To factor the trinomial, you can use the ac method where you multiply the first and last terms together (6 and 9) and add up to 15.
In this case ac= 54
2 numbers that multiply to 54 and add up to 15 would be 6 and 9 so place these numbers where the 15 would be:
![6p^{2} +9p + 6p+9](https://tex.z-dn.net/?f=%206p%5E%7B2%7D%20%2B9p%20%2B%206p%2B9)
Then group terms to simplify:
![(6p^{2} +9p)(6p+9) ](https://tex.z-dn.net/?f=%20%286p%5E%7B2%7D%20%2B9p%29%286p%2B9%29%0A)
Simplify any like terms:
![3p(2p+3) 3(2p+3)](https://tex.z-dn.net/?f=%203p%282p%2B3%29%203%282p%2B3%29)
And group together again:
![(3p+3)(2p+3)](https://tex.z-dn.net/?f=%283p%2B3%29%282p%2B3%29)
You can simplify a 3 out of the first expression one more time to get:
![3(p+1)(2p+3)](https://tex.z-dn.net/?f=3%28p%2B1%29%282p%2B3%29)
Which would be the factored form. I hope this helps :)
Answer:
The answer would be 11 1/3 ft sqared
Let
![x=\arctan y](https://tex.z-dn.net/?f=x%3D%5Carctan%20y)
. Then
![\tan x=y](https://tex.z-dn.net/?f=%5Ctan%20x%3Dy)
, and so as
![x\to\dfrac\pi2^-](https://tex.z-dn.net/?f=x%5Cto%5Cdfrac%5Cpi2%5E-)
, you have
![y\to+\infty](https://tex.z-dn.net/?f=y%5Cto%2B%5Cinfty)
. The limit is then equivalent to
13 units.
Since both are negative, and they’re just asking for distance between, you can just subtract 12 from 25 to get 13.
Answer:
The equation does not have a real root in the interval ![\rm [0,1]](https://tex.z-dn.net/?f=%5Crm%20%5B0%2C1%5D)
Step-by-step explanation:
We can make use of the intermediate value theorem.
The theorem states that if
is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:
- If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
- The image of a continuous function over an interval is itself an interval.
Of course, in our case, we will make use of the first one.
First, we need to proof that our function is continues in
, which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval
, which means to evaluate the equation in 0 and 1:
![f(x)=x^3-3x+8\\f(0)=8\\f(1)=6](https://tex.z-dn.net/?f=f%28x%29%3Dx%5E3-3x%2B8%5C%5Cf%280%29%3D8%5C%5Cf%281%29%3D6)
Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval
. I attached a plot of the equation in the interval
where you can clearly observe how the graph does not cross the x-axis in the interval.