Step-by-step explanation:
Given that,
We have to find the value of m∠E.
Here, two sides are equal, thus it is an isosceles triangle. As the two sides are equal, so their angles must be equal. So, ∠E and ∠D will be equal. Let us assume the measures of both ∠E and ∠D as x.
→ Sum of all the interior angles of ∆ = 180°
→ ∠E + ∠D + ∠F = 180°
→ 116° + x + x = 180°
→ 2x = 180° – 116°
→ 2x = 64°
→ x = 64° ÷ 2
→<u> x = 32°</u>
Henceforth,
→ m∠E = x
→ m∠E = 32°
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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>
The answer is 10 quarters because 10 quarters would be $2.50. You would need 3 more nickles to get $2.65. 10 would be 7 more than 3.