1st number = n
2nd number = n+1
3rd number = n+2
sum of the squares of 3 consecutive numbers is 116
n² + (n+1)² + (n+2)² = 116
n² + (n+1)(n+1) + (n+2)(n+2) = 116
n² + [n(n+1)+1(n+1)] + [n(n+2)+2(n+2)] = 116
n² + n² + n + n + 1 + n² + 2n + 2n + 4 = 116
n² + n² + n² + n + n + 2n + 2n + 1 + 4 = 116
3n² + 6n + 5 = 116 Last option.
Answer:
It would be in this order:
Types of Quads:
Rectangle
Square
Rhombus
Trapezoid
Type of Triangles:
Equaliteral
Isoceles
Scalene
Step-by-step explanation:
Hope it helps
Answer:
Step-by-step explanation:
Here's how you convert:
![\sqrt[n]{x^m}=x^{\frac{m}{n}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Em%7D%3Dx%5E%7B%5Cfrac%7Bm%7D%7Bn%7D)
The little number outside the radical, called the index, serves as the denominator in the rational power, and the power on the x inside the radical serves as the numerator in the rational power on the x.
A couple of examples:
![\sqrt[3]{x^4}=x^{\frac{4}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E4%7D%3Dx%5E%7B%5Cfrac%7B4%7D%7B3%7D)
![\sqrt[5]{x^7}=x^{\frac{7}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E7%7D%3Dx%5E%7B%5Cfrac%7B7%7D%7B5%7D)
It's that simple. For your problem in particular:
![\sqrt[3]{7}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%7D)
is the exact same thing as ![\sqrt[3]{7^1}=7^{\frac{1}{3}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B7%5E1%7D%3D7%5E%7B%5Cfrac%7B1%7D%7B3%7D)
20/28 are female. Since there are 8 male students you just have to subtract 28-8 and you'll get the female total.
Answer: The graph is shifted 2 units to the right.
Step-by-step explanation:
Given a function f(x), we know that one transformation rule is:
If 
then the function is shifted "k" units to the right.
Therefore, for the function 
, when we subtract 2 from the input, then we get the function g(x) in the form:
We can conclude that subtracting 2 from the input of the function , then the graph is shifted 2 units to the right.
