first, consider that 25 = 5^2, and -25 = -1*25 = -1*5^2.
Then, you can write:
![\sqrt[]{-25}=\sqrt[]{(-1)(25)}=\sqrt[]{(-1)(5^2)}=5\sqrt[]{-1}=5i](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B-25%7D%3D%5Csqrt%5B%5D%7B%28-1%29%2825%29%7D%3D%5Csqrt%5B%5D%7B%28-1%29%285%5E2%29%7D%3D5%5Csqrt%5B%5D%7B-1%7D%3D5i)
because i = √-1.
Hence, the answer is 5i
Answer:
figure it out
Step-by-step explanation:
u have to take y2-y1 and x2-x1
Answer:
0
Step-by-step explanation:
m=(y2-y1)/x2-x1), if m is slope and the 2 points are (x1, y1), (x2, y2).
m=(7-7)/(2-5)
m=0/-3
m=0
also, you don't need to know this but if a line has a slope of zero it's a horizontal line
Answer:
a.) 1.38 seconds
b.) 17.59ft
Step-by-step explanation:
h(t) = -16t^2 + 22.08t + 6
if we were to graph this, the vertex of the function would be the point, which if we substituted into the function would give us the maximum height.
to find the vertex, since we are dealing with something called "quadratic form" ax^2+bx+c, we can use a formula to find the vertex
-b/2a
b=22.08
a=-16
-22.08/-16, we get 1.38 when the minuses cancel out. since our x is time, it will be 1.38 seconds
now since the vertex is 1.38, we can substitute 1.38 into the function to find the maximum height.
h(1.38)= -16(1.38)^2 + 22.08t + 6 -----> is maximum height.
approximately = 17.59ft -------> calculator used, and rounded to 2 significant figures.
for c the time can be equal to (69+sqrt(8511))/100, as the negative version would be incompatible since we are talking about time. or if you wanted a rounded decimal, approx 1.62 seconds.
