Answer:
Multiply each term by 4 and simplify.
Inequality Form:
c
<
16
Interval Notation:
(
−
∞
,
16
)
Step-by-step explanation:
We want to factor the following expression: (x 3)^2 14(x 3) 49(x 3) 2 14(x 3) 49(, x, plus, 3, ), squared, plus, 14, (, x, plus,
Nesterboy [21]
Answer:


Step-by-step explanation:
Given

Required
Find U and V
We have:

Expand

Collect like terms


Expand

Group
![[x^2 + 10x] + [10x + 100] = (U + V)^2](https://tex.z-dn.net/?f=%5Bx%5E2%20%2B%2010x%5D%20%2B%20%5B10x%20%2B%20100%5D%20%3D%20%28U%20%2B%20V%29%5E2)
Factorize each group
![x[x + 10] + 10[x + 10] = (U + V)^2](https://tex.z-dn.net/?f=x%5Bx%20%2B%2010%5D%20%2B%2010%5Bx%20%2B%2010%5D%20%3D%20%28U%20%2B%20V%29%5E2)
Factor out x + 10
![[x + 10][x + 10] = (U + V)^2](https://tex.z-dn.net/?f=%5Bx%20%2B%2010%5D%5Bx%20%2B%2010%5D%20%3D%20%28U%20%2B%20V%29%5E2)
So, we have:
![[x + 10]^2 = (U + V)^2](https://tex.z-dn.net/?f=%5Bx%20%2B%2010%5D%5E2%20%3D%20%28U%20%2B%20V%29%5E2)
By comparison


Answer:
The probability would be a 4/6 or 66.67%
Step-by-step explanation:
All of the even numbers (and five) on a six sided die are 2, 4, 5, and 6.
That is four total numbers.
Also, since there are <em>six sides</em> on the cube, you would have a 4/6 chance or 66.67% chance of rolling a 2, 4, 5, or a 6.
Hope this helps!!!!!!
No, the x in 12x2 and 15x could still be factored.