Answer:
a solution is 1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
Step-by-step explanation:
for the equation
(1 + x⁴) dy + x*(1 + 4y²) dx = 0
(1 + x⁴) dy = - x*(1 + 4y²) dx
[1/(1 + 4y²)] dy = [-x/(1 + x⁴)] dx
∫[1/(1 + 4y²)] dy = ∫[-x/(1 + x⁴)] dx
now to solve each integral
I₁= ∫[1/(1 + 4y²)] dy = 1/2 *tan⁻¹ (2*y) + C₁
I₂= ∫[-x/(1 + x⁴)] dx
for u= x² → du=x*dx
I₂= ∫[-x/(1 + x⁴)] dx = -∫[1/(1 + u² )] du = - tan⁻¹ (u) +C₂ = - tan⁻¹ (x²) +C₂
then
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) +C
for y(x=1) = 0
1/2 *tan⁻¹ (2*0) = - tan⁻¹ (1²) +C
since tan⁻¹ (1²) for π/4+ π*N and tan⁻¹ (0) for π*N , we will choose for simplicity N=0 . hen an explicit solution would be
1/2 * 0 = - π/4 + C
C= π/4
therefore
1/2 *tan⁻¹ (2*y) = - tan⁻¹ (x²) + π/4
<span>$2,022.92 would be the answer
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Answer:
Step-by-step explanation:
Approximate the integral 
by dividing the region 
with vertices (0,0),(4,0),(4,2) and (0,2) into eight equal squares.
Find the sum 
Since all are equal squares, so 
for every 
Thus, 
Evaluating the iterate integral ![\int\limits^4_0 \int\limits^2_0 {(x+y)} \, dydx=\int\limits^4_0 {[xy+\frac{y^2}{2} ]}\limits^2_0 \, dx =\int\limits^4_0 {[2x+2]}dx\\\\=[x^2+2x]\limits^4_0=24.](https://tex.z-dn.net/?f=%5Cint%5Climits%5E4_0%20%5Cint%5Climits%5E2_0%20%7B%28x%2By%29%7D%20%5C%2C%20dydx%3D%5Cint%5Climits%5E4_0%20%7B%5Bxy%2B%5Cfrac%7By%5E2%7D%7B2%7D%20%5D%7D%5Climits%5E2_0%20%5C%2C%20dx%20%3D%5Cint%5Climits%5E4_0%20%7B%5B2x%2B2%5D%7Ddx%5C%5C%5C%5C%3D%5Bx%5E2%2B2x%5D%5Climits%5E4_0%3D24.)
Thus, 
9514 1404 393
Answer:
x = 10; WX = 5; HJ = 10
Step-by-step explanation:
The hash marks indicate that points W and X are midpoints of their respective segments. That makes WX a midline of the triangle. The midline is always half the length of the base (HJ). We can use this fact to write an equation relating the lengths.
HJ = 2·WX
x = 2(x -5) . . . .substitute the given expressions
x = 2x -10 . . . . eliminate parentheses
10 = x . . . . . . . add 10-x to both sides
This value of x tells you that ...
HJ = x = 10
WX = x -5 = 5
Answer:
18 ft
Step-by-step explanation:
