<h3>Radius = 14cm</h3>
Step-by-step explanation:
Let O be the centre of the circle. Since P and S are the points of contact of tangents AD and DC respectively, OP ⊥ AD and OS ⊥ DC
Also, AD ⊥ DC (given), therefore, OPDS is a square.
BR = BQ = 27cm...(tangents from an external point to a circle are equal in length)
therefore, CR = CB - BR = (38 - 27)cm = 11cm
SC = CR = 11 cm...(tangents from an external point) therefore, DS = DC - SC = (25 - 11) cm = 14 cm.
therefore,
Radius of a circle = OP = DS = 14 cm...(therefore, OPDS is a square).
<h3>Hope it helps you!! </h3>
Answer:
61
Step-by-step explanation:
just add 3 to the next one
Answer:
- time: t = -0.3
- minimum: v = 0.55
Step-by-step explanation:
For quadratic ax^2 + bx + c, the extreme value is found at x=-b/(2a). For your quadratic, the minimum is found at ...
t = -(3)/(2(5))
t = -0.3 . . . . . time of minimum velocity
__
The value of velocity at that time is ...
v = 5(-0.3)^2 +3(-0.3) +1 = 5(.09) -.9 +1
v = 0.55 . . . . . value of minimum velocity
The number is 8. 8x6=48, 8x5=40, 48-40=8.
