Answer:
Perpendicular would be -7
Parallel would be 7
Because m = slope and the m is connected to the x (mx)
Answer:
X=18
Step-by-step explanation:
Isosceles means two sides are equal. In this case AC and BD are the same. So put both equations equal to each other and solve
(a) It looks like the ODE is
<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)
which is separable:
d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2) => d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>
Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :
∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get
∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>
<em>t</em> = 2<em>x</em> ^2 + <em>C</em>
arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>
<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)
(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.
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A = P ( 1 + r / n ) ^nt
A = value after t periods
P = principal amount (initial investment)
r = annual nominal interest rate (not reflecting the compounding)
n = number of times the interest is compounded per year
t = number of years the money is borrowed for