How do we solve this?
1 answer:
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Answer:
Step-by-step explanation:
We are given that
y(0)=0
y'(0)=1
By comparing with
We get
q(x)=0
p(x),q(x) and g(x) are continuous for all real values of x except 3.
Interval on which p(x),q(x) and g(x) are continuous
and (3,
By unique existence theorem
Largest interval which contains 0=
Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=
Answer:
large area: small area = 100:25 = 4:1
Step-by-step explanation:
Mid-Segment of a triangle is always <u>parallel to the third side</u>, and the length of the midsegment is <u>half the length of the third side</u>.
(x₁,y₁) (x₂,y₂) (x₃,y₃): A(-10,30), B(30,50, C(-40,20)
XY = 1/2 AC , YZ = 1/2 AB , ZX = 1/2 BC
X (10,40) , Y (-5,35) , Z (-25,25) : X(x'₁,y'₁), Y(x'₂,y'₂), Z(x'₃,y'₃)
area of triangle by 3 vertices: (x₁,y₁) (x₂,y₂) (x₃,y₃) is
a = abs[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] / 2 ... this is just a common formula for any triangle
a = abs(-10*30 + 30*-10 + -40*-20) / 2 = 100
a' = abs(10*10 + -5*-15 + -25*5) / 2 = 25
a/a' = 100/25 = 4/1
