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polet [3.4K]
3 years ago
10

Nash is making a framed enlargement of a Mary Cassatt postage stamp to present to a retired postmaster. He enlarges the stamp us

ing a scale of 2 inches to 1 centimeter. If the actual length of the stamp is 3 centimeters and its width is 5 centimeters, what are the dimensions of the scale picture? Please help me
Mathematics
1 answer:
Ulleksa [173]3 years ago
4 0

6 in. x 10 in

Set the problem up as a ratio.

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Find the vertex of y=-x^2+6x-12
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You find the x coordinates using x = -b/2a  = -6/2(-1) = 3

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Vertex (3,-3)
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Find the particular solution of the differential equation that satisfies the initial condition(s). f ''(x) = x−3/2, f '(4) = 1,
sweet [91]

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Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

Step-by-step explanation:

This differential equation has separable variable and can be solved by integration. First derivative is now obtained:

f'' = x - \frac{3}{2}

f' = \int {\left(x-\frac{3}{2}\right) } \, dx

f' = \int {x} \, dx -\frac{3}{2}\int \, dx

f' = \frac{1}{2}\cdot x^{2} - \frac{3}{2}\cdot x + C, where C is the integration constant.

The integration constant can be found by using the initial condition for the first derivative (f'(4) = 1):

1 = \frac{1}{2}\cdot 4^{2} - \frac{3}{2}\cdot (4) + C

C = 1 - \frac{1}{2}\cdot 4^{2} + \frac{3}{2}\cdot (4)

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The first derivative is y' = \frac{1}{2}\cdot x^{2}- \frac{3}{2}\cdot x - 1, and the particular solution is found by integrating one more time and using the initial condition (f(0) = 0):

y = \int {\left(\frac{1}{2}\cdot x^{2}-\frac{3}{2}\cdot x -1  \right)} \, dx

y = \frac{1}{2}\int {x^{2}} \, dx - \frac{3}{2}\int {x} \, dx - \int \, dx

y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x + C

C = 0 - \frac{1}{6}\cdot 0^{3} + \frac{3}{4}\cdot 0^{2} + 0

C = 0

Hence, the particular solution of the differential equation is y = \frac{1}{6} \cdot x^{3} - \frac{3}{4}\cdot x^{2} - x.

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