For question one, the easiest way to start is with t=0, so, everything inside the sine function will equal 0, and sin(0) is 0, so that is our starting point. Then, consider the following:
sin(pi/2) = 1
sin(pi) = 0
sin(3pi/2) = -1
sin(2pi) = 0
(You need to know the unit circle)
Then, we see that t=4 is a solution since the 4 times a half is 2, exactly what we need for 2 (pi). Also because it goes around the unit circle.
Problem 2 you just need the period formula. It is [2pi / b] and b is the number in front of t. So we get (2pi / (2pi / 15)). Simplifying leads to 30pi over 2pi which is just 15. Thus the period is 15 units
Lastly, Problem 3 asks for the inverse of a period since the frequency is 1 over the period. Applying the method from the last problem:
2pi / 524pi which simplifies to 1 over 262 is the period.
Now flipping 1 over 262 gives the frequency, which is just 262 by itself
Hope this long explanation helped
You divided in the wrong order. The resulting image always goes first then the preimage is afterward.
scale factor = (new)/(old) = 10/8 = 1.25
<h3>Answer is choice B</h3>
The scale factor is larger than 1, so this means we have an enlargement going on.
The effect of doubling the length and width have on the perimeter would be 2x bigger
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Based on the situation above the inequality will he use to contradict the assumption is <span>
4:10 ≠ 6:14</span>
if DE is parallel to BC
then
4: (4+5) = 6 : (6 + 8)