<h3>Given :-</h3>
<h3>To Find :-</h3>
<h3>Solution :-</h3>
☼︎ <u>Radius of the Circle</u>;
<h3>Using Formula:</h3>

<u>Putting values in the formula;</u>









henceforth, the Radius of a circle is 14 cm ...!!
Hey there!
We can solve this problem by doing PEMDAS, which lists:

We have in this equation

Let's sp;ve this equation now!
Parentheses ↓

Exponents ↓

Division ↓

Okay now , I show you the steps so it can be easier to solve!

Good luck on your assignment and enjoy your day!
~
Answer:
14.36 AND 9.89 ===> 14 or 10
Step-by-step explanation:
Y = Ax2 Bx C
Enter coefficients here >>> -4 97 -568
Standard Form: y = -4x²+97x-568
-24.25 -12.125 147.015625 -588.0625 20.0625
Grouped Form: No valid Grouping
Graphing Form: y = -4(x-12.13)²+20.06
Factored Form: PRIME
Solution/X-Intercepts: 14.36 AND 9.89
Discriminate =321 is positive, two real solutions
VERTEX: (12.13,20.06) Directrix: Y=20.13
Answer:

Step-by-step explanation:
We need to integrate the given expression. Let I be the answer .
- Let u = 3x , then du = 3dx . Henceforth 1/3 du = dx .
- Now , Rewrite using du and u .
