Answer:
I love algebra anyways
The ans is in the picture with the steps how i got it
(hope this helps can i plz have brainlist :D hehe)
Step-by-step explanation:
As the independant variable increases, the dependant variable increases at a constant rate creating a linear relationship.
Answer:
Step-by-step explanation:
Given: quadrilateral ABCD inscribed in a circle
To Prove:
1. ∠A and ∠C are supplementary.
2. ∠B and ∠D are supplementary.
Construction : Join AC and BD.
Proof: As, angle in same segment of circle are equal.Considering AB, BC, CD and DA as Segments, which are inside the circle,
∠1=∠2-----(1)
∠3=∠4-----(2)
∠5=∠6-------(3)
∠7=∠8------(4)
Also, sum of angles of quadrilateral is 360°.
⇒∠A+∠B+∠C+∠D=360°
→→∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°→→→using 1,2,3,and 4
→→→2∠1+2∠4+2∠6+2∠8=360°
→→→→2( ∠1 +∠6) +2(∠4+∠8)=360°⇒Dividing both sides by 2,
→→→∠B + ∠D=180°as, ∠1 +∠6=∠B , ∠4+∠8=∠B------(A)
As, ∠A+∠B+∠C+∠D=360°
∠A+∠C+180°=360°
∠A+∠C=360°-180°------Using A
∠A+∠C=180°
Hence proved.
credit: someone else
GJ is a tangent
АG is a radius
A tangent line to a circle is perpendicular to the radius drawn to the tangent point ⇒ m∠АGJ = 90° ⇒ ΔАGJ is the right triangle.
<span>By the Pythagorean theorem:
</span>AJ² = АG² + GJ²
AJ² = 9² + 12² = 81 + 144 = 225
AJ = √225 = 15 units.
Answer:
56 + 53pi
Step-by-step explanation:
<u><em>Area of small circles:</em></u>
diameter of small circle: 4cm
forumla to find area of circle: A = pir^2
r is radius = half of diameter -> d/2 = 4 / 2 = 2cm
A = pi (2cm)^2
A = pi (4cm)
A = 4pi
<u><em>Area of large circle:</em></u>
diameter of small circle: 4cm
forumla to find area of circle: A = pir^2
r is radius = half of diameter -> d/2 = 14 / 2 = 7cm
A = pi (7cm)^2
A = pi (49cm)
A = 49pi
<u><em>Area of rectangle:</em></u>
Area = width x length
Area = 14cm x 4cm
Area = 56cm
<u><em>Add all three areas:</em></u>
Area of rectangle + large circle + small circle
56cm + 49pi + 4pi = 56cm + 53pi