Parametrization will work like this:   x=t x+y=1 => y=1-x => y=1-t z=4-√(x^2+y^2) ⇒ z=4-√(t^2+t^2-2t+1) or z=4-√(2t^2-2t+1)   So our parametrization for the curve is <t, 1-t, 4-√(2t^2-2t+1)>   Plug in 3/7...

I know it's been a while since this was posted, but better late than never.  If Kaitong can't benefit from this, another student will.   C'(t)=<6t,e6t,t3)   C(t)=∫C'(t)dt=<3t2,e6t/6,t4/4> + <a,b,c>   C(0)= <-3,-1,4> = <0,1/6,0>+<a,b,c>   a+0=-3...

The power series for 1/(3-x)=(1/3)∑(x/3)n where n goes from 0 to ∞ (See geometric series)   Integrating both sides we get   ∫(1/(3-x))dx=(1/3)∫∑(x/3)n dx + C   Integrating the right hand side term by term, we get   ∫(1/(3-x))dx=(1/3∫∑xn+1/((n+1)3n)...

The double integral will yield the volume under the plane z=2x+1 bounded by the lines x=0 and x=2 and y=0 and y=2 in the xy plane.     Since we are integrating over a rectangular area, the integral simply becomes ∫10∫02 (2x+1)dxdy.  The inner integral ∫02(2x+1)dx...

P=x2y   x+y=63  => y=63-x   P=x2(63-x)   P=63x2-x3   dP/dX = 126x-3x2   For maximum value of P, dP/dx = 0   126x-3x2 = 0   x=0 or x=41   x=0 yields the local minimum for P and...

f(x) = 519ln(x) + 1768   f'(x) = 519/x   The slope (derivative) is positive for all positive values within the domain of x which is [9,12]. A function is increasing when the slope is positive.  Therefore, the answer to your question is "yes, the consumption...

Suppose each side of the original cube has length x inches.  When you cut 5 slices of 0.6 inches each from one side, the side length reduces by 3 inches.  The new dimensions of the loaf are x by x by (x-3).  The volume of this loaf will be x*x*(x-3) cubic inches.  Equating this...

The answer is very simple.   The arc length is given by the formula rθ where θ is the angle the arc covers at the center given in raidans. Since m∠CPA = 920 we have m∠BPD=920 as well (vertical angles).  Since BD covers (subtends is the technical term) 920 at the center and 920...

ƒ(x) = y = ((x-1)/(x+1))³   To find f-1 replace x's with y's and y's with x's and isolate y   x = ((y-1)/(y+1))³   x1/3 = (y-1)/(y+1)   (y+1)(x1/3) = y-1   yx1/3+x1/3 = y-1   y(1-x1/3) = 1+x1/3   y...

The formula in optics relating a lens's focal length f to the object distance u and image distance v is as follows:   1/f = 1/u + 1/v  --Equation 1   The image size to object size ratio should be 4 inches to 1 mile or 1:15840   By the property of similar...

The zeros are obtained by equating the polynomial to 0.  Zeros of a polynomial are the values of the variable in the polynomial (the values of x in this present case) for which the value of the polynomial returns 0.     So we have -(x-3)2(x+1)3(x-7)=0   For...

Slope intercept form of a line is of the form y=mx+b where m is the slope of the line and b the y intercept The problem doesn't specify the initial points.  Let's take points (-4,4) and (6,3).  The slope of a line joining these points is (3-4)/(6+4)=-1/10   We...

Take logarithms on both sides and compare them.     Log10(1993)=93 x Log10(19) which is approximately 274 Log10(1399)=99 x Log10(13) which is approximately 253   if Log(a)>Log(b) then a>b

This I believe is incorrectly stated question.  Satyam, please confirm that the question is ∫√(1+sin(x/4)) and not ∫√((1+sin x)/4)

Use Tan-1 function to find Θ

Sara, Due to lack of complete information, I need to make several guesses in order to answer this question.   Let's suppose the right triangle contains points A, B, C with BC being the horizontal base and C is to the right of B and AC is the hypotenuse of the right triangle.  ∠BCA=θ...

f(x)=2x-1 and g(x)=-2x-1.   The reflection of f(x) over y-axis is f(-x) and f(-x)=-2x-1=g(x)   This tells us that g(x) is a reflection of f(x) over y-axis.     Both functions f and g return real numbers for all values of x.  Therefore the domain for...

The left end point of the interval is not given in the problem.  The right end point is x=1.   Assuming the left end point is 0.   The formula for the volume of revolution of a function between the curve and the x axis about the x axis is V=π∫f(x)2 dx.     f(x)=4x2...

It appears to me that this is a problem on the mean value theorem.  The theorem stipulates that there is point in the OPEN interval where the slope is equal to the average rate of change over the CLOSED interval [a,b].  The average rate of change is f(a)-f(b)/(a-b).  If f(a)=f(b)...