algebra 2
The function does not have rational roots, as it does not factor. If you graph the function, you will find approximate values (3 of them) of the real roots. One of them is positive, and two are negative. To locate them without graphing, you can find f'(x) and equate to zer...
job related
If the slope is 9 inches per foot (run), then the roof is 15" per foot run. I suppose rake factor is the ratio of roof length to rise, which works out to 1.25 in this case. The roof area (for one sloping face) is therefore (15"/9")*10'*30' long = 500 sq....
Math - correction
Correction in an intermediate step (only): ... =M((1+i)^n-1)/(1+i-1) =M((1+i)^n-1)/i
Math
Use the compound interest formula: P=$3690 i=0.21/12=0.0175 n=17 M=amount of monthly payment Amount after 17 months (without payment) = A(1+i)^n Value of his payments at the same interest rate =M+M(1+i)+M(1+i)^2+...+M(1+i)^(n-1) =M((1+i)^n-1)/(1+i-1) =M(1+i)^n/i Equate the two...
Critical Points
Inflection point at (2,3) means f"(2)=0 => f"(x)=6x+2a = 0 at x=2, or a=-6(2)/2=-6 Since f(2)=3 (from given point (2,3) ), we conclude that: f(2)=c+2b-16=3 or 2b+c=19 ....(1) We also know that there is a critical point at (1,5), so f'(1)=0, or b=9 Substituting...
Minimum on a Closed Interval
Note: -3/2 is not in [1,2], so it cannot be the correction solution.
Minimum on a Closed Interval
On a closer look, the first equation f(x)=1/(x sin(πx)) is undefined at x=1 and x=2, so probably not what you want. For equation 2, f(x)=(1/x) sin(πx) you would find f'(x)= (%pi*cos(πx))/x - sin(πx)/x^2 equate f'(x) to zero to find the minimum, and solv...
Minimum on a Closed Interval
Is the function f(x)=1/(x sin(πx)) or f(x)=(1/x) sin(πx) [ as written ]?
Geometry
If centre has (6,8) and one end of the diameter has (4,2), add Δx and Δy to the centre to get the other end of the diameter, namely, (Δx,Δy) = (6-4, 8-2) = (2,6) So the other end of the diameter would be (6+2,8+6)=(8,14). Check: the mid-point of the dia...
math
It could be a rectangle (or any other shape). Use L and W to represent the length and width. Use the area and perimeter formula to construct 2 equations and consequently solve for L and W. The solution consists of integers.
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