Answer:
- 5(x +1.5)^2
- 10(x +1)^2
- 1/4(x +2)^2
- 3(x +5/6)^2
Step-by-step explanation:
When your desired form is expanded, it becomes ...
a(x +b)^2 = a(x^2 +2bx +b^2) = ax^2 +2abx +ab^2
This tells you the overall factor (a) is the leading coefficient of the given trinomial. Factoring that out, you can find b as the root of the remaining constant.
a) 5x^2 +15x +11.25 = 5(x^2 +3x +2.25) = 5(x +1.5)^2
b) 10x^2 +20x +10 = 10(x^2 +2x +1) = 10(x +1)^2
c) 1/4x^2 +x +1 = 1/4(x^2 +4x +4) = 1/4(x +2)^2
d) 3x^2 +5x +25/12 = 3(x^2 +5/3x +25/36) = 3(x +5/6)^2
_____
<em>Additional comment</em>
If you know beforehand that the expressions can be factored this way, finding the two constants (a, b) is an almost trivial exercise. It gets trickier when you're trying to write a general expression in vertex form (this form with an added constant). For that, you must develop the value of b from the coefficient of the linear term inside parentheses.
Answer:
17
Step-by-step explanation:
3(5)+=17 plug 5 in for x then solve hope this helps
Answer:
1/4 - 3 2/5 - (2 1/3 - 1/4) =
-157/30=
-5 7/30
≅ -5.2333333
Step-by-step explanation:
Answer:6minutes
Step-by-step explanation:
3 hours =180min
180min divided by 30 windows is 6
Answer:
The interval [32.6 cm, 45.8 cm]
Step-by-step explanation:
According with the <em>68–95–99.7 rule for the Normal distribution:</em> If 
is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval
![\large [\bar x - s, \bar x +s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-%20s%2C%20%5Cbar%20x%20%2Bs%5D)
around 95% of the data must fall in the interval
![\large [\bar x -2s, \bar x +2s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-2s%2C%20%5Cbar%20x%20%2B2s%5D)
around 99.7% of the data must fall in the interval
![\large [\bar x -3s, \bar x +3s]](https://tex.z-dn.net/?f=%5Clarge%20%5B%5Cbar%20x%20-3s%2C%20%5Cbar%20x%20%2B3s%5D)
So, the range of lengths that covers almost all the data (99.7%) is the interval
[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]
<em>This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%</em>
