the correct question is<span>What value of x would make RQ tangent to Circle P at point Q?
the picture in the attached figure
we know that
If RQ is a tangent to circle P at point Q
then
</span>RQ is perpendicular to the radius PQ at the point of tangency Q.
This means that ΔPRQ would be a right triangle.
Applying Pythagorean theorem
RP²=PQ²+RQ²
(x+9)²=9²+12²
(x+9)²=225
(+/-)(x+9)=√225
(+/-)(x+9)=15
+(x+9)=15-----> x=15-9-----> x=6
-(x+9)=15-----> -x-9=15-----> x=-9-15-----> x=-24
the solution is
x=6
the answer isx=6 units
Answer: place the dot on the graph, is there any question to it?
Step-by-step explanation:
30
−
(
2
3
)
(
x
3
)
=
30
+
−
8
x
3
=
−
8
x
3
+
30
Answer:
n = 2 or n = -3/2
Step-by-step explanation by completing the square:
Solve for n:
2 n^2 - n - 6 = 0
Divide both sides by 2:
n^2 - n/2 - 3 = 0
Add 3 to both sides:
n^2 - n/2 = 3
Add 1/16 to both sides:
n^2 - n/2 + 1/16 = 49/16
Write the left hand side as a square:
(n - 1/4)^2 = 49/16
Take the square root of both sides:
n - 1/4 = 7/4 or n - 1/4 = -7/4
Add 1/4 to both sides:
n = 2 or n - 1/4 = -7/4
Add 1/4 to both sides:
Answer: n = 2 or n = -3/2