Answer:
Options (1), (3), (5) and (6)
Step-by-step explanation:
Option (1)
Both the circles M and H are same in shape. Therefore, both are similar.
True
Option (2)
MH = MS = [Radii of a circle]
MH = MO + OH
MH = ME + OH [Since, MO = ME]
MH = 10 + 25 = 35 cm
Area of circle M = πr²
= π(35)²
= 1225π
Area of circle H = π(25)²
= 625π

Area of M = 1.96(Area of H)
False
Option 3
OH + MS = 25 + 35
= 60 cm
True
Option 4
m∠HMS = 90°
Can't be figured out with the given informations
False
Option 5
Diameter of circle H = 2(radius)
= 2(25)
= 50 cm
True
Option 6
Circumference of circle S = 2πr
= 2π(ES)
Circumference of circle M = 2π(MS)
Ratio of the circumference of circle S and M = 
= 
True.
Options (1), (3), (5) and (6) are correct.
Answer:

Step-by-step explanation:

<em>hope</em><em> </em><em>this</em><em> </em><em>helps</em>
Answer:
( √15 + 8)/7
Step-by-step explanation:
TanA = -√15
.we are to find tan(A-π/4).
In trigonometry
Tan(A-B) = TanA - TanB/1+ tanAtanB
Hence:
tan(A-π/4) = TanA - Tanπ/4/1+ tanAtanπ/4
Substitute tan A value into the formula
tan(A-π/4) = -√15-tanπ/4 / 1+(-√15)(tanπ/4
tan(A-π/4) = -√15-1/1-√15
Rationalize
-√15-1/1-√15 × 1+√15/1+√15
= -√15-√225-1-√15/(1-√225)
= -2√15-15-1/1-15
= -2√15 -16/(-14)
= -2(√15+8)/-14
= √15 + 8/7
Hence the required value is ( √15 + 8)/7
The expression y = 660(0.902)ˣ represents a decay , the rate of decrease is 9.8% .
In the question ,
it is given that ,
the the exponential function is y = 660(0.902)ˣ
we need to determine that , the equation represents a growth or decay .
0.902 is the variable that will determine a growth or decay .
Since, 0.902 is less than 1 , so it represents a decay function .
To find the rate we subtract 0.902 from 1
= 1 - 0.902
= 0.098
= 9.8% decrease rate
Therefore , The expression y = 660(0.902)ˣ represents a decay , the rate of decrease is 9.8% .
The given question is incomplete , the complete question is
Identify if it's Growth or Decay , and determine the percentage rate of increase or decrease , the function is y = 660(0.902)ˣ ?
Learn more about Growth and Decay here
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