3) and cut by two parallel lines 3x + 4y-7 = 0 and 3x + 4Y + 8 = 0 is d 1), find the minimum value of D, 2) when the line L and 2) When the line L is parallel to the X axis, try to find the value of D
- 6. It is known that the integer part of 5 - √ 5 is a, and the decimal part is B. find a & # 178; + AB + B & # 178;
- 7. 6274 times the 5th power of (1 + x) equals 15031. Find out how much x equals! Keep 4 decimal places 1090 times the 5th power of (1 + x) is equal to = 3276. Find out how much x is equal to! Keep 4 decimal places 4233 times the 5th power of (1 + x) is equal to = 10496. Find out how much x is equal to! Keep 4 decimal places 3093 times the 5th power of (1 + x) is equal to = 5503. Find out how much x is equal to! Keep 4 decimal places 11597 times the 5th power of (1 + x) is equal to = 34089. Find out how much x is equal to! Keep 4 decimal places If there is a way, it would be better! 3Q
- 8. Given that the cube of a + the cube of B = 27, the square of a, the square of b-ab = - 6, find (the cube of B - the cube of a) + (the square of a, the square of b-3ab) -2 (cube of B - square of a) for detailed explanation, each step of explanation
- 9. What is the cube of the square of (a + b) (a-b) - (B-A) (a + b)
- 10. If a is a real number not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- If a is not equal to 1, we call 1 / 1-A the difference reciprocal of A. for example, the difference reciprocal of 2 is 1 / 1-2 = - 1, and the difference reciprocal of - 1 is 1 / 1 - (- 1) = 1 / 2. If A1 = - 1 / 3, A2 is the difference reciprocal of A1, A3 is the difference reciprocal of A2, A4 is the difference reciprocal of A3, and so on, then a2013=_______ .
- 11. Given that the line L passes through the point P (2,1), and intersects with the positive half axis of X axis and Y axis at two points a and B respectively, and O is the coordinate origin, then the minimum area of triangle OAB is? Why is 2 / A + 1 / b greater than or equal to 2 √ 2 / AB
- 12. When the inverse scale function y = K / X (k > 0) is known, the image passes through point a (2, m), where ab ⊥ X axis passes through point B, and the area of △ AOB is 3 [1] Find the direct [2] point C [x, y] of K and m on the image of inverse scale function y = K / x, find the value range of function value y when 1 ≤ x ≤ 3, take d [1,0] on the y-axis to connect BD, make a square with BD as the edge in the first quadrant, bdef asks whether e, f fall on the image of y = K / X
- 13. If the line y = KX + 1 and the ellipse X225 + y2m = 1 always have a common point, then the value range of the real number m is______ .
- 14. If the line y = K (x + 1) + 1 and the ellipse x2 / 5 + Y2 / M = 1 always have a common point, and the focus of the ellipse is on the x-axis, then the value range of M is
- 15. 1. It is known that for K ∈ R, the line y = KX + 1 and the ellipse (x ^ 2) / 5 + (y ^ 2) / M = 1 always have a common point, then the value range of the real number m is_______ . How can I figure it out wrong A (0,1) B (0,5) C [1,5] ∪ (5, positive infinity) d [1,5] I figured it out. It's wrong for me
- 16. How to calculate the distance from parabolic focus to hyperbolic asymptote?
- 17. It is known that the left and right focal points of hyperbola are F1F2, the eccentricity is following sign 2 and passing through point (4, - following sign 10) 1 to solve the equation of hyperbola
- 18. It is known that the asymptote equation of hyperbola C is y = ± 3x, and the distance from the right focus f (C, 0) to the asymptote is 3. (1) find the equation of hyperbola C; (2) make a straight line with slope k through F, l intersects the hyperbola at two points a and B, and the central perpendicular of line AB intersects the X axis at D
- 19. A focal coordinate is F1 (0, - 13), and the absolute value of the difference between a point P and two focal distances on the hyperbola is 24
- 20. To solve the hyperbolic equation, it has the same focus as the ellipse x ^ 2 + 4Y ^ 2 = 64, and the absolute value of the distance difference between the point on the hyperbola and the two focuses is 1?