Dr. Hoover allot his time on Tuesday for an annual checkup is 147 minutes and for a sick visit is 42 minutes (Total 189 minutes)
On Wednesday appointment, he allot his time for an annual checkup is 147 minutes and for a sick visit is 21 minutes (Total 168 minutes)
solution
Let us assume, minutes for annual checkup denoted as x and
minutes for sick visit denoted as y
The equation of Tuesday visit is ![3x + 2y = 189 minutes ------ 1](https://tex.z-dn.net/?f=3x%20%2B%202y%20%3D%20189%20minutes%20------%201)
The equation of Wednesday visit is ![3x + 1y = 168 minutes ------ 2](https://tex.z-dn.net/?f=3x%20%2B%201y%20%3D%20168%20minutes%20------%202)
by changing the signs of the equation 2 and subtract it from the equation 1 we will get 1y = 21 minutes
to substitute y =1 in the equation 2 we get
![3x + 2(21) = 189](https://tex.z-dn.net/?f=3x%20%2B%202%2821%29%20%3D%20189)
![3x + 42 = 189](https://tex.z-dn.net/?f=3x%20%2B%2042%20%3D%20189)
![3x = 189 - 42](https://tex.z-dn.net/?f=3x%20%3D%20189%20-%2042)
![3x = 147](https://tex.z-dn.net/?f=3x%20%3D%20147)
![x = 147 ÷ 3 = 49 minutes](https://tex.z-dn.net/?f=x%20%3D%20147%20%C3%B7%203%20%3D%2049%20minutes)
then now we have substitute both x = 49 and y= 21 in equation ----- 2
we will prove that
![147+ 42 = 189 minutes](https://tex.z-dn.net/?f=147%2B%2042%20%3D%20189%20minutes)
so the Tuesday appointment , the time allotted for an annual checkup is 147 minutes and for a sick visit is 42 minutes (Total 189 minutes)
On Wednesday appointment, the time allotted for an annual checkup is 147 minutes and for a sick visit is 21 minutes (Total 168 minutes)
Using a trigonometric identity, the cosine of the angle is given as follows:
![\cos{\theta} = \frac{60}{61}](https://tex.z-dn.net/?f=%5Ccos%7B%5Ctheta%7D%20%3D%20%5Cfrac%7B60%7D%7B61%7D)
<h3>What is the trigonometric identity that relates the sine and the cosine of an angle?</h3>
It is given by:
![\sin^2{\theta} + \cos^2{\theta} = 1](https://tex.z-dn.net/?f=%5Csin%5E2%7B%5Ctheta%7D%20%2B%20%5Ccos%5E2%7B%5Ctheta%7D%20%3D%201)
In this problem, the sine is:
![\sin{\theta} = \frac{11}{61}](https://tex.z-dn.net/?f=%5Csin%7B%5Ctheta%7D%20%3D%20%5Cfrac%7B11%7D%7B61%7D)
Then:
![\cos^2{\theta} = 1 - \sin^2{\theta}](https://tex.z-dn.net/?f=%5Ccos%5E2%7B%5Ctheta%7D%20%3D%201%20-%20%5Csin%5E2%7B%5Ctheta%7D)
![\cos^2{\theta} = 1 - \left(\frac{11}{61}\right)^2](https://tex.z-dn.net/?f=%5Ccos%5E2%7B%5Ctheta%7D%20%3D%201%20-%20%5Cleft%28%5Cfrac%7B11%7D%7B61%7D%5Cright%29%5E2)
![\cos^2{\theta} = \frac{3600}{3721}](https://tex.z-dn.net/?f=%5Ccos%5E2%7B%5Ctheta%7D%20%3D%20%5Cfrac%7B3600%7D%7B3721%7D)
![\cos{\theta} = \pm \sqrt{\frac{3600}{3721}}](https://tex.z-dn.net/?f=%5Ccos%7B%5Ctheta%7D%20%3D%20%5Cpm%20%5Csqrt%7B%5Cfrac%7B3600%7D%7B3721%7D%7D)
On quadrant I, the cosine is positive, hence:
![\cos{\theta} = \frac{60}{61}](https://tex.z-dn.net/?f=%5Ccos%7B%5Ctheta%7D%20%3D%20%5Cfrac%7B60%7D%7B61%7D)
More can be learned about trigonometric identities at brainly.com/question/24496175
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Answer:
15/4 6/5
Step-by-step explanation:
there simplified
Answer:
i think 4.4n+3-3.4n = n+3