It is known that the radius of ⊙ o is 10, the length of chord AB is 103, point C is on ⊙ o, and the distance from point C to the straight line of chord AB is 5, then the area of quadrilateral with O, a, B, C as vertex is 0___ .

It is known that the radius of ⊙ o is 10, the length of chord AB is 103, point C is on ⊙ o, and the distance from point C to the straight line of chord AB is 5, then the area of quadrilateral with O, a, B, C as vertex is 0___ .

As shown in the figure, connect OA and ob, make the radius OE perpendicular to ab through O, and intersect AB with D; in RT △ oad, ad = 12ab = 53, OA = 10; so ∠ AOD = 60 ° od = 5; ① it is easy to know that de = oe-od = 5; so point e meets the requirements of point C; at this time, the diagonal AB and OE of quadrilateral oaeb are perpendicular to each other, so the quadrilateral oaeb is rhombic
If 3x + 5Y + 6 = 0, 3x + y-7z = 0, then the value of X + Y-Z is equal to ()
First correct:
3x + 5Y + 6 = 0 should be 3x + 5Y + Z = 0!
So there is
3x+5y+z=0 ①
3x+y-7z=0 ②
① + 2
6x+6y-6z=0
Divide both sides by 6
x+y-z=0
3x+5y+6=0①
3x+y-7z=0②
①-②
6-7z=0
7z=6
z=6/7
①+②
6x+6y+6-7z=0
6z+6y-6z=z-6
6x+6y-6z=6/7-6
6x+6y-6z=-36/7
X + Y-Z = - 6 / 7: no answer, thank you!!
As shown in the figure, the parabola y = - 0.5x2 + 0.5x + 6 intersects the X axis at two points a and B, and intersects the Y axis at point C
Given point E (0, - 3), take point D on the parabola of the first quadrant, connect De, so that De is bisected by x-axis, try to determine the shape of quadrilateral ACDE, and prove your conclusion
Why is the D coordinate (3,3)
What about the graph? I'll give you a general description. From this equation, we can see that the intersection of the parabola and the X axis is (4,0) and (- 3,0), and the C coordinate is (0,6). Since e (0, - 3), De is bisected by the X axis, the ordinate of point d must be 3. If point D is in the first quadrant, the coordinate is (3,3). You can judge from the graph. It should be a parallelogram
If 2x + 5Y + 4Z = 0, 3x + y-7z = 0, then what is the value of Z + Y-Z?
2x+5y+4z=0-----＞ 6x+15y+12z=0 (1)
3x+y-7z=0-----＞6x=2y-14z=0 (2)
(1) (2) by eliminating 6x, we get the following result:
y+2z=0 (3)
Substituting x + Y-Z, x-3z =?
(3) Substituting (2), we get:
x-3z=0
So x + Y-Z = 0
If a parabola is obtained by the translation of y = 2x & sup2; + 3x-5 and the vertex coordinates (4, - 2), then its analytical formula is
The coefficients of X & sup2; are equal in translation
Vertex (4, - 2)
So y = 2 (x-4) & sup2; - 2
That is y = 2x & sup2; - 16x + 30
If the vertex coordinates of the original parabola are a (- 3 / 4, - 49 / 8), and the vector from a to vertex (4, - 2) is (19 / 4,33 / 8), then the analytical formula of the parabola is y + 33 / 8 = 2 (x + 19 / 4) ^ 2 + 3 (x + 19 / 4) - 5, just sort it out..
Best for you. Thanks
The following equations are solved by proper methods. (1) 6x + 5Y = 253x + 4Y = 20 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; (2) 4 (x − y − 1) = 3 (1 − y) − 2x2 + Y3 = 2
(1) 6X + 5Y = 25, 1 3x + 4Y = 20, 2, 1 - 2 × 2, - 3Y = - 15, the solution is y = 5; substituting y = 5 into 2, 3 x + 20 = 20, the solution is x = 0, so the solution of this equation system is: x = 0y = 5; (2) the original equation system can be reduced to 4x − y = 5, 1 3x + 2Y = 12, 2, 1 × 2 + 2, 11 x = 22, the solution is x = 2
Y = - (x-1) is obtained by translating the parabola to each unit. The axis of symmetry of the parabola after translation is the vertex coordinate. When x = y has the maximum value, its value is
Y = - (x-1) & # 178; is obtained by translating the parabola y = - X & # 178; one unit to the right
The axis of symmetry of the parabola after translation is x = 1, and the vertex coordinate is (1,0). When x = 1, y has the maximum value, which is 0
3x-5y=6 x+4y=-15
3x-5y=6 ①
x+4y=-15 ②
② X 3 - 1, 12Y + 5Y = - 45-6
17y=-51
y=-3
Substituting into 2, the solution is x = - 3
3x-5y=6(1)x+4y=-15(2)
From (2),
x=-4y-15 （3），
Substituting (3) into (1) yields,
3（-4y-15 ）+4y=-15，
The solution is y = - 3,
By substituting y = - 3 into (3), we get
x=-3，
The solution of the original equations is
x=-3y=-3 ．
3x-5y=6
x+4y=-15
3x-5y=6
3x+12y=-45
-17y=51
y=-3
x=-15-4×(-3)=-3
x=-3
y=-3
Parabola y = - 3x ＾ 2-5 opening direction () axis of symmetry is () vertex coordinate is () vertex is the most () point, so the function has the most () value, is ()
Parabola y = - 3x ＾ 2-5, the opening (downward) symmetry axis is (x = 0), the vertex coordinate is (0. - 5), the vertex is the highest (highest) point, so the function has the maximum (maximum) value, which is (- 5)
Is it math homework? Ha ha! Follow up: if you have all the answers, you can send them to me. The point is this question
If the solution of equation 4y-3 (M-Y) = 5y-8 (M-Y) about y is y = 15, try to find the solution of equation 2m (x-1) = (M + 1) (3x-4) - 2 about X
4y-3m+3y=5y-8m+8y
7y-3m=13y-8m
5m=6y
5m=90
m=18
36(x-1)=19(3x-4)-2
36x-36=57x-76-2
36x-36=57x-78
42=21x
X=2
Substituting M = 18 gives x = 2
4y-3(m-y)=5y-8(m-y)
4y-3m+3y=5y-8m+8y
5m=6y
When y = 15, M = 18
2m(x-1)=(m+1)(3x-4)-2
2mx-2m=3mx-4m+3x-4-2
2m-mx-3x=-6
2m-x(m+3)=-6
When m = 18, x = 2
4y-3m+3y=5y-8m+8y
7y-3m=13y-8m
5m=6y
5m=90
m=18
36(x-1)=19(3x-4)-2
36x-36=57x-76-2
36x-36=57x-78
42=21x
X=2
When m = 18, x = 2