QP=24 cm
RS=11.25 cm
QS=18.75 cm
<u>Explanation</u>:
Given that TQ bisects <RTP
(1)
consider ΔRQS and ΔRPT
QS||PT,RP and RT are transversals
(alternate angles)(2)
comparing (1) and (2)
and triangle SQT is isocelus
Therefore SQ=ST(sides opposite to equal angles in an isocelus triangle)
Therefore <RQS=<RPT(corresponding angles)
<RSQ=<RTP(corresponding angles)
therefore by AA criterion for similarity
ΔRQS~ΔRPT
According to the property of similar triangles
Answer:
k can either be
12
or
−
12
.
Step-by-step explanation:
Consider the equation
0=x2+4x+4
. We can solve this by factoring as a perfect square trinomial, so
0=(x+2)2→x=−2 and−2
. Hence, there will be two identical solutions.
The discriminant of the quadratic equation (b2−4ac) can be used to determine the number and the type of solutions. Since a quadratic equations roots are in fact its x intercepts, and a perfect square trinomial will have
2 equal, or 1
distinct solution, the vertex lies on the x axis. We can set the discriminant to 0 and solve:
k2−(4×1×36)=0
k2−144=0
(k+12)(k−12)=0
k=±12
Answer:
GQ=25 units
Step-by-step explanation:
we know that
Point Q is the midpoint of GH
so
GH=GQ+QH and GQ=QH
GH=2GQ -------> equation A
we have
GH=5x-5
GQ=2x+3
substitute in the equation A and solve for x
5x-5=2(2x+3)
5x-5=4x+6
5x-4x=6+5
x=11
Find the length of GQ
GQ=2x+3
substitute the value of x
GQ=2(11)+3
GQ=25 units
<span>Solve for "x":
a/x = x/4a
---
x^2 = 4a^2
x = 2a</span>
If you simplify, it equals 10.82e.
