Answer:
10 servings
Step-by-step explanation:
Since one box contains 5 cups then, two boxes equals <u>10 servings</u>.
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Solve the initial value problem:
dy——— = 2xy², y = 2, when x = – 1. dxSeparate the variables in the equation above:

Integrate both sides:


Take the reciprocal of both sides, and then you have

In order to find the value of
C₁ , just plug in the equation above those known values for
x and
y, then solve it for
C₁:
y = 2, when
x = – 1. So,


Substitute that for
C₁ into (i), and you have

So
y(– 2) is

I hope this helps. =)
Tags: <em>ordinary differential equation ode integration separable variables initial value problem differential integral calculus</em>
Answer:
x+y+3w
Step-by-step explanation:
Y intercept is where the line meets the Y axis
So the points shd be in (0,y) form.
So (0,0) (0,-7) (0,-0.25) are the y-intercepts in the following.