Answer:
To find the original dimensions, write a proportion with the scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve. To find new dimensions, write a proportion with the new scale as the first ratio and the scale dimension compared to the actual dimension as the second ratio. Use cross products to solve.
Y=7-2×(-5/2)
Y=7+5
Y=12
The answer is C
The information given in the triangle shows that the total distance from A to B to C to D to E will be C. 84.
<h3>How to calculate the triangle?</h3>
From the information given, ABF = EDG, GCF are equilateral and AG = 21 while CG = 1/4.
The total distance from A to B to C to D to E will be:
= 21/CG
= 21 / 1/4
= 21/0.25
= 84
In conclusion, the correct option is 84.
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Write the expression below in terms of x and y only:
(I'm going to call it "E")
![\mathsf{E=sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7BE%3Dsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%2Bcos%5E%7B-1%7D%28y%29%5Cright%5D%5Cqquad%5Cquad%28i%29%7D)
Let

so the expression becomes

• Finding

![\mathsf{sin\,\alpha=sin\!\left[sin^{-1}(x)\right]}\\\\ \mathsf{sin\,\alpha=x\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2C%5Calpha%3Dsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%5Cright%5D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Calpha%3Dx%5Cqquad%5Cquad%5Ccheckmark%7D)
• Finding


because

is positive for
![\mathsf{\alpha\in \left[-\frac{\pi}{2},\,\frac{\pi}{2}\right].}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Calpha%5Cin%20%5Cleft%5B-%5Cfrac%7B%5Cpi%7D%7B2%7D%2C%5C%2C%5Cfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D.%7D)
• Finding

![\mathsf{cos\,\beta=cos\!\left[cos^{-1}(y)\right]}\\\\ \mathsf{cos\,\beta=y\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7Bcos%5C%2C%5Cbeta%3Dcos%5C%21%5Cleft%5Bcos%5E%7B-1%7D%28y%29%5Cright%5D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcos%5C%2C%5Cbeta%3Dy%5Cqquad%5Cquad%5Ccheckmark%7D)
• Finding


because

is positive for
![\mathsf{\beta\in [0,\,\pi].}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cbeta%5Cin%20%5B0%2C%5C%2C%5Cpi%5D.%7D)
Finally, you get
![\mathsf{E=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}}\\\\\\ \therefore~~\mathsf{sin\!\left[sin^{-1}(x)+cos^{-1}(y)\right]=x\cdot y +\sqrt{1-y^2}\cdot \sqrt{1-x^2}\qquad\quad\checkmark}](https://tex.z-dn.net/?f=%5Cmathsf%7BE%3Dx%5Ccdot%20y%20%2B%5Csqrt%7B1-y%5E2%7D%5Ccdot%20%5Csqrt%7B1-x%5E2%7D%7D%5C%5C%5C%5C%5C%5C%20%5Ctherefore~~%5Cmathsf%7Bsin%5C%21%5Cleft%5Bsin%5E%7B-1%7D%28x%29%2Bcos%5E%7B-1%7D%28y%29%5Cright%5D%3Dx%5Ccdot%20y%20%2B%5Csqrt%7B1-y%5E2%7D%5Ccdot%20%5Csqrt%7B1-x%5E2%7D%5Cqquad%5Cquad%5Ccheckmark%7D)
I hope this helps. =)
Tags: <em>inverse trigonometric trig function sine cosine sin cos arcsin arccos sum angles trigonometry</em>
Answer:
b
Step-by-step explanation: