ok, so you have problems solving the simultaneous equations? am i to assume that you got the process then? as in, you understand how i formed the two equations
There are two mainstream methods to solve systems of equations (there are more than two ways, but these two get used more often), Elimination, and Substitution.
I will do Elimination, it is your responsibility to look up the rest when you have time.
The objective of Elimination, as its name suggests, is to eliminate on variable by adding (or subtracting) one equation from the other. Of course for this to work, the variable you want to eliminate has to have the same coefficient in both equations. To get that to happen, you may have to multiply one (or both equations) by something. Let's see how this works.
................(1)
..................(2)
Let's eliminate y. To do this i must multiply equation two by 3 (or -3 if you prefer). we get:
....................(1)
.....................(3) = (2)*3
Now we have 3y in both equations, we can therefore, subtract one from the other, and the y's will go away. we obtain:
.................(3) - (1)
Now we solve for x, we get
Now we plug in that value for x into either of the original equations to find y
Hello,
according to the text the equation of the circle is
If the line is a tangent to the circle the distance of the centre to the line must be as large as the radius.
The distance of a point P(x1, y1) to a line Ax + By + c = 0 is calculated by:
Thus circle #3 is the right one
Hello,
Definiton: The circle (line) is the set of points (in a plane) which have the same distance to a fixed point, the centre of the circle.
Definition: The sphere (surface) is the set of points (in a space) which have the same distance to a fixed point, the centre of the sphere.
As you can see there are two answers possible.
Note that, for a cylinder, the surface area, SA, is given by:
since we have:
...........I divided through by
This is a quadratic equation, you should be able to solve for
Now, the volume, V, of a sphere is given by:
use the value you got for in the above formula and you will have your answer
Hello again, Outragexl10!
Now I see that you've solved many of them already.
Oh well . . . maybe I can offer a different approach.
(a) Since is inscribed in a semicircle,5) .[6 points] . In circle is a diameter.
a) Find and
b) If , find the length of the radius to the nearest tenth.
Hence, and are complementary: .
. . which simplifies to:. .
. . which factors: .
. . and has the positive root: .
Therefore: .
(b) Let . .Then: .
In right triangle , we have: .
Therefore: .
7) .[4 points] . is a common external tangent of circle and circle .
The radius of circle is 12 feet and the radius of circle is 5 feet. . ft.
Find the area of quadrilateral .
Note that radii and are perpendicular to tangent .
Note also that:
. . Hence: .
Invert the quadrilateral.Code:S * | * 25 7 | * | * V + - - - - - - - - - - - * R | x | 5 | | 5 | | * - - - - - - - - - - - * P Q
Through draw a line parallel to , meeting at .
. . Let
In right triangle , we have: .
. . Then: .
Quadrilateral is a trapezoid with parallel bases 5 and 12
. . and height 24.
Therefore, its area is: . ft².