Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1 tells us that y '(0) = 1 y(0) = 1 y '(1) = 0 y(1) = 0 Given that the derivative value of y(t) is 3 when t = 2 tells us that y '(3) = 2 y '(0) = 2 y '(2) = 0 y '(2) = 3 (b) Find dy dt = kcos(bt2)·b2t (c) Find the exact values for k and b that satisfy the conditions in part (a). Note: Choose the smallest positive value of b that works.

we have the parent function given by,

1. the function is translated 5 units to the left.

so, we get,   becomes

2. the function is translated 5 units to the right.

so, we get,   becomes

3. the function is translated 5 units upwards.

so, we get,   becomes

4. the function is translated 5 units downwards.

so, we get,   becomes

5. the function is reflected across x-axis.

so, we get,   becomes

6. the function is reflected across y-axis.

so, we get,   becomes .

step-by-step explanation:

mark brainliest.

the arc length is found by finding the circumference of the full circle (2xpixradius) and then dividing by 4 to find just one quarter of the circle (90 degrees).

i believe it is the square root of 122 because when you turn it on it's side, to where the 5 inches is on the bottom, it looks like it is the same length.