This corresponds to Problem 1.21c - f Consider the function f(t) = sin(215t). (c) For what value(s) of t is f(t) = 1? If there are more than one values, give the three values that are closest to t = 0, as a comma separated list. (d) For what value(s) of t is f(t) = 0? If there are more than one values, give the three values that are closest to t = 0, as a comma separated list. (e) (sin(210 Ft)) = (f) / sin(21Ft) = (15 points) This problem corresponds to Problem 2.27 in the text. a) The value of | 8(t)[3e-41 + 1 sin(2110°t)]dt is b) The value of S(t – 6)e-2.5tu(t)dt is J -00 c) The value of 8(t – 1)e-5' [u(t) – u(t + 4)]dt is (20 points) What is the fundamental period of the function s(t) = cos(215000t) + cos(2119000t) in seconds (if t is measures in seconds)?

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

where's the diagram?

step-by-step explanation:

there would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers.