B. First and Second St appear to be parallel
M<1 = M<3
these 2 angles are equal for parallel lines
6x=120
X=120/6=20
We want to solve the Initial Value Problem y' = y + 4xy, with y(0) = 1.
To use Euler's method, define
y(i+1) = y(i) + hy'(i), for i=0,1,2, ...,
where
h = 0.1, the step size.,
x(i) = i*h
1st step.
y(0) = 1 (given) and x(0) = 0.
y(1) ≡ y(0.1) = y(0) + h*[4*x(0)*y(0)] = 1
2nd step.
x(1) = 0.1
y(2) ≡ y(0.2) = y(1) + h*[4*x(1)*y(1)] = 1 + 0.1*(4*0.1*1) = 1.04
3rd step.
x(2) = 0.2
y(3) ≡ y(0.3) = y(2) + h*[4*x(2)*y(2)] = 1.04 + 0.1*(4*0.2*1.04) = 1.1232
4th step.
x(3) = 0.3
y(4) ≡ y(0.4) = y(3) + h*[4*x(3)*y(3)] = 1.1232 + 0.1*(4*0.3*1.1232) = 1.258
5th step.
x(4) = 0.4
y(5) ≡ y(0.5) = y(4) + h*[4*x(4)*y(4)] = 1.258 + 0.1*(4*0.4*1.258) = 1.4593
Answer: y(0.5) = 1.4593
Answer:
V = $3.50t + $90.5....
Step-by-step explanation:
V(t) is a function of t that expresses the value in year 2000+t.
We know that the increase is $3.50 times t.
So,
V(t) = $3.50t + c
where c is the constant.
V(15) = $3.50 (15) + c = $143 [t=15 as mentioned in the question]
and therefore
c = $143 - $3.50 (15)
c= $143 - $52.50
c= $90.5
Now we got the value of c. We can write the equation as
V = $3.50t + $90.5....
(2y-at)(y+2at)
Answer: <span><span><span>−<span><span>2<span>a^2</span></span><span>t^2</span></span></span>+<span><span><span>3a</span>t</span>y</span></span>+<span>2<span>y^<span>2
I hope this helps</span></span></span></span>
