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ehidna [41]
4 years ago
6

Please Help Me

Mathematics
1 answer:
astra-53 [7]4 years ago
5 0
P = 2(L + W)
L = 1/3W - 3

P = 2(1/3W - 3 + W) or P = 2(4/3W - 3)
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Find the Surface Area
Sergeeva-Olga [200]

Answer:

120 km squared.

Step-by-step explanation:

The shape shown has 5 sides: three rectangles and 2 triangles.

If you examine each triangle closely, they are 3-4-5 triangles. The area of one triangle is 3 * 4 * 1/2 = 12 * 1/2 = 12 / 2 = 6. Since there are two triangles, the total area of the triangles is 6 * 2 = 12 km squared.

The three rectangles have the same length, but their widths differ. They all have a side length of 9 km, but the largest rectangle has a width of 5 km, the medium has a width of 4 km, and the smallest has a width of 3 km.

So, the rectangles' areas added up will be (9 * 5) + (9 * 4) + (9 * 3) = 45 + 36 + 27 = 108 km squared.

And so, the total surface area is 12 km squared + 108 km squared, or 120 km squared.

Hope this helps!

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Two bottles are next to each other. Bottle A can hold 1/3 of a Liter and is filled with 2/3 juice.
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Step-by-step explanation:

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Maggie cut out 4 right triangles of equal size from construction paper. Each right triangle has a base of 6 inches and a height
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24

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3 years ago
What is equivalent to 23 over 25
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Read 2 more answers
In Exercises 1 and 2, find the equilibrium solutions of the differential equation specified.1) dy/dt = (y + 3)/(1 - y)2) dy/dt =
Aleks [24]

Answer:

1. Equilibrium solution: y= -3

2. Equilibrium solution y= ±1.414

Step-by-step explanation:

Thinking process:

The equilibrium solution can only be derived when \frac{dy}{dt}  = 0

1. Let's look at the first equation:

\frac{dy}{dt} = \frac{(y+3)}{(1-y)}

equating \frac{dy}{dt}  = 0 to the expression \frac{(y+3)}{(1-y)} gives

y + 3 = 0

     y = -3

Therefore, the equilibrium solution occurs at y = -3

2. Let's look at the second solution:

\frac{dy}{dy} = \frac{(t^{2}- 1) (y^{2}-2) }{y^{2}-4 }

dividing each side by (t²-1) gives

1/(t²-1)\frac{dy}{dt} = (y²-2)/ (y²-4)

factorizing the right hand side gives:

at equilibrium:  \frac{dy}{dt}  = 0, then

y² - 2 = 0

solving for y gives y = ±√2

                                 = ±1.414

5 0
4 years ago
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