Answer:
5
Step-by-step explanation:
cos 60° = y / 10
y = 10 * 1/2 = 5
![\bf f(x)=y=2x+sin(x) \\\\\\ inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x) \\\\\\ \textit{now, the "y" in the inverse, is really just g(x)} \\\\\\ \textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3Dy%3D2x%2Bsin%28x%29%0A%5C%5C%5C%5C%5C%5C%0Ainverse%5Cimplies%20x%3D2y%2Bsin%28y%29%5Cleftarrow%20f%5E%7B-1%7D%28x%29%5Cleftarrow%20g%28x%29%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20the%20%22y%22%20in%20the%20inverse%2C%20is%20really%20just%20g%28x%29%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Bso%2C%20we%20can%20write%20it%20as%20%7Dx%3D2g%28x%29%2Bsin%5Bg%28x%29%5D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
![\bf \textit{let's use implicit differentiation}\\\\ 1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor} \\\\\\ 1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Blet%27s%20use%20implicit%20differentiation%7D%5C%5C%5C%5C%0A1%3D2%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%2Bcos%5Bg%28x%29%5D%5Ccdot%20%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5Cimpliedby%20%5Ctextit%7Bcommon%20factor%7D%0A%5C%5C%5C%5C%5C%5C%0A1%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%5B2%2Bcos%5Bg%28x%29%5D%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B%5B2%2Bcos%5Bg%28x%29%5D%5D%7D%3D%5Ccfrac%7Bdg%28x%29%7D%7Bdx%7D%3Dg%27%28x%29%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D)
now, if we just knew what g(2) is, we'd be golden, however, we dunno
BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)
for inverse expressions, the domain and range is the same as the original, just switched over
so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2
so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2
thus 2 = 2x+sin(x)
![\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2 \\\\\\ -----------------------------\\\\ g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}](https://tex.z-dn.net/?f=%5Cbf%202%3D2x%2Bsin%28x%29%5Cimplies%200%3D2x%2Bsin%28x%29-2%0A%5C%5C%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0Ag%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5Bg%282%29%5D%7D%5Cimplies%20g%27%282%29%3D%5Ccfrac%7B1%7D%7B2%2Bcos%5B2x%2Bsin%28x%29-2%5D%7D)
hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it
<span>8(3x + 6) + 13---expand using distributive property
=24x + 48+13 ...combine like terms
=24x +61....simplify
answer is C.24x + 61
hope it helps</span>
Answer:
BM: <u>y = (2/3) x + 16/3</u> with segment length of 2.77
Step-by-step explanation:
AC formula: m = (6-0)/(0-4) = -3/2
(y-0)/(x-4) = -3/2 y = (-3/2)x + 6 ... (1)
BM slope: BM⊥ AC m = 2/3
BM formula: (y-4) / (x- -2) = (y-4) / (x+2) = 2/3
y-4 = 2/3 x + 4/3
<u>y = (2/3) x + 16/3</u> ... (2) -2≤x≤0.31
intercept of AC and BM (M) from (1) and (2): (-3/2)x + 6 = (2/3) x + 16/3
(13/6) x = 2/3 x = (2/3) / (13/6) = 4/13 ≈ 0.31
y = (2/3) (4/13) + (16/3) = (8/39) + (208/39) = 216/39 = 72/13 ≈ 5.54
M (4/13 , 72/13) or (0.31 , 5.54)
segment BM = √(4/13 - -2)² + (72/13 - 4)² = √1300/169 = 2.77
Answer:
y = 112
Step-by-step explanation:
If y varies directly as the square of x
mathematically;
y ∝ x²
y = kx² where 'k' is the constant of proportionality
let u know the value of k for this equation by making k the subject of the formular from y = kx²
divide both sides by x²
k = y / x²
given that y = 567 x = 9
k = 567/(9)²
k = 567/81
k = 7
now to get the value of y when x = 4
from the equation conneting x, y,
y = kx²
y = 7 × (4)²
y = 7 × 16
y = 112