We are given the quadratic:
, with a=1, b=-8, c=15.
We know that the x-coordinate of the vertex, which is the point where the line of symmetry passes through is
.
Thus, the x-coordinate of the vertex is
.
Thus, the line of symmetry is x=4.
Answer: <span>B. The line of symmetry should have been 4 instead of –4. </span>
<span> 5x + 6y= -12 7) x - 5y = 0 6) x +6y = - 42 8) 4x - 7y = -21 </span>
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Evaluate the indefinite integral:
Trigonometric substitution:
then,
![\begin{array}{lcl} \mathsf{x=sin\,\theta}&\quad\Rightarrow\quad&\mathsf{dx=cos\,\theta\,d\theta\qquad\checkmark}\\\\\\ &&\mathsf{x^2=sin^2\,\theta}\\\\ &&\mathsf{x^2=1-cos^2\,\theta}\\\\ &&\mathsf{cos^2\,\theta=1-x^2}\\\\ &&\mathsf{cos\,\theta=\sqrt{1-x^2}\qquad\checkmark}\\\\\\ &&\textsf{because }\mathsf{cos\,\theta}\textsf{ is positive for }\mathsf{\theta\in \left[\dfrac{\pi}{2},\,\dfrac{\pi}{2}\right].} \end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Blcl%7D%20%5Cmathsf%7Bx%3Dsin%5C%2C%5Ctheta%7D%26%5Cquad%5CRightarrow%5Cquad%26%5Cmathsf%7Bdx%3Dcos%5C%2C%5Ctheta%5C%2Cd%5Ctheta%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3Dsin%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bx%5E2%3D1-cos%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5E2%5C%2C%5Ctheta%3D1-x%5E2%7D%5C%5C%5C%5C%20%26%26%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Csqrt%7B1-x%5E2%7D%5Cqquad%5Ccheckmark%7D%5C%5C%5C%5C%5C%5C%20%26%26%5Ctextsf%7Bbecause%20%7D%5Cmathsf%7Bcos%5C%2C%5Ctheta%7D%5Ctextsf%7B%20is%20positive%20for%20%7D%5Cmathsf%7B%5Ctheta%5Cin%20%5Cleft%5B%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C%5C%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%5D.%7D%20%5Cend%7Barray%7D)
So the integral
becomes
Integrate
by parts:
Substitute back for the variable x, and you get
I hope this helps. =)
Tags: <em>integral inverse sine function angle arcsin sine sin trigonometric trig substitution differential integral calculus</em>
Answer:
Calculate the Macaulay duration of Annuity B at the time of purchase is 1.369.
Step-by-step explanation:
First, we use 0.93 to calculate the v which equals 1/(1+i).
= 0.93
After rearranging the equation, we get 1.07
+ 0.07v - 0.93=0
So, v=0.9
Mac D: 
After substituting the value of v, we get Mac D = 1.369.
You are correct in thinking that the columns of the first matrix must match with the rows of the second matrix.
So for example, we can multiply a 1 x 5 matrix with a 5 x 7 matrix. The two matching '5's are directly what make multiplication possible in this case. For your problem, the first two '3's match and multiplication is possible.
The rows of the first matrix don't need to match with the columns of the second matrix.
