You would:
(4 * 2 - 4) + (3 - 2^2) + (2 * 2^3)
(8 - 4) + (3 - 4) + (16)
4 + -1 + 16 = 19
A function
![f(t)](https://tex.z-dn.net/?f=f%28t%29)
is periodic if there is some constant
![k](https://tex.z-dn.net/?f=k)
such that
![f(t+k)=f(k)](https://tex.z-dn.net/?f=f%28t%2Bk%29%3Df%28k%29)
for all
![t](https://tex.z-dn.net/?f=t)
in the domain of
![f(t)](https://tex.z-dn.net/?f=f%28t%29)
. Then
![k](https://tex.z-dn.net/?f=k)
is the "period" of
![f(t)](https://tex.z-dn.net/?f=f%28t%29)
.
Example:
If
![f(x)=\sin x](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%20x)
, then we have
![\sin(x+2\pi)=\sin x\cos2\pi+\cos x\sin2\pi=\sin x](https://tex.z-dn.net/?f=%5Csin%28x%2B2%5Cpi%29%3D%5Csin%20x%5Ccos2%5Cpi%2B%5Ccos%20x%5Csin2%5Cpi%3D%5Csin%20x)
, and so
![\sin x](https://tex.z-dn.net/?f=%5Csin%20x)
is periodic with period
![2\pi](https://tex.z-dn.net/?f=2%5Cpi)
.
It gets a bit more complicated for a function like yours. We're looking for
![k](https://tex.z-dn.net/?f=k)
such that
![\pi\sin\left(\dfrac\pi2(t+k)\right)+1.8\cos\left(\dfrac{7\pi}5(t+k)\right)=\pi\sin\dfrac{\pi t}2+1.8\cos\dfrac{7\pi t}5](https://tex.z-dn.net/?f=%5Cpi%5Csin%5Cleft%28%5Cdfrac%5Cpi2%28t%2Bk%29%5Cright%29%2B1.8%5Ccos%5Cleft%28%5Cdfrac%7B7%5Cpi%7D5%28t%2Bk%29%5Cright%29%3D%5Cpi%5Csin%5Cdfrac%7B%5Cpi%20t%7D2%2B1.8%5Ccos%5Cdfrac%7B7%5Cpi%20t%7D5)
Expanding on the left, you have
![\pi\sin\dfrac{\pi t}2\cos\dfrac{k\pi}2+\pi\cos\dfrac{\pi t}2\sin\dfrac{k\pi}2](https://tex.z-dn.net/?f=%5Cpi%5Csin%5Cdfrac%7B%5Cpi%20t%7D2%5Ccos%5Cdfrac%7Bk%5Cpi%7D2%2B%5Cpi%5Ccos%5Cdfrac%7B%5Cpi%20t%7D2%5Csin%5Cdfrac%7Bk%5Cpi%7D2)
and
![1.8\cos\dfrac{7\pi t}5\cos\dfrac{7k\pi}5-1.8\sin\dfrac{7\pi t}5\sin\dfrac{7k\pi}5](https://tex.z-dn.net/?f=1.8%5Ccos%5Cdfrac%7B7%5Cpi%20t%7D5%5Ccos%5Cdfrac%7B7k%5Cpi%7D5-1.8%5Csin%5Cdfrac%7B7%5Cpi%20t%7D5%5Csin%5Cdfrac%7B7k%5Cpi%7D5)
It follows that the following must be satisfied:
![\begin{cases}\cos\dfrac{k\pi}2=1\\\\\sin\dfrac{k\pi}2=0\\\\\cos\dfrac{7k\pi}5=1\\\\\sin\dfrac{7k\pi}5=0\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%5Ccos%5Cdfrac%7Bk%5Cpi%7D2%3D1%5C%5C%5C%5C%5Csin%5Cdfrac%7Bk%5Cpi%7D2%3D0%5C%5C%5C%5C%5Ccos%5Cdfrac%7B7k%5Cpi%7D5%3D1%5C%5C%5C%5C%5Csin%5Cdfrac%7B7k%5Cpi%7D5%3D0%5Cend%7Bcases%7D)
The first two equations are satisfied whenever
![k\in\{0,\pm4,\pm8,\ldots\}](https://tex.z-dn.net/?f=k%5Cin%5C%7B0%2C%5Cpm4%2C%5Cpm8%2C%5Cldots%5C%7D)
, or more generally, when
![k=4n](https://tex.z-dn.net/?f=k%3D4n)
and
![n\in\mathbb Z](https://tex.z-dn.net/?f=n%5Cin%5Cmathbb%20Z)
(i.e. any multiple of 4).
The second two are satisfied whenever
![k\in\left\{0,\pm\dfrac{10}7,\pm\dfrac{20}7,\ldots\right\}](https://tex.z-dn.net/?f=k%5Cin%5Cleft%5C%7B0%2C%5Cpm%5Cdfrac%7B10%7D7%2C%5Cpm%5Cdfrac%7B20%7D7%2C%5Cldots%5Cright%5C%7D)
, and more generally when
![k=\dfrac{10n}7](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7B10n%7D7)
with
![n\in\mathbb Z](https://tex.z-dn.net/?f=n%5Cin%5Cmathbb%20Z)
(any multiple of 10/7).
It then follows that all four equations will be satisfied whenever the two sets above intersect. This happens when
![k](https://tex.z-dn.net/?f=k)
is any common multiple of 4 and 10/7. The least positive one would be 20, which means the period for your function is 20.
Let's verify:
![\sin\left(\dfrac\pi2(t+20)\right)=\sin\dfrac{\pi t}2\underbrace{\cos10\pi}_1+\cos\dfrac{\pi t}2\underbrace{\sin10\pi}_0=\sin\dfrac{\pi t}2](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac%5Cpi2%28t%2B20%29%5Cright%29%3D%5Csin%5Cdfrac%7B%5Cpi%20t%7D2%5Cunderbrace%7B%5Ccos10%5Cpi%7D_1%2B%5Ccos%5Cdfrac%7B%5Cpi%20t%7D2%5Cunderbrace%7B%5Csin10%5Cpi%7D_0%3D%5Csin%5Cdfrac%7B%5Cpi%20t%7D2)
![\cos\left(\dfrac{7\pi}5(t+20)\right)=\cos\dfrac{7\pi t}5\underbrace{\cos28\pi}_1-\sin\dfrac{7\pi t}5\underbrace{\sin28\pi}_0=\cos\dfrac{7\pi t}5](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac%7B7%5Cpi%7D5%28t%2B20%29%5Cright%29%3D%5Ccos%5Cdfrac%7B7%5Cpi%20t%7D5%5Cunderbrace%7B%5Ccos28%5Cpi%7D_1-%5Csin%5Cdfrac%7B7%5Cpi%20t%7D5%5Cunderbrace%7B%5Csin28%5Cpi%7D_0%3D%5Ccos%5Cdfrac%7B7%5Cpi%20t%7D5)
More generally, it can be shown that
![f(t)=\displaystyle\sum_{i=1}^n(a_i\sin(b_it)+c_i\cos(d_it))](https://tex.z-dn.net/?f=f%28t%29%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%28a_i%5Csin%28b_it%29%2Bc_i%5Ccos%28d_it%29%29)
is periodic with period
![\mbox{lcm}(b_1,\ldots,b_n,d_1,\ldots,d_n)](https://tex.z-dn.net/?f=%5Cmbox%7Blcm%7D%28b_1%2C%5Cldots%2Cb_n%2Cd_1%2C%5Cldots%2Cd_n%29)
.
<h3>
Answer: Choice D</h3>
=============================
Explanation:
The first figure has 4 sides (quadrilateral)
The second figure has 5 sides (pentagon)
The third figure has 6 sides (hexagon)
Each time we increase the number of sides by 1.
The fourth figure must have 7 sides (heptagon). The only thing that has 7 sides is the figure listed in choice D.
He probably didnt get diabetes
1cm³=1mL
1942cm=x mL
1cm³ 1942cm³
________ = ________
1mL x mL
Cross multiply
1x=1942
divide both sides by 1
x=1942mL
Answer=1942mL