4 years ago

Can you solve this? 1+cot^2θ=csc^2θ

1 answer:

5 0

The equation above is a "pythagorean identity", and therefore, the left-side is indeed equal to the right-side.

what are the values of θ that make the equation true? well, all possible angles.

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From a practice assignment:<br>solve the following differential equation given initial conditions

hodyreva [135]

If $y' = e^y \sin(x)$ and $y(-\pi)=0$ , separate variables in the differential equation to get

$e^{-y} \, dy = \sin(x) \, dx$

Integrate both sides:

$\displaystyle \int e^{-y} \, dy = \int \sin(x) \, dx \implies -e^{-y} = -\cos(x) + C$

Use the initial condition to solve for $C$ :

$-e^{-0} = -\cos(-\pi) + C \implies -1 = 1 + C \implies C = -2$

Then the particular solution to the initial value problem is

$-e^{-y} = -\cos(x) - 2 \implies e^{-y} = \cos(x) + 2$

(A)

4 0

2 years ago

Madelyn has x nickels and y dimes, having a maximum of 15 coins worth no

aliina [53]

9514 1404 393

Answer:

2 nickels, 9 dimes

Step-by-step explanation:

When there are a number of overlapping shaded areas on the graph, I find it convenient to use the reverse of the inequalities. That makes the <em>unshaded</em> area the solution space. Here, the vertices of the triangular solution space are ...

(2, 9), (2, 13), (6, 9)

Any of the grid points within (or on) this triangle is a possible solution. One of them is (2, 9) corresponding to 2 nickels and 9 dimes.

Three solutions are shown:

(x, y) = (2, 9), (3, 10), (4, 11)