Answer:
2. The change in expected height for every one additional centimeter of femur length.
Step-by-step explanation:
<u>1. The expected height for someone with a femur length of 65 centimeters.</u>
<em>Doesn't make sense, that would be height value when centimeters = 65.</em>
<u>2</u><u><em>. </em></u><u>The change in expected height for every one additional centimeter of femur length.</u>
<em>Makes sense, for every increase in one additional centimeter, we can expect the height to be proportional to the slope.</em>
<u>3. The femur length for someone with an expected height of 2.5 centimeters.</u>
<em>Doesn't make sense, the linear relationship relies on the femur length to get the height.</em>
<u>4. The change in expected femur length for every one additional centimeter of height.</u>
<em>Doesn't make sense, again, the linear relationship relies on the femur length.</em>
Answer:
x = 3, y = 4 and z = -5.
Step-by-step explanation:
Use the elimination method:
-3x + 2y + z = -6...........(1)
x + 3y + 2z = 5..............(2)
4x + 4y + 3z = 13..........(3)
Multiply the first equation by -2:
6x - 4y - 2z = 12 Add this to equation (2):
7x - y = 17...........(4)
Now Multiply equation (2) by 3 and equation (3) by -2
3x + 9y + 6z = 15 ............(5)
-8x - 8y - 6z = -26............(6) Adding (5) + (6):
-5x + y = -11..........(7)
Now Add (7) and (4):
2x = 6
x = 3.
Now substitute for x in (4):
7(3) - y = 17
-y = 17 - 21 = -4
y = 4.
Finally substitute for x and y in equation (3) to find z:
4(3) + 4(4) +3z = 13
3z = 13 - 12 - 16 = -15
z = -5.
Notice the picture,
recall your SOH, CAH, TOA
you have,
opposite side, 165
adjacent side, 50
and the angle
that means, we'll need Mrs. tangent
thus
![\bf tan(\theta)=\cfrac{opposite}{adjacent}\implies tan(\theta)=\cfrac{165}{50} \\ \quad \\ tan^{-1}\left[ tan(\theta) \right]=tan^{-1}\left[ \cfrac{165}{50} \right] \\ \quad \\ \theta=tan^{-1}\left[ \cfrac{165}{50}\right] \\ \uparrow \\ \textit{angle of elevation}\iff\textit{angle of depression}](https://tex.z-dn.net/?f=%5Cbf%20tan%28%5Ctheta%29%3D%5Ccfrac%7Bopposite%7D%7Badjacent%7D%5Cimplies%20tan%28%5Ctheta%29%3D%5Ccfrac%7B165%7D%7B50%7D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%20tan%5E%7B-1%7D%5Cleft%5B%20tan%28%5Ctheta%29%20%5Cright%5D%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%20%5Cright%5D%0A%5C%5C%20%5Cquad%20%5C%5C%0A%5Ctheta%3Dtan%5E%7B-1%7D%5Cleft%5B%20%5Ccfrac%7B165%7D%7B50%7D%5Cright%5D%0A%5C%5C%20%5Cuparrow%20%20%5C%5C%0A%20%5Ctextit%7Bangle%20of%20elevation%7D%5Ciff%5Ctextit%7Bangle%20of%20depression%7D)
