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leva [86]
3 years ago
14

Double bubble/ triple 100 points !!!!!!

Mathematics
2 answers:
viktelen [127]3 years ago
4 0

Answer:

Diagonal < Diameter

Step-by-step explanation:

The square will fit inside without touching if the diagonal length is less than the diameter

Diagonal = sqrt(7² + 7²)

7sqrt(2)

Diameter = 11

7sqrt(2) < 11

Because

(7sqrt(2))² < 11²

98 < 121

balu736 [363]3 years ago
3 0

Answer:

see below

Step-by-step explanation:

We need to find the diameter of the square

We can find this using the Pythagorean theorem

a^2+b^2 = c^2  where the legs are 7 and 7 and the diameter is c

7^2 +7^2 = c^2

49+49 = c^2

98 = c^2

Taking the square root of each side

sqrt(98) = sqrt(c^2)

9.899494937 = c

Since 9.9 is less than 11 which is the diameter of the circle it will never touch the circle.

Since the longest part of the square is less than the diameter of the circle, the square will fit  inside the circle without touching

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Put the differential equation 9ty+ety′=yt2+81 into the form y′+p(t)y=g(t) and find p(t) and g(t). p(t)= help (formulas) g(t)= he
alexgriva [62]

Answer:

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And

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Step-by-step explanation:

Given that,

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9ty + e^{t}y' = \frac{y}{t^{2} + 81 }

e^{t}y' + (9t - \frac{1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t(t^{2} + 81 ) - 1}{t^{2} + 81 } )y = 0\\e^{t}y' + (\frac{9t^{3} + 729t  - 1}{t^{2} + 81 } )y = 0\\y' + [\frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) } ]y = 0

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p(t) = \frac{9t^{3} + 729t  - 1}{e^{t}(t^{2} + 81) }

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The differential equation 9ty + e^{t}y' = \frac{y}{t^{2} + 81 } is  linear and homogeneous.

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