Answer:
Let P be the external point. O be the origin. join O and P get OP and nearest point on the circle from P be A.
Let Q be the point onthe circle in which, tangent make 90° with radius at Q.
PQ = 8 and OQ = 6
we get a right angled triangle PQO right angled at Q.
so, OP^2 = OQ^2 + PQ^2= 8^2 + 6^2 = 64 + 36 =1==
therefore OP =10cm
we need nearest point from P, which is PA
PA = OP - OA= 10 -6=4cm
Answer:
Time to land on 11ft is 4.25 seconds
Step-by-step explanation:
Given
Required
Solve for t when height is 11ft
This implies that 
So: substitute 11 for y in
Collect Like Terms
Express both sides as squares
Take square roots of both sides (ignore negative)
Solve for t
Hence, time to land on 11ft is 4.25 seconds
Answer:
a. 15/23
b. 13/27
c. 400g
Step-by-step explanation:
a. When the denominators are the same, you can just sum the numerators.
Which becomes, 13+2=15--> 15/23
b. Same, when the denominator is the same, you can just minus the numerators. Which becomes, 25-12=13--> 13/27
c. 1kg=1000g. 1000/5=200✖️2=400
Answer:
56/125
Step-by-step explanation:
