answer:The least squares vector ( mathbf{v} ) can be calculated as mathbf{v} = (mathbf{A}^top mathbf{A})^{-1} mathbf{A}^top mathbf{b} After performing the calculations, the resulting least squares vector ( mathbf{v} ) is mathbf{v} = left( begin{array}{c} -0.114 -1.113 end{array} right)

answer:To evaluate the function at ( x = -3 ), we substitute the value of ( x ) into the function: [ f(-3) = sin(7 - 2(-3)) sin(cos(4(-3) + 8)) ] Calculating each part: [ f(-3) = sin(7 + 6) sin(cos(-12 + 8)) ] [ f(-3) = sin(13) sin(cos(-4)) ] Now, we compute the individual trigonometric values: [ sin(13) approx -0.255 ] [ cos(-4) = cos(4) approx 0.766 ] Since ( sin(cos(-4)) = sin(cos(4)) ), we have: [ f(-3) = (-0.255) cdot (sin(0.766)) ] However, we made a mistake in the intermediate step. The value of ( sin(0.766) ) should be computed, not ( cos(0.766) ): [ sin(0.766) approx 0.725 ] Multiplying these: [ f(-3) = (-0.255) cdot (0.725) approx -0.184 ] Therefore, the evaluation of the function at ( x = -3 ) is approximately ( -0.184 ).

answer:First, we can simplify the equation using the property of exponents that states a^b cdot a^c = a^{b+c}. 8^{2x+10} cdot 8^{13-25x} = 64^{22x-6} 8^{2x+10+13-25x} = 64^{22x-6} 8^{-23x+23} = 64^{22x-6} Now, we can rewrite the equation with the same base on both sides: 8^{-23x+23} = (8^2)^{22x-6} 8^{-23x+23} = 8^{44x-12} Equating the exponents, we get: -23x+23 = 44x-12 67x = 35 x = frac{35}{67} Therefore, the only real solution to the equation is x = frac{35}{67}. The answer is left{ frac{35}{67} right}

answer:Using the MacLaurin expansion of ln(1+t), substituting t=x^2, dividing by x^w, and integrating term by term, we obtain: #int ln(1+x^2)/x^w = C+sum_(n=0)^oo (-1)^n/(n+1)x^(2n+3-w)/(2n+3-w)# This series is valid for all values of w except for odd integers other than w=1.