Answer:
<h2>
X=1</h2><h2>
y=2</h2>
please see the attached picture for full solution..
hope this helps...
Good luck on your assignment..
The 'perfect square' method is basically saying that see if the number inside the square root is a factor of a perfect square (4, 9, 16, 25, etc). But, square root of 22 is not a factor of one of these, so you just have to approximate this. We known that square root of 25 is 5, and 16 is 4. This has to be between 4 and 5 and a little bit more to the 5 side. So, we can approximately say that:
![\sqrt{22} = 4.71](https://tex.z-dn.net/?f=%20%5Csqrt%7B22%7D%20%3D%204.71%20)
Hope this helps!
Answer:
(7, -27)
Step-by-step explanation:
Use the midpoint formula.
M = ( x₂ + x₁ , y₂ + y₁) / 2
(5, -11) = (x₂ + 3 , y₂ + 5) / 2 Substitute the midpoint and point U.
(10 , -22) = (x₂ + 3 , y₂ + 5) Multiply by 2 on both sides
x₂ + 3 = 10 y₂ + 5 = -22 Split for each point and solve.
x₂ = 10 - 3 y₂ = -22 - 5
x₂ = 7 y₂ = -27
Point V is (7, -27).
Answer:
B
Step-by-step explanation:
Given the function ![h(t)=(t+3)^2+5.](https://tex.z-dn.net/?f=h%28t%29%3D%28t%2B3%29%5E2%2B5.)
Find the average rate of change on interval ![[t_1,t_2]:](https://tex.z-dn.net/?f=%5Bt_1%2Ct_2%5D%3A)
![\dfrac{h(t_2)-h(t_1)}{t_2-t_1}=\\ \\=\dfrac{((t_2+3)^2+5)-((t_1+3)^2+5)}{t_2-t_1}=\\ \\=\dfrac{(t_2^2+6t_2+9+5)-(t_1^2+6t_1+9+5)}{t_2-t_1}=\\ \\=\dfrac{t_2^2-t_1^2+6t_2-6t_1}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1)+6(t_2-t_1)}{t_2-t_1}=\\ \\=\dfrac{(t_2-t_1)(t_2+t_1+6)}{t_2-t_1}=\\ \\=t_2+t_1+6](https://tex.z-dn.net/?f=%5Cdfrac%7Bh%28t_2%29-h%28t_1%29%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%28%28t_2%2B3%29%5E2%2B5%29-%28%28t_1%2B3%29%5E2%2B5%29%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%28t_2%5E2%2B6t_2%2B9%2B5%29-%28t_1%5E2%2B6t_1%2B9%2B5%29%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7Bt_2%5E2-t_1%5E2%2B6t_2-6t_1%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%28t_2-t_1%29%28t_2%2Bt_1%29%2B6%28t_2-t_1%29%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%28t_2-t_1%29%28t_2%2Bt_1%2B6%29%7D%7Bt_2-t_1%7D%3D%5C%5C%20%5C%5C%3Dt_2%2Bt_1%2B6)
This value is negative only for
so correct option is B