The <em>echo</em> number 20222022202220222022 is the <em>perfect</em> square of 4496890281.
<h3>What echo number is a perfect square</h3>
An <em>echo</em> number has a <em>perfect</em> square if its square root is also a <em>natural</em> number. After some iterations we found that <em>echo</em> number 20222022202220222022 is a <em>perfect</em> square:
![\sqrt{20222022202220222022} = 4496890281](https://tex.z-dn.net/?f=%5Csqrt%7B20222022202220222022%7D%20%3D%204496890281)
The <em>echo</em> number 20222022202220222022 is the <em>perfect</em> square of 4496890281. ![\blacksquare](https://tex.z-dn.net/?f=%5Cblacksquare)
To learn more on natural numbers, we kindly invite to check this verified question: brainly.com/question/17429689
Use PEMDAS and you should get 225
The answer is 3/2
Explanation: the formula is y-y /x-x
So it’ll be 2-(-4) /3-(-1) which will give 6 /4 and that simplifies will give 2/ 3
Answer:
![n^3+6n^2+8n=3315](https://tex.z-dn.net/?f=n%5E3%2B6n%5E2%2B8n%3D3315)
Step-by-step explanation:
Let n be the first odd number.
The 2nd consecutive odd number would be
and 3rd consecutive odd number would be
.
We have been given that the product of 3 consecutive odd numbers is 3,315. The product of 3 consecutive odd numbers would be
.
Now we will equate the product with 3315 as:
Let us simplify the left side of equation.
![(n\cdot n+n\cdot 2)(n+4)=3315](https://tex.z-dn.net/?f=%28n%5Ccdot%20n%2Bn%5Ccdot%202%29%28n%2B4%29%3D3315)
![(n^2+2n)(n+4)=3315](https://tex.z-dn.net/?f=%28n%5E2%2B2n%29%28n%2B4%29%3D3315)
Now, we will apply FOIL to find the product of left side as:
![n^2\cdot n+n^2\cdot 4+2n\cdot n+2n\cdot4=3315](https://tex.z-dn.net/?f=n%5E2%5Ccdot%20n%2Bn%5E2%5Ccdot%204%2B2n%5Ccdot%20n%2B2n%5Ccdot4%3D3315)
![n^3+4n^2+2n^2+8n=3315](https://tex.z-dn.net/?f=n%5E3%2B4n%5E2%2B2n%5E2%2B8n%3D3315)
![n^3+6n^2+8n=3315](https://tex.z-dn.net/?f=n%5E3%2B6n%5E2%2B8n%3D3315)
Therefore, our required equation would be
.
Answer:
<h3> A. the graph of a quadratic function y = (1/4)(x+3)(x-4)</h3>
Step-by-step explanation:
(1/4)(x+3)(x-4) = 0
x+3 = 0 or x-4 = 0
x = -3 or x = 4
For the others:
B. -(1/4)(x-3)(x+4) = 0
x-3 = 0 or x+4 = 0
x = 3 or x = -4
C. -(x+3)(x+4) = 0
x+3 = 0 or x+4 = 0
x = -3 or x = -4
D. (x-3)(x-4) = 0
x-3 = 0 or x-4 = 0
x = 3 or x = 4