9/16
7 of the squares are 4 so we just substract that from the total of 16 and we get 9
Answer:

Step-by-step explanation:
We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.
We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is
.
Upon substituting our given values, we will get:

Let us solve for r.


We know that area of an equilateral triangle is equal to
, where s represents side length of triangle.

The area within circle and outside the triangle would be difference of area of circle and triangle as:

We can make a common denominator as:


Therefore, our required expression would be
.
Answer:
![7\sqrt[11]{x^{5}}](https://tex.z-dn.net/?f=7%5Csqrt%5B11%5D%7Bx%5E%7B5%7D%7D)
Step-by-step explanation:
we know that
![a^{\frac{n}{m}}=\sqrt[m]{a^{n}}](https://tex.z-dn.net/?f=a%5E%7B%5Cfrac%7Bn%7D%7Bm%7D%7D%3D%5Csqrt%5Bm%5D%7Ba%5E%7Bn%7D%7D)
In this problem we have

therefore
![7x^{\frac{5}{11}}=7\sqrt[11]{x^{5}}](https://tex.z-dn.net/?f=7x%5E%7B%5Cfrac%7B5%7D%7B11%7D%7D%3D7%5Csqrt%5B11%5D%7Bx%5E%7B5%7D%7D)
Answer:
This is an experiment because a treatment was applied to a group.
Step-by-step explanation:
This is because it cannot be a survey since the children are taking a test, which has a hypothesis. Hope you get this right. I took the test and got right.