TheThe area of the shaded region if the radius of the outer circle is 4 and the radius of the inner circle is 2 is 12π.
<h3>Area of the shaded region</h3>
Area of a circle = πr²
r =radius of the circle
Area of the outer circle:
Area of the outer circle = π(4)²
Area of the outer circle = 16π
Area of the inner circle:
Area of the inner circle = π (2)²
Area of the inner circle = 4 π
Area of the shaded Region :
Area of the shaded Region = 16π - 4 π
Area of the shaded Region = 12π
Therefore the area of the shaded region if the radius of the outer circle is 4 and the radius of the inner circle is 2 is 12π.
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Answer:
Step-by-step explanation:
Since this is a system with two equations, we can easily use a graphing calculator to find the solution
The intersection between the graphs of the equations
y = 3x + 10
2y = 6x – 4 (if we divide by two)
y = 3x - 2
However, we can see that the graphs have the same slope m = 3
Since both equations are not equal, there is no solution for this system
Please see attached graph. The lines are parallel and they never touch each other.
Answer:
-x + 1
Step-by-step explanation:
-10x^2 - x + 6 + 10x^2 - 5 = -x + 1
The 10x^2 cancels out with the -10x^2
Answer:
yes
yes
no
yes
yes
no
Step-by-step explanation:
<h3>Answer to Question 1:</h3>
AB= 24cm
BC = 7cm
<B = 90°
AC = ?
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
AC^2 = 24^2 + 7^2
AC^2 = 576 + 49
AC^2 = √625
AC = 25
<h3>Answer to Question 2 :-</h3>
sin A = 3/4
CosA = ?
TanA = ?
<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>
(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
4² = AB² + 3²
16 = AB + 9
AB = √7cm
<h3>CosA = Adjacent side/Hypotenuse</h3>
= AB/AC
= √7/4
<h3>TanA= Opp. side/Adjacent side</h3>
=BC/AB
= 3/√7
