Answer:
The area of circle with radius half of original circle is 7 π cm² .
Step-by-step explanation:
Given as :
The Area of original circle = 28 π square centimeter
Let The radius of original circle = R
Let The area of circle with radius half of original circle = A square centimeters
Let The radius of circle with radius half of original circle = R'
Now, According to question
∵ Area of original circle = π × radius²
So, 28 π cm² = π × R²
Or, R² = 
Or, R² = 28 cm²
∴ R =
cm
i.e R = 2
cm
So, The radius of original circle = R = 2
cm
<u>Again</u>
∵ The radius of circle with radius half of original circle = R'
So, R' = 
i.e R' = 
∴ R' =
cm
So, The area of circle with radius half of original circle = π × R'²
i.e A = π × R'²
Or, A = π ×(
) ²
or, A = π × 7
So, The area of circle with radius half of original circle = 7 π cm²
Hence The area of circle with radius half of original circle is 7 π cm² . Answer
the question is confusing but the ? angle is 90 degrees don't know if that helps at all though sorry
Answer:
infinite solutions
Step-by-step explanation:
the lines are coincident, so there are infinite solutions
Answer:
sqrt(2)/2
Step-by-step explanation:
Given tan(x)=2-cot(x), find sin(x).
Rewrite in terms of sine and cosine:
sin(x)/cos(x)=2-cos(x)/sin(x)
Multiply both sides by cos(x)sin(x):
sin^2(x)=2sin(x)cos(x)-cos^2(x)
Rewrite cos^2(x) using the identity sin^2(x)+cos^2(x)=1:
sin^2(x)=2sin(x)cos(x)-(1-sin^2(x))
Distribute:
sin^2(x)=2sin(x)cos(x)-1+sin^2(x)
Subtracting sin^2(x) on both sides:
0=2sin(x)cos(x)-1
Add 1 on both sides:
1=2sin(x)cos(x)
Use identity sin(2x)=2sin(x)cos(x) to rewrite right:
1=sin(2x)
Since sin(pi/2)=1, then 2x=pi/2.
Dividing both sides by 2 gives x=pi/4.
So sin(pi/4)=sqrt(2)/2