<span>Solution
m = {-13, -1}</span>
Answer:
7/10
Step-by-step explanation:
Area=1/2 times base times height
note:bh=base times height
a=1/2bh
b=width
h=-4+2w
h=2w-4
subsitute
a=1/2w(2w-4)
a=1/2(2s^2-4w)
a=w^2-2w
a=63
63=w^2-2w
subtract 63 from both sdies
0=w^2-2w-63
factor
find what 2 numbers multiply to get -63 and add to get -2
the numbers are -9 and 7
so
0=(w-9)(w+7)
if xy=0 then x and/or y=0
so
w-9=0
w+7=0
solve each
w-9=0
add 9 to both sdies
w=9
w+7=0
subtract 7 from both sides
w=-7
width cannot be negative so this can be discarded
width=9
subsitute
l=2w-4
l=2(9)-4
l=18-4
l=14
legnth=14 in
width/base=9 in
Answer: amplitude is 5 and Period is pi/2
Step-by-step explanation:
yes ♀️
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
