![x^ \frac{m}{n}= \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7Bm%7D%7Bn%7D%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%20%20)
pemdas, so exponent first before multiply
4(x^1/2)=4x^2
this is different from
(4x)^1/2
so
![x^ \frac{1}{2}= \sqrt[2]{x^1}](https://tex.z-dn.net/?f=x%5E%20%5Cfrac%7B1%7D%7B2%7D%3D%20%5Csqrt%5B2%5D%7Bx%5E1%7D%20%20)
times that by 4
4√x
Answer:
34
Step-by-step explanation:
Answer:
y = 3x + 2
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Here m = 3 and c = 2 , then
y = 3x + 2 ← equation of line
B. Up
Because a positive ‘a’ value will make the parabola open upwards, and because the power is 2 (even) both sides will open up in a ‘U’ shape
9514 1404 393
Answer:
- 3n+6 (n = smallest)
- 3n (n = middle)
Step-by-step explanation:
The usual method for doing this is to let n represent the smallest one. Then the three integers are ...
n, n+2, and n+4
and their sum is ...
(n) +(n+2) +(n+4) = 3n+6 . . . . sum of 3 consecutive odd integers (n = smallest)
_____
Personally, for consecutive number problems, I prefer to let the variable represent the average value. If n is the average value of 3 consecutive odd integers, is is the middle integer. Of course, the sum will be 3 times the average:
(n-2) +(n) +(n+2) = 3n . . . . sum of 3 consecutive odd integers (n = middle one)
