Answer:
Step-by-step explanation:
Given that,
The arc length is four times the radius
Let he radius be 'r'
Then, the arc length be 's'
The arc of a sector can be calculated using
s=θ/360 × 2πr
Then, given that s=4r
So, 4r = θ × 2πr / 360
Divide both side r
4 = θ × 2π/360
Then, make θ subject of formula
θ × 2π = 360 × 4
θ = 360 × 4 / 2π
θ = 720 / π
So, area of the sector can be determine using
A = θ / 360 × πr²
Since r = ¼s
Then,
A = (θ/360) × π × (¼s)²
A = (θ/360) × π × (s²/16)
A = θ × π × s² / 360 × 16
Since θ = 720 / π
A = (720/π) × π × s² / 360 × 16
A = 720 × π × s² / 360 × 16 × π
A = s² / 8
Then,
s² = 8A
Then,
s= √(8A)
s = 2 √2•A
Answer:
Isosceles triangle
Step-by-step explanation:
There are three different kinds of triangle: equilateral, isosceles, and scalene.
-Equilateral triangles are triangles where all 3 sides and angles are the same. (First image)
-Isosceles triangles are triangles where 2 of the sides and angles are the same, while the third side and angle is different. (Second image)
-Scalene triangles are triangles where all 3 sides and angles are different. (Third image)
In the case of this triangle, two of the sides have the same 52 degree angle. They would also have the same side lengths, and <u>thus be a isosceles triangle. </u>
Hope this helps! :3
![~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chspace%7B10em%7D%5Ctextit%7Bfunction%20transformations%7D%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%20A%28%20Bx%2B%20C%29%5E2%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%5Csqrt%7B%20Bx%2B%20C%7D%2B%20D%20%5C%5C%5C%5C%20f%28x%29%3D%20A%28%5Cmathbb%7BR%7D%29%5E%7B%20Bx%2B%20C%7D%2B%20D%20%5Cend%7Barray%7D%5Cqquad%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20f%28x%29%3D%5Ccfrac%7B1%7D%7BA%28Bx%2BC%29%7D%2BD%20%5C%5C%5C%5C%5C%5C%20f%28x%29%3D%20A%20sin%5Cleft%28%20B%20x%2B%20C%20%5Cright%29%2B%20D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)


keeping in mind that template, let's take a looksie
