Answer:
The x-coordinate of the solution is x=3
Step-by-step explanation:
we have
------> equation A
------> equation B
Solve the system of equations by substitution
Substitute equation B in equation A and solve for x
Find the value of y
------> 
The solution of the system of equations is the point (3,2)
therefore
The x-coordinate of the solution is x=3
Answer:
F'(x)=4x3+16x
Step-by-step explanation:
Answer:
Step-by-step expla3
x
−
2
y
<
10
Solve for y
.
Tap for more steps...
y
>
−
5+
3
x
2
Find the slope and the y-intercept for the boundary line.
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Slope:
3
2
y-intercept:
(
0
,
−
5
)
Graph a dashed line, then shade the area above the boundary line since
y
is greater than
−
5
+
3
x
2
.
y
>
−
5
+3
x2
In order to do that we would need to see the list of potential answers.
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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>
