The answer is (x - 1)(3x + 5)
Let's substitute (x + 2) with a:
3(x + 2)² - 10(x + 2) + 3 = 3a² - 10a + 3 =
= 3a² - 9a - a + 3 =
= 3a * a - 3a *3 - (a - 3) =
= 3a(a - 3) - (a - 3) =
= (a - 3)(3a - 1)
Now, substitute a with x + 2:
(a - 3)(3a - 1) = (x + 2 - 3)(3(x + 2) - 1) =
= (x - 1)(3x + 6 - 1) =
= (x - 1)(3x + 5)
Option B:
![-4.9t^2+20t+1.2=0](https://tex.z-dn.net/?f=-4.9t%5E2%2B20t%2B1.2%3D0)
General formula for the height(feet) of a projectile over time:
![h(t)=-16t^2+v_0t+h_0](https://tex.z-dn.net/?f=h%28t%29%3D-16t%5E2%2Bv_0t%2Bh_0)
where h represents height (feet)
t represents time
represents initial velocity
represents initial height(feet)
General formula for the height(metres) of a projectile over time:
![h(t)=-4.9t^2+v_0t+h_0](https://tex.z-dn.net/?f=h%28t%29%3D-4.9t%5E2%2Bv_0t%2Bh_0)
where h represents height (feet)
t represents time
represents initial velocity
represents initial height(metres)
<u>Given data in the question:</u>
metre
m/s
To determine the equation to find time:
Substitute
and
in general formula for the height(metres) of a projectile over time:
![h(t)=-4.9t^2+20t+1.2](https://tex.z-dn.net/?f=h%28t%29%3D-4.9t%5E2%2B20t%2B1.2)
![\Rightarrow -4.9t^2+20t+1.2=0](https://tex.z-dn.net/?f=%5CRightarrow%20-4.9t%5E2%2B20t%2B1.2%3D0)
Therefore option B is the correct answer.
![-4.9t^2+20t+1.2=0](https://tex.z-dn.net/?f=-4.9t%5E2%2B20t%2B1.2%3D0)
dam what Answer:
Step-by-step explanation:dom
no it’s not working